Internal Energy

Every object contains energy at the microscopic level, even when the object is not moving as a whole. The total energy stored by the particles inside a substance is called its internal energy.

Internal energy is especially important when studying temperature, heating, changes of state, gases, and thermodynamics.

For a substance, internal energy is mainly associated with:

• the kinetic energy of its particles
• the potential energy associated with interactions between its particles

This means that heating a substance can change its internal energy even when the entire object remains stationary.

What Is Internal Energy?

Internal energy is the total microscopic kinetic and potential energy of the particles in a system.

We commonly represent internal energy using the symbol:

U

For a simple particle model:

Internal energy = microscopic kinetic energy + microscopic potential energy

The word internal is important. We are considering energy associated with particles inside the system, rather than the motion or position of the whole system.

For example, imagine a sealed container of gas sitting on a table. The container is not moving, so its macroscopic kinetic energy is zero relative to the table.

However, the gas molecules inside are moving rapidly in random directions.

The gas therefore has internal energy.

Microscopic and Macroscopic Energy

Energy can be considered on different scales.

Macroscopic energy describes the energy of an object or system as a whole.

Examples include:

• a moving car having kinetic energy
• a book raised above the floor having gravitational potential energy
• a rotating wheel having kinetic energy

Microscopic energy describes energy associated with the enormous number of particles inside matter.

Examples include:

• molecules moving randomly
• atoms vibrating in a solid
• particles rotating or vibrating
• potential energy associated with forces between particles

Internal energy concerns these microscopic forms of energy.

Particle Kinetic Energy

Particles in matter are constantly moving.

Their motion depends partly on the state of matter.

Solids

Particles mainly vibrate around fixed positions.

Liquids

Particles move around and slide past one another while remaining relatively close together.

Gases

Particles move rapidly and randomly through the available space.

The motion of these particles gives them microscopic kinetic energy.

In general, faster-moving particles have greater kinetic energy.

Particle Potential Energy

Particles can also store potential energy because forces act between them.

The potential energy depends on factors such as:

• the distance between particles
• their arrangement
• the attractive or repulsive forces between them

This becomes particularly important during changes of state.

For example, when a solid melts, energy is transferred into the substance. During the phase change, some of this energy changes the arrangement and separation of particles rather than increasing their average speed.

The potential-energy part of the internal energy therefore changes.

Internal Energy and Temperature

Temperature is closely related to the microscopic motion of particles.

For many introductory particle models:

higher temperature → greater average kinetic energy of the particles

If a gas is heated, its particles generally move faster.

If it is cooled, they generally move more slowly.

However, temperature and internal energy are not the same thing.

Temperature is related to the average kinetic energy of the particles.

Internal energy depends on the total microscopic energy of all the particles in the system.

This distinction is extremely important.

Temperature Is Not Total Energy

Consider two containers of water:

• Cup A contains 100 mL of water at 50°C.
• Container B contains 10 L of water at 50°C.

Both have the same temperature, so the average kinetic energy per particle is approximately the same.

However, Container B contains far more particles.

It therefore has much more total internal energy.

So:

same temperature does not necessarily mean same internal energy.

How Internal Energy Changes

The internal energy of a system can change when energy is transferred into or out of it.

Two particularly important transfer mechanisms are:

• heating
• doing work

We can represent the change as:

ΔU = energy transferred by heating + energy transferred by work done on the system

Here:

ΔU means "change in internal energy."

If more energy enters the system than leaves it, internal energy increases.

If more energy leaves than enters, internal energy decreases.

Increasing Internal Energy by Heating

Suppose a metal block is placed on a hot plate.

Energy is transferred from the hot plate to the block.

The particles in the block vibrate more vigorously.

Their average kinetic energy increases.

As a result:

• internal energy increases
• temperature usually increases

The energy has been transferred into the microscopic energy stores of the material.

Decreasing Internal Energy by Cooling

Now suppose the hot metal block is placed in cooler surroundings.

Energy is transferred from the block to the surroundings.

Its particles gradually vibrate less vigorously.

Therefore:

• microscopic kinetic energy decreases
• internal energy decreases
• temperature decreases

Eventually, the block and its surroundings may reach the same temperature.

Internal Energy Can Change Without a Temperature Change

Heating does not always cause the temperature to rise.

During a change of state, energy can increase the internal energy of a substance without increasing its temperature.

Consider melting ice.

As energy is transferred to ice at its melting point, the temperature can remain constant while the ice changes into liquid water.

The transferred energy changes the potential energy associated with the particle arrangement.

Therefore:

internal energy increases even though temperature remains constant during the phase change.

This is one reason temperature cannot be used as a direct measure of total internal energy.

Internal Energy in Gases

Gases provide an especially useful way to study internal energy.

Gas particles:

• move continuously
• travel in random directions
• collide with each other
• collide with the walls of their container

In an idealized gas model, much of our attention is placed on the microscopic kinetic energy of the particles.

Higher temperature means greater average kinetic energy.

Therefore, increasing the temperature of a fixed amount of an ideal gas increases its internal energy.

Heating a Gas in a Rigid Container

Imagine gas inside a strong, sealed container whose volume cannot change.

If energy is transferred to the gas by heating:

1. the gas particles gain kinetic energy
2. they move faster
3. collisions with the container walls become more frequent and/or more forceful
4. the temperature increases
5. the pressure increases

The gas cannot expand because the container is rigid.

The added energy therefore increases the gas's internal energy.

Gas Expansion

Now imagine gas inside a cylinder with a movable piston.

When the gas is heated, it may expand and push the piston outward.

In this situation, some transferred energy may:

• increase the internal energy of the gas
• be transferred as the gas does work on the piston

This is an important thermodynamic idea:

Energy supplied to a gas does not necessarily remain entirely as internal energy.

Some can be transferred from the gas by mechanical work.

Compressing a Gas

Internal energy can also be increased by doing work on a gas.

Imagine pushing down on a piston and compressing the gas inside.

Work is done on the gas.

This transfers energy into the gas.

If the compression happens rapidly enough that little energy escapes to the surroundings, the temperature of the gas rises.

This occurs because the internal energy of the gas increases.

A familiar example is a bicycle pump. When air is compressed repeatedly, the pump can become warm.

Expansion Can Cool a Gas

The reverse can also occur.

If a gas expands and does work on its surroundings without receiving enough energy by heating to replace the transferred energy, its internal energy can decrease.

Its temperature may then decrease.

So:

compression can increase internal energy

while:

expansion can decrease internal energy

depending on the energy transfers taking place.

Internal Energy and Amount of Matter

Internal energy depends partly on how much matter is present.

Imagine two samples of the same substance at the same temperature:

• Sample A has a mass of 1 kg.
• Sample B has a mass of 5 kg.

Sample B contains more particles.

Although the particles have similar average kinetic energies, Sample B has a greater total microscopic energy.

Therefore, it generally has greater internal energy.

Internal energy is therefore an extensive property: it depends on the amount of matter in the system.

Worked Example

A gas is heated while contained in a rigid container.

What happens to its internal energy?

Energy is transferred into the gas by heating.

The particles move faster, increasing their average kinetic energy.

Because the container is rigid, the gas cannot expand and do significant boundary work.

Therefore:

the internal energy of the gas increases.

Its temperature also rises.

Worked Example: Compression

A gas is compressed quickly using a piston.

What happens?

Work is done on the gas.

Energy is transferred into the gas.

Its internal energy increases.

The average kinetic energy of its particles can increase.

Therefore:

the temperature of the gas can rise.

Energy Transfer vs. Energy Storage

A useful distinction is between energy stores and energy transfers.

Internal energy is an energy store within a system.

Heating is an energy-transfer process.

Mechanical work is also an energy-transfer process.

So it is better to say:

Energy is transferred to the object's internal energy store by heating.

rather than:

The object contains heat.

This distinction helps describe energy changes more accurately.

Connecting Internal Energy to the Particle Model

The particle model gives us a microscopic explanation for thermal behaviour.

When a substance is heated:

• particles may move faster
• particles may vibrate more strongly
• particle spacing may change
• particle arrangement may change
• microscopic kinetic energy may increase
• microscopic potential energy may increase

All of these effects can contribute to changes in internal energy.

This explains why the concept of internal energy connects macroscopic observations, such as temperature and state, with microscopic particle behaviour.

Real-World Connections

Car Engines

Hot gases inside an engine have internal energy. Expanding gases push against pistons and transfer energy mechanically.

Refrigerators

Refrigerators use compression and expansion of a refrigerant to transfer energy from inside the refrigerator to the surroundings.

Bicycle Pumps

Rapidly compressing air can raise its temperature because work transfers energy into the gas.

Aerosol Cans

Gas escaping from a pressurized container can expand and cool.

Steam Turbines

High-temperature steam expands and does work on turbine blades, transferring some of its internal energy into mechanical energy.

Common Mistakes

Thinking temperature and internal energy are the same

Temperature is related to the average kinetic energy of particles. Internal energy is the total microscopic kinetic and potential energy.

Thinking a stationary object has no energy

An object can have zero macroscopic kinetic energy while its particles still possess microscopic kinetic and potential energy.

Thinking heating always raises temperature

During a phase change, internal energy can increase while temperature remains constant.

Thinking gases only store kinetic energy in all situations

The ideal gas model emphasizes molecular kinetic energy, but real gases can also involve intermolecular potential energy.

Calling internal energy "heat"

Heat is not best treated as something stored inside an object. Heating describes an energy-transfer process.

Ignoring the amount of matter

Two objects at the same temperature can have very different internal energies if they contain different amounts or types of matter.

Key Terms

Internal energy — the total microscopic kinetic and potential energy of the particles in a system.

Microscopic energy — energy associated with individual particles and their interactions.

Macroscopic energy — energy associated with the motion, position, or behaviour of a system as a whole.

Temperature — a measure related to the average kinetic energy of particles in a substance.

Microscopic kinetic energy — energy associated with the random motion or vibration of particles.

Microscopic potential energy — energy associated with particle positions, arrangements, and interactions.

System — the object or collection of matter being studied.

Heating — transfer of energy resulting from a temperature difference.

Work — transfer of energy when a force causes displacement or through processes such as gas compression and expansion.

Compression — reducing the volume of a gas, often by doing work on it.

Expansion — an increase in gas volume, during which the gas may do work on its surroundings.

Ideal gas — a simplified model of a gas in which particles are treated as having negligible volume and intermolecular interactions are neglected except during collisions.

Key Takeaways

• Internal energy is the total microscopic energy stored within a system.
• It includes microscopic kinetic energy and potential energy.
• Macroscopic motion is different from microscopic particle motion.
• Temperature is related to the average kinetic energy of particles.
• Temperature and internal energy are not the same quantity.
• Objects at the same temperature can have different internal energies.
• Internal energy can change through heating or mechanical work.
• Internal energy can increase without temperature increasing during a change of state.
• Heating a gas generally increases its internal energy.
• Compressing a gas can transfer energy into its internal energy store.
• An expanding gas can do work on its surroundings and lose internal energy.
• The particle model connects microscopic behaviour with macroscopic observations such as temperature and pressure.

Check Your Understanding

1. Define internal energy.
2. What two microscopic forms of energy contribute to the internal energy of a substance?
3. Explain the difference between microscopic and macroscopic kinetic energy.
4. A sealed container of gas is sitting motionless on a table. Explain why the gas still has internal energy.
5. Explain why a 10 L container of water can have more internal energy than a 100 mL cup of water at the same temperature.
6. What generally happens to the average kinetic energy of particles when temperature increases?
7. Explain how internal energy can increase during melting even though the temperature remains constant.
8. A gas is heated inside a rigid container. Describe what happens to its particle motion, internal energy, temperature, and pressure.
9. Explain why rapidly compressing air in a bicycle pump can make the pump warmer.
10. A gas expands and does work on its surroundings without receiving energy by heating. Predict what may happen to its internal energy and temperature.
11. Explain why saying "an object contains heat" is less accurate than describing heating as an energy-transfer process.
12. A moving container of hot gas suddenly stops. Explain the difference between the change in the container's macroscopic kinetic energy and the gas's internal energy.

Heat and Work

Energy can move into or out of a system in different ways. Two of the most important energy-transfer processes in thermodynamics are heating and work.

The key idea is:

Heat and work are ways of transferring energy. They are not substances stored inside an object.

An object can store energy as internal energy, but energy can be transferred into or out of that internal energy store by heating or by doing work.

Heat as an Energy Transfer

In everyday language, we often say that an object "contains heat." In physics, it is more accurate to describe heating as a transfer of energy caused by a temperature difference.

If two objects at different temperatures interact, energy tends to transfer:

from the higher-temperature object → to the lower-temperature object

For example, place a hot metal block into cool water.

Energy transfers from the metal to the water.

As a result:

• the metal loses internal energy
• the water gains internal energy
• the metal cools
• the water warms

Eventually, they may reach the same temperature.

At that point there is no net energy transfer by heating between them.

Heat and Temperature Are Different

Temperature tells us about the thermal state of a system and is related to the average kinetic energy of its particles.

Heating describes energy being transferred because of a temperature difference.

They are therefore not the same thing.

For example, when a pan is placed on a hot stove:

• the stove has a higher temperature
• energy is transferred to the pan by heating
• the internal energy of the pan increases
• the temperature of the pan usually rises

How Energy Is Transferred by Heating

Energy can be transferred because of temperature differences through several mechanisms.

Conduction

Conduction occurs through interactions between neighbouring particles.

For example, when one end of a metal rod is heated, particles in that region vibrate more vigorously. Energy is transferred through the material.

Metals are particularly good thermal conductors.

Convection

Convection occurs in fluids such as liquids and gases through the movement of the fluid.

Warmer, less dense regions may rise while cooler, denser regions sink, producing convection currents.

Thermal Radiation

Energy can also be transferred by electromagnetic radiation, especially infrared radiation.

Unlike conduction and convection, radiation does not require matter and can travel through a vacuum.

Energy from the Sun reaches Earth mainly by radiation.

Work as an Energy Transfer

Work is another way energy can be transferred.

Mechanical work occurs when a force causes displacement.

For a constant force acting in the direction of motion:

W = Fd

where:

• W = work done in joules (J)
• F = force in newtons (N)
• d = displacement in metres (m)

If you push a box across the floor, you do work on the box.

Energy is transferred from you to the box and its surroundings.

Work and Internal Energy

Work can also change the internal energy of a system.

A useful example is compressing a gas.

Imagine air inside a cylinder fitted with a movable piston.

If you push the piston inward:

• you exert a force
• the piston moves
• work is done on the gas
• energy is transferred into the gas
• the internal energy of the gas can increase

The gas may therefore become warmer.

This demonstrates that internal energy can change without energy being transferred by heating.

Compression of a Gas

Suppose a gas is trapped inside a cylinder.

When the piston is pushed inward, the gas is compressed.

The surroundings do work on the gas.

Energy is transferred into the gas's internal energy store.

If little energy escapes to the surroundings:

compression → internal energy increases → temperature increases

This is why a bicycle pump can become warm when air is compressed repeatedly.

Expansion of a Gas

The reverse process occurs when a gas expands against an external pressure.

Imagine a gas pushing a piston outward.

The gas exerts a force on the piston, and the piston moves.

Therefore, the gas does work on its surroundings.

Energy is transferred from the gas to the surroundings.

If this energy is not replaced by heating:

expansion → internal energy decreases → temperature may decrease

Work Done by an Expanding Gas

For a gas expanding against a constant external pressure, the work done by the gas can be calculated using:

W = pΔV

where:

• W = work done by the gas in joules (J)
• p = external pressure in pascals (Pa)
• ΔV = change in volume in cubic metres (m³)

The change in volume is:

ΔV = V₂ − V₁

where:

• V₁ = initial volume
• V₂ = final volume

For expansion:

V₂ > V₁

so ΔV is positive.

Why W = pΔV?

Consider a gas pushing a piston of area A through a distance d.

Mechanical work is:

W = Fd

Pressure is:

p = F/A

so:

F = pA

Substitute this into the work equation:

W = pAd

But:

Ad = ΔV

because the piston area multiplied by the distance moved gives the change in volume.

Therefore:

W = pΔV

This connects ordinary mechanical work with the work done by an expanding gas.

Worked Example: Expanding Gas

A gas expands from 0.020 m³ to 0.035 m³ against a constant pressure of 100,000 Pa.

First calculate the change in volume:

ΔV = V₂ − V₁

ΔV = 0.035 − 0.020

ΔV = 0.015 m³

Now calculate the work:

W = pΔV

W = 100000 × 0.015

W = 1500 J

The gas does:

1,500 J of work on its surroundings.

This means 1,500 J of energy has been transferred from the gas through mechanical work.

Another Worked Example

A gas expands from 2.0 × 10⁻³ m³ to 5.0 × 10⁻³ m³ against a constant external pressure of 2.0 × 10⁵ Pa.

Calculate the volume change:

ΔV = 5.0 × 10⁻³ − 2.0 × 10⁻³

ΔV = 3.0 × 10⁻³ m³

Now:

W = pΔV

W = (2.0 × 10⁵)(3.0 × 10⁻³)

W = 600 J

Therefore, the gas transfers 600 J of energy to its surroundings by doing work.

Unit Check

The pressure-volume equation produces joules because:

1 Pa = 1 N/m²

Therefore:

Pa × m³ = (N/m²)(m³)

= N·m

and:

1 N·m = 1 J

So:

Pa·m³ = J

This is useful for checking that the units in a calculation are correct.

Be Careful with Volume Units

Gas volumes are often given in litres, but pressure is commonly measured in pascals.

When using:

W = pΔV

volume should normally be converted to m³.

Remember:

1 m³ = 1000 L

Therefore:

1 L = 0.001 m³

or:

1 L = 1 × 10⁻³ m³

Example

A gas expands by 4.0 L.

Convert to cubic metres:

4.0 L × 0.001 = 0.0040 m³

If the pressure is 150,000 Pa:

W = 150000 × 0.0040

W = 600 J

Expansion vs. Compression

It is useful to distinguish who is doing the work.

Gas Expands

The gas pushes the surroundings.

The gas does work on the surroundings.

Energy is transferred out of the gas by work.

Gas Is Compressed

The surroundings push the gas.

Work is done on the gas.

Energy is transferred into the gas by work.

This distinction becomes important when analysing changes in internal energy.

Heat and Work Together

A system can experience both heating and work at the same time.

Consider a gas in a cylinder being heated.

Energy is transferred into the gas by heating.

As the gas expands, it pushes the piston.

Energy is transferred out of the gas by work.

The change in internal energy depends on the balance between these transfers.

For example, suppose:

• 900 J is transferred into a gas by heating
• the gas does 350 J of work on its surroundings

The remaining energy increases the gas's internal energy:

ΔU = 900 − 350

ΔU = 550 J

The internal energy increases by 550 J.

Energy Accounting

A useful way to think about thermodynamics is as energy accounting.

Energy entering a system must either:

• increase an energy store, or
• leave the system through another transfer pathway

Suppose 2,000 J is transferred into a gas by heating and the gas does 800 J of work.

Then:

Energy in = 2000 J

Energy out by work = 800 J

Increase in internal energy:

2000 − 800 = 1200 J

Energy is conserved throughout the process.

Identifying Heat and Work in Everyday Situations

Hot Drink Cooling

Energy transfers from the drink to the surroundings because of a temperature difference.

Main process: heating

Bicycle Pump

You push the piston and compress the gas.

Main process: work done on the gas

Gas Expanding Against a Piston

The gas pushes the piston outward.

Main process: work done by the gas

Metal Spoon in Hot Soup

Energy transfers from the hotter soup into the cooler spoon.

Main process: heating

Rubbing Your Hands Together

Forces act through distances, and mechanical energy is ultimately transferred into internal energy.

Work contributes to increasing internal energy.

Heating Gas That Expands

Energy enters by heating while energy also leaves as the gas does work.

Both heating and work are involved.

Heat and Work Are Not Stored

This is an important thermodynamic distinction.

A system can possess:

• internal energy
• kinetic energy
• gravitational potential energy
• other energy stores

But heat and work describe energy being transferred across the boundary of a system.

Once the transfer has occurred, we describe the resulting change in the system's energy rather than saying that the system "stores work" or "stores heat."

Real-World Applications

Internal Combustion Engines

Hot gases expand inside an engine cylinder and do work on a piston. This transfers energy from the gas into mechanical motion.

Steam Turbines

High-pressure steam expands and transfers energy by doing work on turbine blades.

Refrigerators

A refrigerant is repeatedly compressed and expanded. Work and heating processes allow energy to be transferred from inside the refrigerator to the surroundings.

Air Compressors

Mechanical work is done on a gas during compression, which can significantly increase its temperature.

Power Stations

Many power stations use heating to increase the internal energy of a working fluid, which then expands and performs mechanical work.

Common Mistakes

Thinking heat is stored inside an object

Internal energy is stored. Heating describes an energy transfer caused by a temperature difference.

Thinking temperature is heat

Temperature and heating are different concepts. Temperature describes the thermal state of a system; heating describes energy transfer.

Forgetting who does the work

During expansion, the gas does work on the surroundings.

During compression, the surroundings do work on the gas.

Using litres directly in W = pΔV

If pressure is measured in pascals, convert litres to cubic metres.

Using the final volume instead of the change in volume

Use:

ΔV = V₂ − V₁

not simply V₂.

Assuming all energy transferred by heating increases internal energy

If a gas expands, some energy may leave the system as work.

Key Terms

Heating — transfer of energy caused by a temperature difference.

Work — transfer of energy when a force acts through a displacement or through processes such as gas compression and expansion.

Internal energy — the total microscopic kinetic and potential energy of the particles in a system.

System — the object or collection of matter being studied.

Surroundings — everything outside the system that can interact with it.

System boundary — the dividing surface between a system and its surroundings.

Conduction — energy transfer through particle interactions within or between materials.

Convection — energy transfer involving bulk movement of a fluid.

Thermal radiation — transfer of energy by electromagnetic radiation.

Expansion — an increase in the volume of a gas.

Compression — a decrease in the volume of a gas.

Pressure — force per unit area.

Volume change — the difference between final and initial volume: ΔV = V₂ − V₁.

Work done by a gas — energy transferred from a gas when it expands against an external pressure.

Key Takeaways

• Heat and work are energy-transfer processes.
• Heating occurs because of a temperature difference.
• Work transfers energy through forces and displacement.
• Heating and work can both change the internal energy of a system.
• Compressing a gas involves work being done on the gas.
• An expanding gas can do work on its surroundings.
• For constant external pressure, work done by an expanding gas can be calculated using W = pΔV.
• Pressure should be in Pa and volume change in m³ to obtain work in joules.
• 1 L = 1 × 10⁻³ m³.
• A gas can receive energy by heating while simultaneously transferring energy away by doing work.
• Energy conservation allows us to account for changes in internal energy.
• Identifying the system and surroundings makes heat-and-work problems easier to understand.

Check Your Understanding

1. Explain the difference between internal energy and heating.
2. In which direction does energy transfer by heating between objects at different temperatures?
3. Give one example of energy transfer by conduction, convection, and radiation.
4. Explain how doing work on a gas can increase its internal energy.
5. Why can a bicycle pump become warm when air is compressed?
6. A gas expands from 0.030 m³ to 0.050 m³ against a pressure of 80,000 Pa. Calculate the work done by the gas.
7. A gas expands by 6.0 L against a constant pressure of 120,000 Pa. Calculate the work done.
8. A gas expands from 3.0 × 10⁻³ m³ to 8.0 × 10⁻³ m³ against a pressure of 2.5 × 10⁵ Pa. Calculate the work done.
9. A gas receives 1,400 J of energy by heating and does 500 J of work on its surroundings. By how much does its internal energy increase?
10. A gas receives 2,500 J by heating while doing 1,800 J of work. Calculate its change in internal energy.
11. Explain why W = pV would generally be incorrect for calculating the work done during an expansion.
12. A student says, "The gas contains 500 J of heat." Explain why this statement is not an accurate thermodynamic description.

The First Law of Thermodynamics

The First Law of Thermodynamics is an application of the law of conservation of energy to thermodynamic systems.

It tells us that energy cannot be created or destroyed. Energy can be transferred between a system and its surroundings or converted from one form to another, but the total amount of energy is conserved.

For a thermodynamic system, energy can commonly be transferred:

• by heating
• by doing work

These transfers can change the internal energy of the system.

Stating the First Law

One useful form of the First Law is:

ΔU = Q − W

where:

• ΔU = change in internal energy of the system
• Q = energy transferred to the system by heating
• W = work done by the system on its surroundings

All three quantities are measured in joules (J).

This equation is an energy balance:

change in stored internal energy = energy transferred in − energy transferred out

Understanding the Signs

Using the convention:

ΔU = Q − W

the signs have physical meanings.

Heating the System

If energy is transferred into the system by heating:

Q is positive

If energy leaves the system by heating:

Q is negative

Work Done by the System

If the system expands and does work on its surroundings:

W is positive

Energy leaves the system.

If the surroundings do work on the system, such as compressing a gas:

W is negative

Energy enters the system.

The sign convention is extremely important when solving First Law problems.

Why Sign Conventions Matter

You may encounter textbooks that define W as work done on the system. With that convention, the First Law is written:

ΔU = Q + W

Both forms describe the same conservation law.

In these notes, we will consistently use:

ΔU = Q − W

with W meaning work done by the system.

Conservation of Energy

Imagine a gas inside a cylinder with a movable piston.

If 1,000 J of energy is transferred into the gas by heating, that energy cannot simply disappear.

It might:

• increase the internal energy of the gas
• leave the gas when it does work on the piston
• be divided between both processes

Suppose:

Q = 1000 J

W = 300 J

Then:

ΔU = Q − W

ΔU = 1000 − 300

ΔU = 700 J

The gas's internal energy increases by 700 J.

The energy accounting is:

1000 J in = 700 J stored + 300 J transferred out by work

Internal Energy

Recall that internal energy is the total microscopic kinetic and potential energy associated with the particles of a system.

The First Law deals with the change in internal energy:

ΔU = U_final − U_initial

If:

ΔU > 0

the internal energy has increased.

If:

ΔU < 0

the internal energy has decreased.

If:

ΔU = 0

there has been no overall change in internal energy.

Worked Example: Heating Without Work

A rigid container of gas receives 750 J of energy by heating.

Because the container is rigid, the gas does no expansion work.

Therefore:

Q = +750 J

W = 0 J

Use:

ΔU = Q − W

ΔU = 750 − 0

ΔU = +750 J

The internal energy increases by 750 J.

This makes sense: all the transferred energy remains in the system's internal energy store.

Worked Example: Heating and Expansion

A gas receives 1,500 J by heating and does 600 J of work on its surroundings.

Q = +1500 J

W = +600 J

Therefore:

ΔU = 1500 − 600

ΔU = +900 J

The internal energy increases by 900 J.

Although 1,500 J entered the gas, 600 J was transferred back out through work.

Worked Example: More Work Than Heating

A gas receives 800 J by heating but does 1,200 J of work.

Q = +800 J

W = +1200 J

Therefore:

ΔU = 800 − 1200

ΔU = −400 J

The negative sign means the gas's internal energy decreases by 400 J.

This can happen because the gas transfers more energy out through work than it receives by heating.

Work Done on a Gas

Now suppose a gas is compressed.

The surroundings do 500 J of work on the gas, and no energy is transferred by heating.

Using our sign convention:

Q = 0

W = −500 J

Therefore:

ΔU = 0 − (−500)

ΔU = +500 J

The internal energy increases.

This helps explain why gases can become warmer when compressed.

Cooling a System

Suppose 350 J of energy leaves a gas by heating and no work is done.

Because energy leaves by heating:

Q = −350 J

W = 0

Therefore:

ΔU = −350 − 0

ΔU = −350 J

The internal energy decreases by 350 J.

Energy Transfer Diagrams

Energy-transfer diagrams can help visualize the First Law.

Imagine a box representing the system.

Arrows pointing into the box represent energy entering the system.

Arrows pointing out represent energy leaving.

For example:

800 J heating → [ GAS ] → 250 J work

The First Law tells us:

ΔU = 800 − 250

ΔU = +550 J

So 550 J remains as an increase in the gas's internal energy.

When interpreting a diagram, ask:

1. What is the system?
2. Which arrows represent energy entering?
3. Which arrows represent energy leaving?
4. Is the transfer caused by heating or work?
5. What is the resulting change in internal energy?

Example Energy Diagram

Suppose an energy diagram shows:

1200 J heating → [ SYSTEM ] → 700 J work

Energy entering:

1200 J

Energy leaving:

700 J

Change in internal energy:

ΔU = 1200 − 700

ΔU = +500 J

The system stores 500 J more internal energy than it did initially.

Finding an Unknown Heat Transfer

The First Law can be rearranged.

Starting with:

ΔU = Q − W

To find Q:

Q = ΔU + W

Example

A gas's internal energy increases by 600 J while it does 250 J of work.

Find the energy transferred to the gas by heating.

Q = ΔU + W

Q = 600 + 250

Q = 850 J

Therefore, 850 J was transferred to the gas by heating.

Finding Unknown Work

Starting with:

ΔU = Q − W

Rearrange:

W = Q − ΔU

Example

A gas receives 2,000 J by heating, while its internal energy increases by 1,300 J.

W = Q − ΔU

W = 2000 − 1300

W = 700 J

The gas does 700 J of work on its surroundings.

First Law and Gas Expansion

For expansion against a constant external pressure:

W = pΔV

This can be combined with the First Law:

ΔU = Q − pΔV

Example

A gas receives 1,000 J by heating and expands by 0.004 m³ against a constant pressure of 150,000 Pa.

First calculate the work:

W = pΔV

W = 150000 × 0.004

W = 600 J

Now apply the First Law:

ΔU = Q − W

ΔU = 1000 − 600

ΔU = +400 J

Therefore, the gas's internal energy increases by 400 J.

Constant-Volume Process

If a gas is held inside a rigid container:

ΔV = 0

Therefore:

W = pΔV = 0

The First Law becomes:

ΔU = Q

So any net energy transferred by heating directly changes the internal energy.

For example, if 900 J enters by heating:

ΔU = +900 J

No Heat Transfer

A process in which no energy is transferred by heating is called an adiabatic process.

For such a process:

Q = 0

The First Law becomes:

ΔU = −W

If a gas expands and does work:

W > 0

therefore:

ΔU < 0

Its internal energy decreases.

If a gas is compressed:

W < 0

therefore:

ΔU > 0

Its internal energy increases.

This explains why rapid compression can heat a gas and rapid expansion can cool it.

No Change in Internal Energy

Suppose:

ΔU = 0

Then:

0 = Q − W

so:

Q = W

This means the energy entering the system by heating equals the energy leaving through work.

Example

A gas receives 500 J by heating and does 500 J of work.

ΔU = 500 − 500

ΔU = 0

Its internal energy does not change.

This does not mean that no energy was transferred. Energy entered and left at equal rates overall.

Comparing Simple Processes

Real-World Connection: Car Engine

Inside a combustion engine, hot gases contain significant internal energy.

As the gas expands:

• it pushes against a piston
• the gas does work
• energy is transferred into mechanical motion

The First Law allows us to account for where the energy goes.

Not all the energy released during combustion becomes useful mechanical work. Some remains as internal energy, and some is transferred to the surroundings.

Real-World Connection: Bicycle Pump

When you rapidly compress air inside a bicycle pump:

• you do work on the gas
• energy enters the gas
• its internal energy increases
• its temperature may rise

This is a practical example of the First Law.

Real-World Connection: Power Stations

Many power stations transfer energy to a working fluid by heating.

The hot fluid or steam then expands and does work on a turbine.

The First Law describes the energy balance:

energy transferred in → change in internal energy + work transferred out

Common Mistakes

Adding work when the gas does work

With our convention:

ΔU = Q − W

If the gas does work on its surroundings, energy leaves the system, so W is subtracted.

Forgetting the negative sign during compression

If work is done on the gas, W is negative using our convention.

Therefore:

ΔU = Q − (negative W)

which increases internal energy.

Assuming ΔU = Q

This is only true when no work is done.

Thinking ΔU = 0 means nothing happened

A system can receive energy by heating and transfer the same amount by work, producing no net change in internal energy.

Confusing internal energy with temperature

Internal energy and temperature are related but are not the same quantity.

Switching sign conventions halfway through a problem

Choose one convention and use it consistently. In these notes:

Q > 0: energy enters by heating.

W > 0: the system does work.

Key Terms

First Law of Thermodynamics — the conservation of energy applied to thermodynamic systems.

Internal energy (U) — the total microscopic kinetic and potential energy associated with the particles in a system.

Change in internal energy (ΔU) — the difference between the final and initial internal energy: ΔU = U_final − U_initial.

Heat transfer (Q) — energy transferred because of a temperature difference.

Work (W) — energy transferred mechanically; in the convention used here, positive W represents work done by the system.

System — the object or collection of matter being studied.

Surroundings — everything outside the system.

Energy conservation — the principle that energy cannot be created or destroyed, only transferred or transformed.

Expansion — an increase in gas volume, during which a gas may do work.

Compression — a decrease in gas volume, usually involving work being done on the gas.

Adiabatic process — a process in which no energy is transferred by heating.

Constant-volume process — a process in which the system's volume does not change.

Key Takeaways

• The First Law of Thermodynamics is based on conservation of energy.
• Using the convention in these notes: ΔU = Q − W.
• Q is positive when energy enters the system by heating.
• Q is negative when energy leaves the system by heating.
• W is positive when the system does work on its surroundings.
• W is negative when work is done on the system.
• Positive ΔU means internal energy increases.
• Negative ΔU means internal energy decreases.
• At constant volume, no expansion work is done, so ΔU = Q.
• In an adiabatic process, Q = 0, so ΔU = −W.
• If ΔU = 0, energy entering by heating can equal energy leaving by work.
• Energy-transfer diagrams can be interpreted by tracking energy entering, leaving, and remaining in the system.

Check Your Understanding

1. State the First Law of Thermodynamics in words.
2. Using the sign convention in these notes, write the First Law as an equation.
3. A gas receives 900 J by heating and does 300 J of work. Calculate ΔU.
4. A gas receives 1,200 J by heating and does 1,500 J of work. Calculate ΔU and explain what the sign means.
5. A gas loses 400 J by heating and does no work. Calculate ΔU.
6. A gas is compressed, with 650 J of work done on it. No energy is transferred by heating. Calculate ΔU.
7. A gas's internal energy increases by 800 J while it does 250 J of work. How much energy was transferred to it by heating?
8. A gas receives 1,600 J by heating and its internal energy increases by 1,000 J. Calculate the work done by the gas.
9. A gas receives 2,400 J by heating and expands by 0.010 m³ against a constant pressure of 1.5 × 10⁵ Pa. Calculate the work done and the change in internal energy.
10. Explain how a system can have ΔU = 0 even though energy is being transferred both into and out of it.
11. A gas expands adiabatically and does 450 J of work. Determine its change in internal energy.
12. Draw a simple energy-transfer diagram for a gas that receives 1,000 J by heating, does 350 J of work, and increases its internal energy by 650 J.

P–V Diagrams

A pressure–volume diagram, or P–V diagram, is a graph used to show how the pressure and volume of a gas change during a thermodynamic process.

On a P–V diagram:

• pressure (P) is plotted on the vertical axis
• volume (V) is plotted on the horizontal axis

Each point on the graph represents a particular state of the gas.

By following a path from one point to another, we can determine how the gas changes and, importantly, how much work is done during expansion or compression.

Reading a P–V Diagram

Suppose a gas begins at point A and moves to point B.

If B is to the right of A:

volume increases → expansion

If B is to the left of A:

volume decreases → compression

If B is above A:

pressure increases

If B is below A:

pressure decreases

The direction of the arrow on the path is therefore important.

The same curve travelled in opposite directions represents a different physical process.

State vs. Process

A state describes the condition of the gas at a particular moment.

For example:

P = 200 kPa and V = 0.004 m³

A process describes the change from one state to another.

On a P–V diagram:

• a point represents a state
• a line or curve represents a process
• an arrow shows the direction of the process

A gas can sometimes move between the same initial and final states using different paths.

This becomes important when calculating work.

Constant-Pressure Processes

A process occurring at constant pressure is called an isobaric process.

Because pressure remains constant, it appears as a horizontal line on a P–V diagram.

If the gas moves to the right:

volume increases → gas expands

If it moves to the left:

volume decreases → gas is compressed

For constant pressure, the work done by the gas is:

W = pΔV

where:

ΔV = V₂ − V₁

Therefore:

W = p(V₂ − V₁)

Worked Example: Constant-Pressure Expansion

A gas expands at a constant pressure of 150,000 Pa from:

V₁ = 0.002 m³

to:

V₂ = 0.006 m³

First calculate the change in volume:

ΔV = 0.006 − 0.002

ΔV = 0.004 m³

Now:

W = pΔV

W = 150000 × 0.004

W = 600 J

The gas does 600 J of work on its surroundings.

Area Under a P–V Graph

One of the most important features of a P–V diagram is:

The area under the process curve represents the work done by the gas.

For a constant-pressure expansion, the area forms a rectangle.

Area = height × width

Therefore:

W = pΔV

This gives the same equation we derived from mechanical work.

Why Does Area Represent Work?

Pressure is measured in pascals:

Pa = N/m²

Volume is measured in:

m³

Therefore:

Pa × m³ = (N/m²)(m³)

= N·m

But:

N·m = J

Therefore:

Pa·m³ = J

The area under a pressure-volume graph naturally has units of energy.

Expansion and Compression

The direction of the process determines whether the gas does work or has work done on it.

Expansion

The gas moves from smaller volume to larger volume:

left → right

The gas pushes against its surroundings.

Using our previous First Law convention:

W > 0

The gas does work on the surroundings.

Compression

The gas moves from larger volume to smaller volume:

right → left

The surroundings compress the gas.

Therefore:

W < 0

when W represents work done by the gas.

Work is being done on the gas.

Constant-Volume Processes

A process occurring at constant volume is called an isochoric process or isovolumetric process.

On a P–V diagram, it appears as a vertical line.

Because:

ΔV = 0

the gas does no pressure-volume work.

Therefore:

W = 0

Even though pressure may change substantially, there is no expansion or compression.

Example

A gas is heated inside a rigid sealed container.

Its temperature rises and its pressure increases.

However, the container cannot expand.

Therefore:

ΔV = 0

and:

W = 0

Constant-Temperature Processes

A process occurring at constant temperature is called an isothermal process.

For an ideal gas:

PV = constant

Therefore, as volume increases, pressure decreases.

An isothermal process appears as a curved line on a P–V diagram.

During isothermal expansion:

• volume increases
• pressure decreases
• the gas does work
• the temperature remains constant

For an ideal gas, internal energy depends on temperature. Therefore, during an isothermal process:

ΔU = 0

From the First Law:

ΔU = Q − W

so:

0 = Q − W

Therefore:

Q = W

Energy transferred into the gas by heating equals the work done by the gas.

Adiabatic Processes

An adiabatic process occurs when no energy is transferred by heating between the system and its surroundings.

Therefore:

Q = 0

Using the First Law:

ΔU = Q − W

we get:

ΔU = −W

During adiabatic expansion:

• the gas does work
• internal energy decreases
• temperature decreases

During adiabatic compression:

• work is done on the gas
• internal energy increases
• temperature increases

An adiabatic curve on a P–V diagram is generally steeper than an isothermal curve for the same ideal gas starting from the same state.

Recognizing Common Processes

Being able to recognize these shapes makes P–V diagrams much easier to interpret.

Work Along a Curved Path

If pressure changes as the gas expands, we cannot generally use one pressure value in:

W = pΔV

Instead, the work corresponds to the area under the curve.

In more advanced mathematics this is written:

W = ∫P dV

The integral means that we add many very small pressure × volume-change contributions across the entire process.

At this level, the important idea is:

larger area under the P–V curve → more work done by the gas

Estimating Work from a Graph

Suppose pressure decreases approximately linearly from 300 kPa to 100 kPa while volume increases from 0.002 m³ to 0.006 m³.

Because the graph forms a trapezoid, we can use average pressure:

Average pressure = (300000 + 100000) / 2

Average pressure = 200000 Pa

Volume change:

ΔV = 0.006 − 0.002

ΔV = 0.004 m³

Work:

W = average pressure × ΔV

W = 200000 × 0.004

W = 800 J

The gas does approximately 800 J of work.

Comparing Different Thermodynamic Paths

Suppose a gas moves from the same initial state A to the same final state B using two different paths.

Path 1 has a higher pressure during most of the expansion.

Path 2 has a lower pressure.

Because work is the area under the P–V curve:

Path 1 has the greater area → more work is done.

This demonstrates an important thermodynamic idea:

Work depends on the path taken between two states.

Internal Energy Is Different

Although work depends on the path, the change in internal energy depends only on the initial and final states.

Suppose a gas moves from state A to state B.

Whether it follows Path 1 or Path 2:

ΔU is the same

provided the initial and final states are identical.

However:

W may be different

and therefore, from:

ΔU = Q − W

the amount of energy transferred by heating must also be different.

This distinction between state quantities and path-dependent quantities is fundamental in thermodynamics.

State Functions and Path Functions

Internal energy is a state function.

Its change depends only on:

• initial state
• final state

It does not depend on the route taken.

Heat transfer and work are path dependent.

Their values depend on how the system moves between the two states.

A useful analogy is climbing a mountain.

Your change in altitude depends only on your starting and finishing elevations.

But the distance you walk depends on the route you take.

Similarly:

ΔU → depends on endpoints

Q and W → depend on path

Multi-Step Paths

Some P–V diagrams contain several separate processes.

Suppose a gas follows:

A → B → C

From A to B, it expands at constant pressure.

From B to C, its pressure decreases at constant volume.

A → B

Horizontal movement means:

constant pressure

Volume increases, so the gas does work.

B → C

Vertical movement means:

constant volume

Therefore:

W = 0

The total work for the complete process is therefore just the area under the horizontal A → B section.

Worked Multi-Step Example

A gas begins at:

A: P = 200 kPa, V = 0.002 m³

It expands at constant pressure to:

B: P = 200 kPa, V = 0.005 m³

It then drops in pressure at constant volume to:

C: P = 100 kPa, V = 0.005 m³

Work from A to B

Convert pressure:

200 kPa = 200000 Pa

Volume change:

ΔV = 0.005 − 0.002

ΔV = 0.003 m³

Work:

W = 200000 × 0.003

W = 600 J

Work from B to C

Volume does not change.

Therefore:

W = 0

Total Work

W_total = 600 + 0

W_total = 600 J

Cyclic Processes

Sometimes a system eventually returns to its original state.

This is called a cycle.

On a P–V diagram, the path forms a closed loop.

Because the gas returns to its original state:

ΔU = 0 for the complete cycle

However, the gas may still do net work.

The area enclosed by the loop represents the net work done during the cycle.

This idea is extremely important in understanding:

• engines
• refrigerators
• heat pumps
• power stations

Clockwise and Counterclockwise Cycles

For the sign convention used in these notes:

Clockwise Cycle

Expansion generally occurs at higher pressure than compression.

The gas does more work during expansion than is done on it during compression.

Therefore:

net work is done by the gas

Counterclockwise Cycle

Compression occurs at higher pressure.

Therefore:

net work is done on the gas

The direction of the loop therefore provides useful information about the overall energy transfer.

Connecting P–V Diagrams to the First Law

Recall:

ΔU = Q − W

A P–V diagram gives us information about W.

Once we determine the work and know either Q or ΔU, we can calculate the remaining quantity.

Example

A gas expands and the area under its P–V curve represents 700 J of work.

During the process, 1,100 J is transferred to the gas by heating.

Therefore:

ΔU = Q − W

ΔU = 1100 − 700

ΔU = +400 J

The internal energy increases by 400 J.

Interpreting Physical Changes

A P–V diagram is more than a mathematical graph. Each movement corresponds to something physically happening to the gas.

Moving right

Gas expands.

Moving left

Gas is compressed.

Moving upward

Pressure increases.

Moving downward

Pressure decreases.

Horizontal path

Constant pressure.

Vertical path

Constant volume.

Curved path

Pressure and volume are both changing.

By combining this information with the First Law and the ideal gas model, we can build a much more complete description of what is happening.

Units on P–V Diagrams

Always inspect the axis units carefully.

Pressure may be shown in:

• Pa
• kPa
• MPa

Remember:

1 kPa = 1000 Pa

1 MPa = 1,000,000 Pa

Volume may be shown in:

• m³
• L
• cm³

Remember:

1 L = 1 × 10⁻³ m³

When calculating work in joules using pressure and volume:

use Pa and m³

because:

Pa·m³ = J

Common Mistakes

Thinking the slope represents work

Work is related to the area under the curve, not the slope.

Using the area beside the curve

For expansion or compression, work corresponds to the area between the process path and the volume axis over the relevant volume interval.

Calculating work for a vertical line

A vertical process has:

ΔV = 0

so:

W = 0

Ignoring the direction of the process

The same path can represent expansion or compression depending on the arrow direction.

Assuming work depends only on the endpoints

Work depends on the path taken.

Forgetting unit conversions

Using kPa and m³ without accounting for the prefix can produce an answer off by a factor of 1,000.

Confusing work with internal energy

The area under a P–V curve represents work, not the total internal energy of the gas.

Key Terms

P–V diagram — a graph showing pressure against volume for a thermodynamic system.

Thermodynamic state — the condition of a system described by variables such as pressure, volume, and temperature.

Thermodynamic process — a change from one state to another.

Isobaric process — a process occurring at constant pressure.

Isochoric process — a process occurring at constant volume.

Isothermal process — a process occurring at constant temperature.

Adiabatic process — a process in which no energy is transferred by heating.

Expansion — an increase in the volume of a gas.

Compression — a decrease in the volume of a gas.

Work — energy transferred mechanically; on a P–V diagram it is represented by the area under a process curve.

State function — a quantity whose change depends only on the initial and final states.

Path dependent — describing a quantity whose value depends on the particular process followed.

Thermodynamic cycle — a sequence of processes that returns a system to its original state.

Key Takeaways

• A P–V diagram plots pressure vertically and volume horizontally.
• A point represents a thermodynamic state.
• A line or curve represents a thermodynamic process.
• Movement to the right represents expansion.
• Movement to the left represents compression.
• A horizontal line represents constant pressure.
• A vertical line represents constant volume.
• The area under a P–V curve represents work done by the gas.
• A constant-volume process has W = 0.
• Different paths between the same states can involve different amounts of work.
• Internal-energy change depends only on the initial and final states.
• Work and heat transfer depend on the path.
• For a complete cycle, the area enclosed by the loop represents net work.
• P–V diagrams connect graphical analysis with the First Law of Thermodynamics.

Check Your Understanding

1. Which quantity is plotted on the horizontal axis of a P–V diagram?
2. What does movement from left to right represent physically?
3. What type of process appears as a horizontal line?
4. What type of process appears as a vertical line?
5. Explain why no pressure-volume work is done during a constant-volume process.
6. A gas expands at 120,000 Pa from 0.002 m³ to 0.007 m³. Calculate the work done.
7. A gas expands at constant pressure of 250 kPa from 3.0 L to 7.0 L. Calculate the work done in joules.
8. A gas expands while its pressure decreases linearly from 400 kPa to 200 kPa. Its volume increases from 0.002 m³ to 0.006 m³. Calculate the work using the area of the trapezoid.
9. Two paths connect the same initial and final states. Path A lies entirely above Path B on a P–V diagram during expansion. Which path involves more work done by the gas? Explain.
10. A gas follows a vertical path from P = 100 kPa to P = 300 kPa. How much pressure-volume work is done?
11. A P–V diagram shows that a gas does 900 J of work while receiving 1,400 J by heating. Calculate its change in internal energy.
12. Explain why two processes connecting exactly the same initial and final states can have the same ΔU but different values of Q and W.

Energy Conservation Applications

The First Law of Thermodynamics allows us to analyse real systems by tracking how energy enters, leaves, and is stored.

For a thermodynamic system, energy may:

• enter or leave by heating
• enter or leave through work
• change the internal energy of the system

Using the sign convention from the previous notes:

ΔU = Q − W

where:

• ΔU = change in internal energy
• Q = energy transferred to the system by heating
• W = work done by the system on its surroundings

The First Law is fundamentally an energy balance. Energy cannot simply appear or disappear.

Thinking in Terms of a System

Before solving an energy-conservation problem, we must decide what we are studying.

This is called the system.

Everything outside the system is called the surroundings.

For example, if we study the gas inside a piston-cylinder:

System: gas

Surroundings: piston, cylinder and everything outside the gas

Energy can cross the system boundary.

Once the system has been identified, ask:

• What energy enters?
• What energy leaves?
• Is work done by or on the system?
• Does the internal energy change?

This approach is often more useful than immediately searching for an equation.

Energy Balance

A simple energy balance can be thought of as:

energy entering − energy leaving = increase in stored energy

For the simple thermodynamic systems considered here:

ΔU = Q − W

Suppose 2,500 J enters a gas by heating and the gas does 900 J of work.

Then:

ΔU = 2500 − 900

ΔU = +1600 J

The energy accounting is:

2500 J in = 1600 J stored + 900 J out

Nothing has been created or destroyed.

Energy-Flow Diagrams

An energy-flow diagram can make an energy balance easier to understand.

Suppose:

1800 J heating → [ SYSTEM ] → 700 J work

The change in internal energy is:

ΔU = 1800 − 700

ΔU = +1100 J

The system gains 1,100 J of internal energy.

If instead:

600 J heating → [ SYSTEM ] → 1000 J work

then:

ΔU = 600 − 1000

ΔU = −400 J

The system loses 400 J of internal energy.

A Systematic Problem-Solving Method

For an energy-balance problem, use the following process:

1. Define the system.
2. Identify all known energy transfers.
3. Determine which transfers enter and which leave.
4. Choose and maintain a consistent sign convention.
5. Write the First Law.
6. Substitute the known values.
7. Solve for the unknown quantity.
8. Interpret the sign and units of the answer.
9. Check whether the answer is physically reasonable.

This method works for many thermodynamic systems.

Worked Example: Simple Energy Balance

A gas receives 3,200 J of energy by heating and does 1,400 J of work on its surroundings.

Calculate its change in internal energy.

Use:

ΔU = Q − W

Substitute:

ΔU = 3200 − 1400

ΔU = +1800 J

The positive result means the internal energy increases by 1,800 J.

Finding an Unknown Heat Transfer

A gas does 600 J of work while its internal energy increases by 900 J.

Find Q.

Start with:

ΔU = Q − W

Rearrange:

Q = ΔU + W

Substitute:

Q = 900 + 600

Q = 1500 J

Therefore, 1,500 J of energy was transferred into the gas by heating.

Finding Unknown Work

A gas receives 4,000 J by heating while its internal energy increases by 2,500 J.

Find the work done by the gas.

Start with:

ΔU = Q − W

Rearrange:

W = Q − ΔU

Substitute:

W = 4000 − 2500

W = 1500 J

The gas does 1,500 J of work on its surroundings.

Engines

A heat engine is a device that converts part of an energy transfer from a high-temperature source into useful mechanical work.

Examples include:

• internal combustion engines
• steam engines
• gas turbines
• steam turbines

A simplified engine involves:

high-temperature source → working substance → useful work + energy transferred to cooler surroundings

Not all of the energy entering an engine becomes useful work.

Energy Balance in an Engine

Suppose an engine receives:

Q_in = 5000 J

and transfers:

Q_out = 3200 J

to the cooler surroundings.

If the engine completes a cycle, it returns to its original thermodynamic state.

Therefore:

ΔU = 0

The energy balance becomes:

energy in = useful work + energy out

Therefore:

W = 5000 − 3200

W = 1800 J

The engine produces 1,800 J of mechanical work per cycle.

Why Engines Cannot Simply Convert All Input Energy to Work

Real heat engines must transfer some energy to their surroundings.

They also experience effects such as:

• friction
• heating of engine components
• exhaust losses
• sound
• incomplete combustion
• other irreversible processes

Consequently, only part of the supplied energy becomes useful mechanical work.

Engine Efficiency

The efficiency of an engine compares useful energy output with energy input.

efficiency = useful energy output / total energy input

As a percentage:

efficiency = (useful energy output / total energy input) × 100%

Example

An engine receives 8,000 J and produces 2,800 J of useful work.

Efficiency:

efficiency = 2800 / 8000 × 100%

efficiency = 35%

Therefore, 35% of the input energy becomes useful work.

The remaining energy is transferred elsewhere.

Worked Engine Problem

A heat engine receives 12,000 J from a high-temperature source during each cycle and transfers 7,500 J to its surroundings.

Calculate the useful work.

W = Q_in − Q_out

W = 12000 − 7500

W = 4500 J

Now calculate the efficiency:

efficiency = 4500 / 12000 × 100%

efficiency = 37.5%

So the engine produces 4,500 J of work per cycle with an efficiency of 37.5%.

Compressors

A compressor uses mechanical work to increase the pressure of a gas.

Instead of obtaining work from an expanding gas, we do work on the gas.

Examples include compressors in:

• refrigerators
• air conditioners
• compressed-air systems
• gas pipelines
• industrial equipment
• heat pumps

Energy Transfer in a Compressor

When a gas is compressed:

• work is done on the gas
• energy enters the gas
• its internal energy may increase
• its temperature may increase

Using our convention, W represents work done by the system.

Therefore, when work is done on the gas:

W is negative

Example

800 J of work is done on a gas, and no energy is transferred by heating.

Therefore:

Q = 0

W = −800 J

Using:

ΔU = Q − W

ΔU = 0 − (−800)

ΔU = +800 J

The gas gains 800 J of internal energy.

Compressor with Energy Loss

Suppose 2,000 J of work is done on a gas while 500 J of energy leaves the gas by heating.

Using our sign convention:

Q = −500 J

W = −2000 J

Therefore:

ΔU = Q − W

ΔU = −500 − (−2000)

ΔU = +1500 J

The gas's internal energy increases by 1,500 J.

The energy accounting can also be viewed more simply:

2000 J in by work − 500 J out by heating = 1500 J stored

Both approaches give the same result.

Engines and Compressors Compared

The First Law applies to both. The main difference is the direction of energy transfer.

Engineering Energy Systems

The First Law can be applied to much more than a sealed gas cylinder.

Engineers use energy balances to analyse systems such as:

• engines
• turbines
• compressors
• pumps
• boilers
• heat exchangers
• refrigerators
• power stations
• air-conditioning systems

The details can become more complex, but the basic question remains:

Where does the energy come from, where does it go, and how much is stored?

Turbines

A turbine extracts energy from a moving or expanding fluid.

For example, high-temperature steam can enter a turbine and expand.

As it passes through the turbine:

• the steam transfers energy
• turbine blades rotate
• mechanical work is produced
• the steam leaves with less available energy

Power Stations

A simplified thermal power station involves several energy transfers:

fuel or other energy source → heating of working fluid → expanding steam or gas → turbine → generator

At each stage, conservation of energy applies.

For example, if a turbine receives an effective energy input of 1,000 MJ and produces 420 MJ of useful mechanical output, the remaining energy has not disappeared.

It must be accounted for through:

• energy remaining in the working fluid
• energy transferred to cooling systems
• friction
• sound
• other losses to the surroundings

Energy conservation applies even when a machine is inefficient.

Refrigerators and Heat Pumps

A refrigerator does not "create cold."

Instead, it uses work to transfer energy from a cooler region to a warmer region.

A simplified energy balance is:

energy removed from refrigerator + electrical work input = energy transferred to room

For example, suppose a refrigerator removes 400 J from its interior while receiving 150 J of electrical work.

The energy transferred to the room is:

400 + 150 = 550 J

This explains why the back of a refrigerator releases more energy to the room than is removed from the refrigerated compartment.

Assumptions in Thermodynamic Calculations

Real engineering systems are complicated. To make calculations manageable, we often make assumptions.

An assumption simplifies the system while keeping the model useful.

Common assumptions include the following.

Negligible Heat Loss

We may assume:

Q ≈ 0

This is approximately reasonable when a process occurs rapidly or the system is well insulated.

Such a process may be treated as adiabatic.

Constant Pressure

For some expansion problems, we assume the external pressure remains constant.

This allows:

W = pΔV

Ideal Gas Behaviour

We may assume the gas behaves as an ideal gas.

This neglects molecular volume and intermolecular forces under conditions where these effects are relatively unimportant.

Negligible Friction

We may ignore friction in an idealized engine or piston.

In a real machine, friction transfers energy into internal energy and reduces useful mechanical output.

No Change in Kinetic Energy

If the system as a whole is not accelerating significantly, changes in macroscopic kinetic energy may be ignored.

No Change in Gravitational Potential Energy

If height changes are small, gravitational potential-energy changes may be treated as negligible.

Why Assumptions Matter

Every mathematical model is based on assumptions.

Consider a piston problem where we calculate:

W = pΔV

This assumes that the relevant pressure can be treated appropriately over the expansion.

If pressure changes significantly, simply using one pressure value may not accurately represent the work.

Similarly, assuming no heat loss in an engine may make a classroom calculation easier, but real engines always transfer some energy to their surroundings.

A good solution should therefore state important assumptions when they affect the answer.

Ideal vs. Real Systems

An ideal system is a simplified model.

A real system includes additional effects.

For example:

Ideal Compressor

We might assume:

• no energy loss by heating
• no friction
• perfect insulation

Real Compressor

In reality:

• components become warm
• friction occurs
• sound is produced
• energy transfers to the surroundings

The First Law still holds in both cases.

The difference is that the real system requires more energy transfers to be included in the balance.

Worked Engineering Problem

A compressor receives 5,000 J of mechanical work. During compression, 1,200 J of energy is transferred from the gas to the surroundings by heating.

Find the change in internal energy.

Energy entering by work:

5000 J

Energy leaving by heating:

1200 J

Therefore:

ΔU = 5000 − 1200

ΔU = +3800 J

The internal energy of the gas increases by 3,800 J.

Using the formal sign convention gives the same result:

Q = −1200 J

W = −5000 J

ΔU = Q − W

ΔU = −1200 − (−5000)

ΔU = +3800 J

Worked Engineering Problem: Engine

An engine receives 20,000 J of energy from combustion during one cycle. It produces 6,000 J of useful mechanical work and returns to its initial state.

Because it completes a cycle:

ΔU = 0

Therefore, the remaining energy must leave the system:

Q_out = 20000 − 6000

Q_out = 14000 J

The engine transfers 14,000 J to the surroundings.

Its efficiency is:

efficiency = 6000 / 20000 × 100%

efficiency = 30%

Multi-Step Energy Balance

Sometimes several transfers occur.

Suppose a system:

• receives 4,000 J by heating
• receives 1,500 J through work done on it
• loses 800 J by heating

Instead of worrying about signs immediately, we can first organize the transfers.

Total energy entering:

4000 + 1500 = 5500 J

Energy leaving:

800 J

Net energy gained:

5500 − 800 = 4700 J

Therefore:

ΔU = +4700 J

This "energy in minus energy out" method can be particularly useful when several transfers occur.

Connecting Energy Balances to P–V Diagrams

A P–V diagram can provide information about the work term in the First Law.

Recall:

area under a P–V curve = work done by the gas

Suppose the area under an expansion curve represents:

W = 1200 J

and 2,000 J enters by heating.

Then:

ΔU = Q − W

ΔU = 2000 − 1200

ΔU = +800 J

This connects graphical analysis directly to conservation of energy.

Energy Conservation in a Complete Cycle

If a thermodynamic system completes a cycle, it returns to its original state.

Therefore:

U_final = U_initial

and:

ΔU = 0

From:

ΔU = Q − W

we obtain:

Q_net = W_net

So over a complete cycle, the net energy transferred to the system by heating equals the net work done by the system.

This principle is central to the operation of heat engines.

Checking an Energy Balance

After completing a calculation, check that:

total energy entering = total energy leaving + increase in stored energy

For example:

Energy entering = 6,000 J

Work leaving = 2,200 J

Increase in internal energy = 3,800 J

Check:

6000 = 2200 + 3800

6000 = 6000

The energy balance is satisfied.

This is one of the best ways to catch mistakes.

Common Mistakes

Not defining the system

Whether energy is entering or leaving depends on what has been chosen as the system.

Confusing work done by and work done on the gas

Using our convention:

• work done by gas → W positive
• work done on gas → W negative

Ignoring energy losses

Real engineering systems transfer energy through heating, friction, sound and other pathways.

Assuming inefficiency violates conservation of energy

Energy is never "lost" in the conservation sense. It is transferred into less useful forms or to the surroundings.

Mixing sign conventions

If using:

ΔU = Q − W

then W means work done by the system. Keep this convention throughout the calculation.

Ignoring assumptions

A calculated result may only be valid if assumptions such as constant pressure, negligible heat loss or ideal-gas behaviour are reasonable.

Key Terms

Energy balance — an accounting of all energy entering, leaving, and being stored within a system.

System — the object or collection of matter selected for analysis.

Surroundings — everything outside the system.

System boundary — the boundary separating the system from its surroundings.

Heat engine — a device that converts part of an energy transfer from a high-temperature source into mechanical work.

Compressor — a device that uses mechanical work to increase the pressure of a gas.

Turbine — a machine that extracts energy from a moving or expanding fluid to produce mechanical work.

Efficiency — the ratio of useful energy output to total energy input.

Assumption — a simplifying condition adopted when constructing or using a model.

Adiabatic process — a process in which no energy is transferred by heating.

Ideal gas — a simplified model of a gas in which molecular volume and intermolecular interactions are neglected.

Energy loss — an informal description of energy transferred away from the desired output; the energy itself is still conserved.

Key Takeaways

• The First Law provides an energy balance for thermodynamic systems.
• Energy entering, leaving, and being stored must all be accounted for.
• Using our convention, ΔU = Q − W.
• Engines transfer part of their energy input into useful mechanical work.
• Compressors require mechanical work to be transferred into a gas.
• Turbines extract energy from expanding or moving fluids.
• Refrigerators use work to transfer energy from a cooler region to warmer surroundings.
• Efficiency measures the fraction of input energy converted into useful output.
• An inefficient machine still obeys conservation of energy.
• Engineering calculations often rely on simplifying assumptions.
• Important assumptions can include negligible heat transfer, constant pressure, ideal-gas behaviour, and negligible friction.
• P–V diagrams can provide the work term needed for an energy balance.
• For a complete thermodynamic cycle, ΔU = 0.
• Every realistic energy analysis should ask: Where does the energy enter, where does it leave, and where is it stored?

Check Your Understanding

1. A gas receives 2,800 J by heating and does 900 J of work. Calculate ΔU.
2. A gas does 1,200 J of work while its internal energy decreases by 300 J. Determine Q.
3. A compressor does 2,500 J of work on a gas while 600 J leaves by heating. Calculate the change in internal energy.
4. A heat engine receives 10,000 J per cycle and transfers 6,500 J to its surroundings. Calculate its useful work output.
5. Calculate the efficiency of the engine in Question 4.
6. An engine receives 15,000 J and has an efficiency of 32%. Calculate its useful work output.
7. A refrigerator removes 600 J from its interior and receives 250 J of electrical work. How much energy is transferred to the room?
8. Explain why a compressor can become hot while operating.
9. A gas receives 3,000 J by heating. The area under its P–V expansion curve represents 1,100 J of work. Calculate ΔU.
10. Explain why an engine with an efficiency of 30% does not violate conservation of energy.
11. A calculation assumes that a gas expansion is adiabatic. Explain what this assumption means and give one situation in which it might be approximately reasonable.
12. An engineer models a compressor as frictionless and perfectly insulated. Identify two ways in which a real compressor would differ from this idealized model.