First Law of Thermodynamics
3. The First Law of Thermodynamics
Learning outcomes
- I can state the First Law.
- I can explain conservation of energy in thermodynamic systems.
- I can calculate changes in internal energy.
- I can interpret energy transfer diagrams.
- I can apply the First Law to simple processes.
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.
| Process | Sign |
|---|---|
| Heating system | Q > 0 |
| Cooling system | Q < 0 |
| System does work | W > 0 |
| Work done on system | W < 0 |
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:
- What is the system?
- Which arrows represent energy entering?
- Which arrows represent energy leaving?
- Is the transfer caused by heating or work?
- 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
| Process | Q | W | Effect on ΔU |
|---|---|---|---|
| Heating rigid gas | + | 0 | Increases |
| Cooling rigid gas | − | 0 | Decreases |
| Gas expands without heating | 0 | + | Decreases |
| Gas compressed without heating | 0 | − | Increases |
| Heating exceeds work output | + | + | Increases |
| Work output exceeds heating | + | + | Decreases |
| Heating equals work output | + | + | No change |
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
- State the First Law of Thermodynamics in words.
- Using the sign convention in these notes, write the First Law as an equation.
- A gas receives 900 J by heating and does 300 J of work. Calculate ΔU.
- A gas receives 1,200 J by heating and does 1,500 J of work. Calculate ΔU and explain what the sign means.
- A gas loses 400 J by heating and does no work. Calculate ΔU.
- A gas is compressed, with 650 J of work done on it. No energy is transferred by heating. Calculate ΔU.
- 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?
- A gas receives 1,600 J by heating and its internal energy increases by 1,000 J. Calculate the work done by the gas.
- 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.
- Explain how a system can have ΔU = 0 even though energy is being transferred both into and out of it.
- A gas expands adiabatically and does 450 J of work. Determine its change in internal energy.
- 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.