The First Law for a Closed System
Joule's experiment: raise the same internal energy twice, once by heat and once by work, and watch the bar land in the same place.
Skip to the animationInternal energy is a property, so the change in it between two states is fixed regardless of whether the energy arrived as heat or as work — which is what ΔU = Q − W says, and what Joule's experiment established.
Joule's experiment, which is the whole argument
- 1Heat a rigid sealed vessel of water with 100 kJ. The boundary cannot move, so
W = 0and all of it becomes internal energy. - 2Now take an identical insulated vessel and stir it with a paddle driven by a falling weight, adding 100 kJ of work.
Q = 0. - 3The temperature rise is identical. The two final states are indistinguishable.
So the system does not record *how* the energy arrived. Internal energy is a property — fixed by the state, independent of history. Heat and work are transfers, not things a system contains.
The law, and its sign convention
ΔU = Q − W, with Q positive into the system and W positive when done by it. The convention is engine-shaped: you buy heat and sell work, so each is positive in its useful direction.
With a movable boundary both terms appear at once. Heat 100 kJ into a gas under a piston that does 30 kJ of work, and ΔU = 70 kJ. The rise of the piston is the work term, W = ∫p dV — the area under the process on a p-V diagram.
The standard processes
| Process | Constant | Consequence |
|---|---|---|
| Isochoric | Volume | W = 0, so ΔU = Q |
| Isobaric | Pressure | W = pΔV, so Q = ΔH |
| Isothermal | Temperature | ΔU = 0 for an ideal gas, so Q = W |
| Adiabatic | No heat crosses | ΔU = −W |
| Polytropic | pVⁿ | The general case; the others are special values of n |
Adiabatic does not mean isolated. No heat crosses, but work still may — an adiabatic compression raises internal energy substantially with Q = 0 throughout. This is the most persistent confusion in the subject.
Round a cycle
Every property returns to its starting value round a closed cycle, so ∮dU = 0 and therefore ∮δQ = ∮δW: net heat in equals net work out. The enclosed area on a p-V diagram is that net work.
This is what makes heat engine analysis possible at all, and it is a direct consequence of U being a property rather than a separate assumption.
What the law forbids, and what it does not
- Forbidden: a perpetual motion machine of the first kind — one producing work with no energy input. It would need
∮δW > 0with∮δQ = 0. - Not forbidden: a device taking 100 kJ of heat and producing 100 kJ of work. The books balance perfectly.
Nobody has ever built the second one. The first law counts energy and says nothing whatsoever about direction — which is precisely the gap the second law exists to close.
The numbers you will be asked for
- First law, closed system
ΔU = Q − W
Q positive in, W positive out.
- Displacement work
W = ∫ p dV
The area under the process line on a p-V diagram.
- Cyclic form
∮ δQ = ∮ δW
Because ∮dU = 0 for any property.
- Specific heats
c_v = (∂u/∂T)_v · c_p = (∂h/∂T)_p
And c_p − c_v = R for an ideal gas.
- Adiabatic ideal gas
pVᵞ = constant
γ = c_p/c_v.
Advantages and disadvantages
Advantages
- One equation covers every closed-system process there is.
- Internal energy is path-independent, so only the end states matter.
- Round a cycle it reduces to net heat equals net work.
- It rules out the creation of energy immediately and unconditionally.
Disadvantages
- It says nothing about direction, so it permits processes that never occur.
- It gives no efficiency limit at all.
- Properties are defined only in equilibrium, so rapid processes must be approximated as quasi-static.
- Two sign conventions are in circulation, differing by a minus sign.
Watch it work
Check yourself
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One question at a time. Pick an answer to see why it is right or wrong, then move on — there is no score to keep and nothing is saved.