# Work and the Joule Experiment

Physics II · Thermodynamics · https://tryals.app/learn/physics-ii/work-and-the-joule-experiment

## Work Depends on How You Get There

For a fluid the elementary work done **on** the system is $\delta W = -P\,dV$, so a finite process gives

$$W = -\int_{V_1}^{V_2} P\,dV$$

and the integral needs the whole path, not just the endpoints. Different systems have their own forms, $\delta W = \sigma\,dA$ for a surface, $\mathcal{E}\,dq$ for a cell, $-\mathbf{m}\cdot d\mathbf{B}$ for a magnetic moment, but all share the pattern *intensive variable times change in extensive variable*.

Comparing two paths between the same endpoints makes the path dependence concrete:

| Path | Work done by the gas |
|---|---|
| Isobaric at $P_1$, then isochoric | $P_1(V_2 - V_1)$ |
| Isochoric, then isobaric at $P_2$ | $P_2(V_2 - V_1)$ |
| Isothermal | $nRT\ln(V_2/V_1)$ |
| Free expansion into vacuum | Zero |

Same start, same finish, different answers. That is why $\delta W$ is written with a $\delta$ rather than a $d$, it is an inexact differential with no function behind it.

**Joule's experiment** is what rescues energy from this. He performed **adiabatic** work on water in three different ways — a falling weight turning a paddle, electrical heating, mechanical compression — and found that the work needed to move between two given states was *always the same*, regardless of method. If adiabatic work is path-independent, it defines a state function:

$$\Delta U = W_{adiabatic}$$

Once $U$ exists, heat is defined as the difference for a non-adiabatic path: $Q = \Delta U - W$. Heat is not a separate substance but a residual, whatever energy crossed the boundary that the work terms do not account for.

> **Common pitfall:** reading free expansion into a vacuum as "the gas did work by expanding". There is nothing to push against, so $P_{ext} = 0$ and $W = 0$ exactly. The gas expands, but does no work and, for an ideal gas, does not cool either.

## Practice questions

6 of this lesson's 11 practice questions, with answers. The full set is in the app.

### 1. Why does a gas expanding freely into an evacuated chamber do no work at all?

A. Its internal energy is fully converted into heat
B. The expansion happens too fast for work to occur
C. Its temperature remains constant during the process
D. None, because there is nothing to push against

**Answer:** D. None, because there is nothing to push against

**Why:** Work is $-\int P_{ext}dV$, and $P_{ext} = 0$ throughout a free expansion, so $W = 0$ exactly. The gas expands and does nothing, and for an ideal gas it does not cool either, since $U$ depends on temperature alone.

Page: https://tryals.app/practice/physics-ii/work-and-the-joule-experiment/why-does-a-gas-expanding-freely-into-an-evacuated-chamber-do-no-work

### 2. Arrange the steps of Joule’s reasoning in the order they establish internal energy.

**Answer:**

1. Perform adiabatic work on a system by several different methods
2. Observe that the work needed between two given states is always the same
3. Conclude that adiabatic work is path-independent and defines a state function
4. Define heat as the difference between the energy change and the work for a non-adiabatic path

**Why:** The measurement comes first: adiabatic work turned out path-independent no matter how it was delivered. Only that fact licenses defining $U$, and only once $U$ exists can heat be defined as the residual $Q = \Delta U - W$.

Page: https://tryals.app/practice/physics-ii/work-and-the-joule-experiment/arrange-the-steps-of-joules-reasoning-in-the-order-they-establish

### 3. A gas is compressed at a constant external pressure of 200 kPa from 0.050 m$^3$ to 0.020 m$^3$. Compute the work done on the gas, in kJ.

**Answer:** 6 (within ±0.2)

**Why:** $W = -P\Delta V = -200 \times (-0.030) = +6.0$ kJ. Compression means the surroundings do work on the gas, so the sign is positive from the system’s point of view.

Page: https://tryals.app/practice/physics-ii/work-and-the-joule-experiment/a-gas-is-compressed-at-a-constant-external-pressure-of-200-kpa-from

### 4. Sort each expression by the kind of system whose work it describes.

**Answer:**

- A fluid changing volume: Minus pressure times change in volume
- A surface changing area: Surface tension times change in area
- An electrical cell: EMF times charge transferred

**Why:** Every work term pairs an intensive variable with the change in its extensive partner: $P$ with $V$, surface tension with area, EMF with charge. The pattern is universal even though the physics differs.

Page: https://tryals.app/practice/physics-ii/work-and-the-joule-experiment/sort-each-expression-by-the-kind-of-system-whose-work-it-describes

### 5. Which processes do zero work on or by a gas?

A. A rigid sealed container being warmed
B. A constant-volume heating
C. Expansion into an evacuated chamber
D. An isothermal expansion against a piston

**Answer:** A. A rigid sealed container being warmed; B. A constant-volume heating; C. Expansion into an evacuated chamber

**Why:** Free expansion has no opposing pressure and constant-volume processes have no $dV$, so all three give zero. An isothermal expansion against a piston has both ingredients and does $nRT\ln(V_2/V_1)$.

Page: https://tryals.app/practice/physics-ii/work-and-the-joule-experiment/which-processes-do-zero-work-on-or-by-a-gas

### 6. Work is path-dependent, yet Joule used mechanical work to establish a unique internal energy function. How does restricting measurements to adiabatic boundaries resolve this contradiction?

A. Work becomes identical across all paths once heat flow is banned
B. Heat and work convert into each other at a fixed mechanical ratio
C. Adiabatic work depends solely on endpoints, defining a state function
D. Any irreversible losses are eliminated by using thermally rigid walls

**Answer:** C. Adiabatic work depends solely on endpoints, defining a state function

**Why:** Adiabatic work uniquely fixes energy changes because isolating the system prevents boundary leakage, whereas non-adiabatic routes allow variable heat exchanges. Equating adiabaticity with irreversibility confuses boundary constraints with process dynamics.

Page: https://tryals.app/practice/physics-ii/work-and-the-joule-experiment/work-is-path-dependent-yet-joule-used-mechanical-work-to-establish-a