# 5.5: The Joule Experiment

Going back to the expression for changes in internal energy that stems from assuming that $$U$$ is a function of $$V$$ and $$T$$ (or $$U(V, T)$$ for short)

$dU = \left( \dfrac{\partial U}{\partial V} \right)_TdV+ \left( \dfrac{\partial U}{\partial T} \right)_V dT$

one quickly recognizes one of the terms as the constant volume heat capacity, $$C_V$$. And so the expression can be re-written

$dU = \left( \dfrac{\partial U}{\partial V} \right)_T dV + C_V dT$

But what about the first term? The partial derivative is a coefficient called the “internal pressure”, and given the symbol $$\pi_T$$.

$\pi_T = \left( \dfrac{\partial U}{\partial V} \right)_T$

James Prescott Joule (1818-1889) recognized that $$\pi_T$$ should have units of pressure (Energy/volume = pressure) and designed an experiment to measure it.

He immersed two copper spheres, A and B, connected by a stopcock. Sphere A is filled with a sample of gas while sphere B was evacuated. The idea was that when the stopcock was opened, the gas in sphere A would expand ($$\Delta V > 0$$) against the vacuum in sphere B (doing no work since $$p_{ext} = 0$$. The change in the internal energy could be expressed

$dU = \pi_T dV + C_V dT$

But also, from the first law of thermodynamics

$dU = dq + dw$

Equating the two

$\pi_T dV + C_V dT = dq + dw$

and since $$dw = 0$$

$\pi_T dV + C_V dT = dq$

Joule concluded that $$dq = 0$$ (and $$dT = 0$$ as well) since he did not observe a temperature change in the water bath which could only have been caused by the metal spheres either absorbing or emitting heat. And because $$dV > 0$$ for the gas that underwent the expansion into an open space, $$\pi_T$$ must also be zero! In truth, the gas did undergo a temperature change, but it was too small to be detected within his experimental precision. Later, we (once we develop the Maxwell Relations) will show that

$\left( \dfrac{\partial U}{\partial V} \right)_T = T \left( \dfrac{\partial p}{\partial T} \right)_V -p \label{eq3}$

## Application to an Ideal Gas

For an ideal gas $$p = RT/V$$, so it is easy to show that

$\left( \dfrac{\partial p}{\partial T} \right)_V = \dfrac{R}{V} \label{eq4}$

so combining Equations \ref{eq3} and \ref{eq4} together to get

$\left( \dfrac{\partial U}{\partial V} \right)_T = \dfrac{RT}{V} - p \label{eq5}$

And since also becuase $$p = RT/V$$, then Equation \ref{eq5} simplifies to

$\left( \dfrac{\partial U}{\partial V} \right)_T = p -p = 0$

So while Joule’s observation was consistent with limiting ideal behavior, his result was really an artifact of his experimental uncertainty masking what actually happened.

## Appliation to a van der Waals Gas

For a van der Waals gas,

$p = \dfrac{RT}{V-b} - \dfrac{a}{V^2} \label{eqV1}$

so

$\left( \dfrac{\partial p}{\partial T} \right)_V = \dfrac{R}{V-b} \label{eqV2}$

and

$\left( \dfrac{\partial U}{\partial V} \right)_T = T\dfrac{R}{V-b} - p \label{eqV3}$

Substitution of the expression for $$p$$ (Equation \ref{eqV1}) into this Equation \ref{eqV3}

$\left( \dfrac{\partial U}{\partial V} \right)_T = \dfrac{a}{V^2}$

In general, it can be shown that

$\left( \dfrac{\partial p}{\partial T} \right)_V = \dfrac{\alpha}{\kappa_T}$

And so the internal pressure can be expressed entirely in terms of measurable properties

$\left( \dfrac{\partial U}{\partial V} \right)_T = T \dfrac{\alpha}{\kappa_T}-p$

and need not apply to only gases (real or ideal)!