I'm currently reviewing Ball's physical chemistry in preparation for the physical chemistry ACS exam.

On page 47 he makes the claim that $ \frac{\partial U}{\partial V}|_{T}=0 $ for an ideal gas. This seems like a reasonable assumption given that temperature is a measure of energy content. However, I cannot find a way to justify this mathematically, nor can I find a way to invoke the ideal gas law. On the page where this happens, he's talking about expanding into a vacuum.

If we let $ dU=\frac{\partial U}{\partial V}|_{T}dV+\frac{\partial U}{\partial T}|_{V}dT $ we can then say $ dU=\frac{\partial U}{\partial V}|_{T}dV+\frac{\partial U}{\partial T}|_{V}dT=dQ+dW $

Since $\frac{\partial U}{\partial T}|_{V}$ is constant volume heat capacity we can say that this cancels out with dQ so we are left with:

$ \frac{\partial U}{\partial V}|_{T}dV=dW $ or

$\frac{\partial U}{\partial V}|_{T}dV=-P_{external}dV$

Which tells us the only way this is zero is if the gas is expanding into a vacuum. So what's going on? On page 53 of the text he uses in a situation where there isn't a vacuum.

Thanks for any help!

  • $\begingroup$ If z = x + y = u + v = 0 it does not generally mean x = u and y = v. It just means x + y = 0 and u + v = 0 // In your case it means dU = 0 + 0 = (dW) + (-dW) = 0 // dQ <> Cv .dT = 0 // dW <> (dU/dV)_T . dV = 0 $\endgroup$
    – Poutnik
    May 14 at 9:31
  • 3
    $\begingroup$ Note that (dU/dV)_T=0 is not an assumption, but the consequence of the ideal gas definition. $\endgroup$
    – Poutnik
    May 14 at 10:26

2 Answers 2


The main idea is to obtain a mathematical expression only involving the measurable variables $p$, $T$, and $V$.

We start with the internal energy as a function of its canonical variables. Differentiating at both sides with respect to volume at constant temperature we get \begin{equation} \mathrm{d}U = -p\mathrm{d}V + T\mathrm{d}S \to \left(\frac{\partial U}{\partial V}\right)_T = -p + T\color{blue}{\left(\frac{\partial S}{\partial V}\right)_T} \tag{1} \end{equation} We need an expression for the last partial derivative in blue in Eq. (1). In here, we can use the Maxwell's relationship applied to Helmholt'z free energy in terms of its canonical variables. Writing this equation we have \begin{equation} \mathrm{d}A = -p\mathrm{d}V - S\mathrm{d}T \tag{2} \end{equation} and from Eq. (2) we get the desired relationship \begin{equation} \color{blue}{\left(\frac{\partial S}{\partial V}\right)_T} = \left(\frac{\partial p}{\partial T}\right)_V \tag{3} \end{equation} Combining Eqs. (1) and (3) we get \begin{equation} \left(\frac{\partial U}{\partial V}\right)_T = -p + T\left(\frac{\partial p}{\partial T}\right)_V \tag{4} \end{equation} For an ideal gas we have $p = RT/V$ and Eq. (4) then yields the result you are seeking \begin{equation} \left(\frac{\partial U}{\partial V}\right)_T = -p + T\left(\frac{R}{V}\right) = -p + \left(\frac{RT}{V}\right) = -p + p = 0 \therefore \boxed{\left(\frac{\partial U}{\partial V}\right)_T = 0} \tag{5} \end{equation}


The internal energy depends on the random kinetic energy and the potential energy of interaction of the molecules. The latter is equal to zero when the specific volume is very high and the molecules are far apart. This is the ideal gas limit, and here, U no longer depends on specific volume. The former is the part of U which depends on temperature.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.