$$\left(\frac{\partial U}{\partial V}\right)_{\!T}$$

I know that this partial derivative is equal to zero for an ideal gas, but how do I determine that? Do I need to use the fundamental thermodynamic relations?


1 Answer 1


In fact, you need to use two of the fundamental relations.

For an ideal gas, the internal energy $U$ can be considered to be a function of $V$ and $T$. So, the total differential of $U$ would read

$$\mathrm{d}U = \left(\frac{\partial U}{\partial V}\right)_{\!T}\,\mathrm{d}V + \left(\frac{\partial U}{\partial T}\right)_{\!V}\,\mathrm{d}T \tag{1}$$

The first partial derivative is the internal pressure, which we are trying to determine.

The other thing that we know is that (assuming only expansion work and no net reaction)

$$\mathrm{d}U = T\,\mathrm{d}S - p\,\mathrm{d}V \tag{2}$$

The idea here is that we want to change the term involving $\mathrm{d}S$ into a combination of $\mathrm{d}T$ and $\mathrm{d}V$, so that we can compare coefficients with equation $(1)$. So, we just express $S$ as a function of $T$ and $V$:

$$\mathrm{d}S = \left(\frac{\partial S}{\partial V}\right)_{\!T}\,\mathrm{d}V + \left(\frac{\partial S}{\partial T}\right)_{\!V}\,\mathrm{d}T \tag{3}$$

Substituting $(3)$ into $(2)$

$$\begin{align} \mathrm{d}U &= T\left[\left(\frac{\partial S}{\partial V}\right)_{\!T}\,\mathrm{d}V + \left(\frac{\partial S}{\partial T}\right)_{\!V}\,\mathrm{d}T\right] - p\,\mathrm{d}V \tag{4} \\ &= \left[T\left(\frac{\partial S}{\partial V}\right)_{\!T} - p\right]\mathrm{d}V + \left[T\left(\frac{\partial S}{\partial T}\right)_{\!V}\right]\mathrm{d}T \tag{5} \end{align}$$

Comparing the coefficient of the $\mathrm{d}V$ term with $(1)$ we find that1

$$\left(\frac{\partial U}{\partial V}\right)_{\!T} = T\left(\frac{\partial S}{\partial V}\right)_{\!T} - p \tag{6}$$

Equation $(6)$ is entirely general and applies to all substances, not just ideal gases. Now, we turn to the Helmholtz free energy:

$$\mathrm{d}A = -S\,\mathrm{d}T - p\,\mathrm{d}V \tag{7}$$

From this we obtain the Maxwell relation

$$\left(\frac{\partial S}{\partial V}\right)_{\!T} = \left(\frac{\partial p}{\partial T}\right)_{\!V} \tag{8}$$

For an ideal gas $p = nRT/V$, so

$$\left(\frac{\partial p}{\partial T}\right)_{\!V} = \frac{nR}{V} \tag{9}$$

Finally, substituting $(8)$ and $(9)$ into $(6)$ gives

$$\begin{align} \left(\frac{\partial U}{\partial V}\right)_{\!T} &= T\left(\frac{nR}{V}\right) - p \tag{10} \\ &= \frac{nRT}{V} - p \tag{11} \\ &= p - p \tag{12} \\ &= 0 \tag{13} \end{align}$$

as desired.


1 Some texts may write that you can obtain $(6)$ from $(2)$ simply by "dividing" by $\mathrm{d}V$ and imposing conditions of constant temperature:

$$\begin{align} \mathrm{d}U &= T\,\mathrm{d}S - p\,\mathrm{d}V & & \tag{2} \\ \frac{\mathrm{d}U}{\mathrm{d}V} &= T\frac{\mathrm{d}S}{\mathrm{d}V} - p\frac{\mathrm{d}V}{\mathrm{d}V} & &\text{(dividing by }\mathrm{d}V\text{)} \tag{14} \\ \left(\frac{\partial U}{\partial V}\right)_{\!T} &= T\left(\frac{\partial S}{\partial V}\right)_{\!T} - p\left(\frac{\partial V}{\partial V}\right)_{\!T} & &\text{(constant }T\text{)} \tag{15} \\ &= T\left(\frac{\partial S}{\partial V}\right)_{\!T} - p & &\text{(derivative of }V\text{ with respect to itself is }1\text{)} \tag{6} \end{align}$$

This gives the correct result, and in general this "shortcut" works, but it is mathematically not rigorous.

  • $\begingroup$ Why the shortcut is not rigorous ? is it because $dV$ can be zero ? $\endgroup$
    – user31607
    Feb 11, 2017 at 19:50
  • 1
    $\begingroup$ That's just how calculus works. dx/dy is not actually a fraction of infinitesimals and you cannot "divide through" by infinitesimal quantities. Sometimes it works, but sometimes it does not. The best example is: $$\left(\frac{\partial x}{\partial y}\right)_{\!z} \left(\frac{\partial y}{\partial z}\right)_{\!x} \left(\frac{\partial z}{\partial x}\right)_{\!y} = -1$$ (if you simply cancelled out everything, you would arrive at $+1$ - the wrong answer). $\endgroup$ Feb 11, 2017 at 19:52
  • $\begingroup$ @A---B It doesn't matter whether dV is zero or not. This goes back to the very fundamentals of calculus: although we write it like a fraction, the derivative is not a fraction. $\endgroup$ Feb 11, 2017 at 19:54
  • $\begingroup$ But, I think in non standard analysis we can do that, where derivative is defined as $$y^\prime = \operatorname{st}\left({\Delta y \over \Delta x}\right)$$; $\Delta x$, $\Delta y \in \mathbb{^*R}$ and are +ve infinitesimals and $\operatorname{st}$ is the standard part. I don't know about partial differentiation under non-standard analysis, so I won't comment on that. I have just studied a bit of normal differentiation under non-standard analysis. $\endgroup$
    – user31607
    Feb 11, 2017 at 19:58
  • $\begingroup$ Right; you've lost me. I don't know this stuff. $\endgroup$ Feb 11, 2017 at 20:00

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