In an exercise, I have derived that

$$\mathrm dS_\mathrm{m} = \frac{3R}{2T}\mathrm dT + \frac{R}{V}\mathrm dV$$

using the 1st and 2nd law of thermodynamics (1 mol ideal gas).

Now, I want to write $\mathrm dS$ as a function of $T$ and $P$ instead of $T$ and $V$. How can I do that?

I have tried to do this:

$$\frac{R}{V} = \frac{P}{T}$$


$$\mathrm dV = \frac{\mathrm dV}{\mathrm dP}\mathrm dP = -\frac{RT}{P^2}\mathrm dP$$

This gives me

$$\frac{R}{V}\mathrm dV = -\frac{R}{P}\mathrm dP,$$

which is wrong according to my T.A.

How can I write $\mathrm dS$ as a function of $T$ and $P$?

  • 1
    $\begingroup$ When differentiating, are you sure $T$ was supposed to be treated as constant? I think this is correct: $$dV=\frac{R.dT}{P}-\frac{RT.dP}{P^2}$$ (Obtained by directly differentiating $V=\frac{RT}{P}$) $\endgroup$ May 8, 2019 at 9:50

1 Answer 1


The short way is to proceed as noted in the comments. The equivalent long way is to start from the total differential for $S_m$ in terms of $p$ and $T$:

$$dS_m = \left(\frac{\partial S_m}{\partial p}\right)_Tdp + \left(\frac{\partial S_m}{\partial T}\right)_pdT$$

The second term is given by

$$ \left(\frac{\partial S_m}{\partial T}\right)_p = \frac{C_p}{T}=\frac{5R}{2T}$$

The first can be determined using the identity

$$ \left(\frac{dS_m}{dp}\right)_T = -\left(\frac{\partial V_m}{\partial T}\right)_p=-\frac{R}{p}$$


$$dS_m = \left(-\frac{R}{p}\right)dp + \left(\frac{5R}{2T}\right)dT$$

It is not difficult to show that this is equivalent to the result obtained by substituting


into the expansion in terms of $V$ and $T$ shown in the OP.


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