# Entropy and total differential

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:

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

and

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

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 $$P$$?

• 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 $V=\frac{RT}{P}$) – William R. Ebenezer May 8 '19 at 9:50

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

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

Substituting,

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

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

$$dV=\frac{R.dT}{P}-\frac{RT.dP}{P^2}$$

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