Atkin's Physical Chemistry Book; Chapter 3; question 3.3(a):
Calculate $\Delta S$ when the state of $\mathrm{3~mol}$ of perfect gas atoms, for which $C_{p,m}$ = $\frac{5}{2}R$, is changed from $\mathrm{298~K}$ and $\mathrm{1~atm}$ to $\mathrm{398~K}$ and $\mathrm{5~atm}$. How do you rationalize the sign of $\Delta S$?


Since this is at a constant pressure, $dq = dH = C_pdT$. Then I used the equation $$dS = \frac{C_pdT}{T}$$ I integrated both sides to get the following equation: $$\Delta S = C_p \ln\frac{398}{298}$$

Subbing in the values I got that $\Delta S = \mathrm{+18.3~J}$. However, the answer in the book is $\mathrm{-21~J}$. Shouldn't that be definitely wrong because the temperature of the system increased and hence the total entropy should have increased?

  • 2
    $\begingroup$ Notice that your assumption that you are at constant pressure is incorrect. You are given the constant pressure heat capacity, but the pressure increases from $\text{1 atm}$ to $\text{5 atm}$. $\endgroup$
    – Ben Norris
    Commented Aug 9, 2015 at 11:33
  • $\begingroup$ So how should I attempt this problem? $\endgroup$
    – Nanoputian
    Commented Aug 9, 2015 at 11:34
  • $\begingroup$ Your energy change involves PV work. Try calculating the volume change with IGL, then incorporate it into your calculation of DeltaS $\endgroup$
    – Dan Burden
    Commented Aug 9, 2015 at 17:36

2 Answers 2


The pressure changes, so we cannot use the simplified equation you use. From the perfect gas law, we can calculate the volume change, but we don't need to go there. The big problem is that in addition to the heat transfer, pressure-volume work is done. Work is easy to determine at constant pressure. When pressure changes also, it gets harder. However, let's use enthalpy to get rid of work (and thus the change in volume), leaving us with $dS=f\left(\frac{dT}{T},\frac{dP}{P}\right)$, which is the data we are given.

Here's the derivation.

Second Law of Thermodynamics:

$$dS = \frac{\delta Q}{T}$$

First Law of Thermodynamics:

$$dU = \delta Q + \delta W = \delta Q - PdV_m$$

Definition of Enthalpy:

$$dH = dU + d(PV_m) = dU +PdV_m + V_mdP$$


$$dH = \delta Q + PdV_m + V_mdP - PdV_m = \delta Q +V_mdP$$ $$\delta Q = dH - V_mdP$$

Perfect gas equation (notice that I am using molar volume $V_m$, so we will be calculating molar entropy change):

$$PV_m = RT$$ $$V_m = \frac{RT}{P}$$

At constant pressure (Yes, I know pressure changes, bear with me. We'll put that back in).

$$dH = C_p dT$$

Substitute to get rid of $V_m$, since we don't have volume data.

$$dQ = C_p dT - \frac{RT}{P}dP$$

Substitute into the second law:

$$dS = C_p \frac{dT}{T} - R\frac{dP}{P}$$

Integrate (now we account for change in pressure:

$$\int_{S_1}^{S_2} dS = C_p \int_{T_1}^{T_2} \frac{dT}{T} - R \int_{P_1}^{P_2} \frac{dP}{P} $$

$$\Delta S = C_p \ln \left(\frac{T_2}{T_1}\right) - R \ln \left(\frac{P_2}{P_1} \right)$$

Using this method, I get the correct answer after I multiply the molar entropy change by the number of moles.

Note, that if you are given volume change instead, you can use a similar derivation to reach a similar equation:

$$\Delta S = C_v \ln \left(\frac{T_2}{T_1}\right) + R \ln \left(\frac{V_{m,2}}{V_{m,1}} \right)$$



Since the change in entropy does not depend on path, you can do it as a two step process. First you heat it at constant pressure to the final temperature, then you compress it at constant temperature to the final pressure. The change in entropy for each of these steps is easy to determine. You already got the result for the first step.


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