## 9.7

Volume: $\Delta S = nR\ln \frac{V_{2}}{V_{1}}$
Temperature: $\Delta S = nC\ln \frac{T_{2}}{T_{1}}$

sofiakavanaugh
Posts: 58
Joined: Thu Jul 13, 2017 3:00 am
Been upvoted: 1 time

### 9.7

Hi,

The question asks: assuming the heat capacity of an ideal gas is independent of temperature, calculate deltaS associated w/ raising temp of 1.00 mol of ideal gas atoms reversibly from 37.6 to 157.9 degrees C.
a) at constant pressure
b) at constant volume

My question is that for example in part a we assume constant pressure, but that doesn't necessarily mean constant volume. So when we calculate the change in entropy for part a, why wouldn't we have to account for the change in volume? Why can we just use CplnT2/T1 and not have to add nRlnV2/V1? And my question goes for part b too, we are assuming constant volume this time but not constant pressure so why don't we have to account for that in our calculations? Doesn't the fact that we are calculating these two things separately imply that there is a change in volume at constant p, and a change in p at constant v?

Peter Dis1G
Posts: 97
Joined: Fri Sep 29, 2017 7:04 am

### Re: 9.7

Hi, it's because according to ideal gas, PV=nRT. It says independent of temperature, which means that there is no relation between pressure and volume for that equation and therefore we shouldn't have to account for the volume.