## 9.7

$\Delta S = \frac{q_{rev}}{T}$

Haocheng Zhang 2A
Posts: 52
Joined: Fri Sep 29, 2017 7:04 am

### 9.7

Assuming that the heat capacity of an ideal gas is independent of temperature, calculate the entropy change associated with rising the temperature of 1.00mol of ideal gas atoms reversibly from 37.6℃ to 157.9℃ at (a)constant pressure and (b)constant volume.

I have no idea how to do this question. Can someone help me? Thanks!

Vasiliki G Dis1C
Posts: 53
Joined: Fri Sep 29, 2017 7:04 am

### Re: 9.7

For part a, you want to use dS=dq/T. Since we know dq=nCdT, we can substitute it in to have dS=(nCdT)/T. When you integrate this, you get ΔS=nCln(T2/T1). From here, you can plug in everything to find ΔS. (for an ideal gas C=(5/2)R) You end up with ΔS= (1)(5/2(8.314))(ln(431/310.8))=6.80 J/K
For part b, the same equation is true where ΔS=nCln(T2/T1), but the only thing that changes is C. C is now 3/2R for an ideal gas at constant volume. Plugging in what we know, we get ΔS=(1)(3/2(8.314)(ln(431/310/8))=4.08 J/K
Hope this helps!

Emily Mei 1B
Posts: 50
Joined: Fri Sep 29, 2017 7:04 am

### Re: 9.7

Example 9.2 on page 322 of the textbook also talks about this if you want a textbook reference!