## Ch. 9 #7 HW

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

Elias Ruben 1O
Posts: 47
Joined: Wed Sep 21, 2016 2:56 pm

### Ch. 9 #7 HW

For this one, I know that for constant pressure, you would use (5/2)R for heat capacity, and for constant volume it would be (3/2)R. For constant pressure, I used the equation (1.00 mol)(20.785 J/mol x K)[ln(157.9/37.6)]. I got that the change in entropy was 29.8 J/K, but the book says it's 6.80 J/K. Likewise, I got 17.9 J/K for the constant volume problem, but it's 4.08 in the book. What am I doing wrong?

Michelle_Li_1H
Posts: 32
Joined: Wed Sep 21, 2016 3:00 pm

### Re: Ch. 9 #7 HW

Because there is a temperature change, to calculate the change in entropy, you would use the equation n*Cp*ln(T2/T1) for part (a). n in this case would be 1 mol, while the the Cp would be 5/2*R, which would be 5/2*(8.314 J/K*mol). Lastly, you would convert the temperature to K and take the natural log of (T2/T1) to get ln(430.9/310.6). After multiplying all this out, you should get 6.804 J/K.

For part (b) it would be the same process, except with Cv instead of Cp.

Hope this helps!