## 6th edition 8.27 part b

$w=-P\Delta V$
and
$w=-\int_{V_{1}}^{V_{2}}PdV=-nRTln\frac{V_{2}}{V_{1}}$

Tyra Nguyen 4H
Posts: 74
Joined: Fri Sep 28, 2018 12:25 am

### 6th edition 8.27 part b

8.27 Calculate the work for each of the following processes beginning with a gas sample in a piston assembly with
T = 305 K, P = 1.79 atm, and V = 4.29 L: (a) irreversible expansion against a constant external pressure of 1.00 atm
to a final volume of 6.52 L; (b) isothermal, reversible expansion to a final volume of 6.52 L.

Someone already asked for clarification on this question, but I really do not understand how to do part b, as I cannot calculate the moles. I have n=(PV)/(RT) from the ideal gas law equation. I plugged in P=1.79 atm, V=6.52 L, R= 8.206x102 L.atm.K-1mol-1, and T=305 K. My answer is .00406. what am I doing wrong?

Felicia1E
Posts: 31
Joined: Fri Sep 28, 2018 12:22 am

### Re: 6th edition 8.27 part b

The volume should be 4.29L, not the final volume, since you want to find the number of moles at the start of the expansion.

Tyra Nguyen 4H
Posts: 74
Joined: Fri Sep 28, 2018 12:25 am

### Re: 6th edition 8.27 part b

Felicia1E wrote:The volume should be 4.29L, not the final volume, since you want to find the number of moles at the start of the expansion.

I plug this in and my answer is 3.068 x 10-5 mols. Plugging in the rest of the numbers, I get the work to be -0.3J which is very off from the -326 it is supposed to be. What am I doing incorrectly

Felicia1E
Posts: 31
Joined: Fri Sep 28, 2018 12:22 am

### Re: 6th edition 8.27 part b

Hmmm, well for the number of mols you get from PV=nRT using the values discussed should be .307 mols. I am not sure how you got 3.068 x 10^-5 mols. After getting the mols though, since it is a reversible isothermal expansion, you use w=-nRTln(Vf/Vi). If n=.307 mols, R=8.314 J/(mol K), T=305 K and Vf=6.52 and Vi=4.29, w=-326 J.