## Pizza Rolls 6

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

Ellen Amico 2L
Posts: 101
Joined: Thu Sep 19, 2019 12:16 am

### Pizza Rolls 6

Can someone explain pizza rolls question 6? I'm confused on why ΔS = 0 and ΔU = 0. Also, I added up 2 calculations for the heat, but I can't seem to get the right answer. If anyone has tips on how to approach the problem please let me know!

Haley Dveirin 1E
Posts: 101
Joined: Sat Jul 20, 2019 12:17 am

### Re: Pizza Rolls 6

delta S and delt U are 0 because the final and initial state of the system is the same (ends up being 100.0L and 1.0atm in the end again) and because U and S are state functions it means the change is 0. As for q and w, I have no idea i need help with those too

Amy Xiao 1I
Posts: 101
Joined: Sat Jul 20, 2019 12:15 am

### Re: Pizza Rolls 6

I also can't get the answer for q and w; it seems like he subtracted 23.3 and 9.12 to get 14.2, but I don't understand why you wouldn't add the work values you get for each part of the problem together instead.

Edit: it seems that work will be positive if the system is being compressed, like in the first part of the problem

Vuong_2F
Posts: 90
Joined: Sat Sep 14, 2019 12:17 am

### Re: Pizza Rolls 6

Amy Xiao 1I wrote:I also can't get the answer for q and w; it seems like he subtracted 23.3 and 9.12 to get 14.2, but I don't understand why you wouldn't add the work values you get for each part of the problem together instead.

Edit: it seems that work will be positive if the system is being compressed, like in the first part of the problem

He did add the work values from each part of the problem, it's just that for the second part it's an isochoric process (constant volume) and no work is done.