# Calculate area of PV diagram. Two isotherms, two isobars

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1. Jan 26, 2017

### llatosz

1. The problem statement, all variables and given/known data
Everything is in attached file. Given the PV diagram with P2, P1, V2, V1.

2. Relevant equations
PV=nRT
W=nRT*ln(Vf/Vi)

3. The attempt at a solution
Attempt in attached file is very organized. I showed 2 of my peers and they are getting the same answer as well. Anybody have any ideas?

It makes sense that the two isothermal processes should cancel when connected by isobars. Isobars simply shift the same function along the X-axis, yielding the same area under both curves... or at least so i think.

It definitely doesn't make sense why the two isobaric processes mathematically cancel. They shouldn't.
This is very strange and is really stumping my friends and I.

2. Jan 26, 2017

### TSny

Are the temperatures the same for the two isothermal processes?

3. Jan 26, 2017

### llatosz

OH RIGHT! Thank you very much! So that should affect the work from the isothermal processes. nRT is not the same value for both processes, meaning I could not combine the logarithms the way I did, so the work from the isotherms actually do not cancel?

But as for the work from the isobaric processes, different temperatures shouldn't matter because Temperature just acts as a dummy variable to set one point on an isotherm equal to another point. Is this a correct understanding?

4. Jan 26, 2017

### TSny

Yes, that's right.

I'm note sure I follow. But, you are right that the work cancels for the two constant P processes.

5. Jan 26, 2017

### llatosz

Alright, that is good news.
My last confusion is that I dont see how the work could cancel for the two constant P processes because one of the constant P processes is at higher pressure, and since the differences in volume look very similar, the magnitude of work from the high P process should be quite greater than the magnitude from the low P processes.

6. Jan 26, 2017

### TSny

Your diagram is not drawn to scale very accurately. If P2 is 5 times P1, then ΔV for the isobaric process at P1 is 5 times ΔV for the isobaric process at P2.

7. Jan 26, 2017

### llatosz

Oh that is true. Alright great, I fully understand! Thank you very much, I really appreciate it!