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### Quiz 1 preparation, winter 2013 #3

Posted: **Wed Jan 21, 2015 3:56 pm**

by **Shannon Han 2B **

For #3 on the first quiz 1 preparation in the chem 14B workbook, I don't understand the calculations for the work for the 2 pathways. I understand that you use w = -P ∆V, but how do you know what values of P to use for each pathway? Why do you use 5 atm for pathway 1 and 2 atm for pathway 2?

### Re: Quiz 1 preparation, winter 2013 #3

Posted: **Wed Jan 21, 2015 5:22 pm**

by **Regina Chi 2K**

I'm not really sure if my guess is correct, but there is a definite change in volume. And if you see the respective volumes for both pathways, the pressure is constant when there is a volume change. And therefore, you would use 5.0 atm for pathway one and 2.0 atm for pathway two.

### Re: Quiz 1 preparation, winter 2013 #3

Posted: **Wed Jan 21, 2015 5:42 pm**

by **Shannon Han 2B **

Well I noticed the change in volume but why don't you use 5 atm for both or 2 atm for both? The pressures for both change too and they both start at 5 atm and end at 2 atm.

### Re: Quiz 1 preparation, winter 2013 #3

Posted: **Wed Jan 21, 2015 8:23 pm**

by **Niharika Reddy 1D**

If you look at the graph you plot in the beginning of the problem, it might make more sense. Work is not a state function, so it is path dependent. For path 1, there are 2 segments: one with a change in volume from 1.0 to 4.0 L at a constant pressure of 5.0 atm, and the second with a change in pressure from 5.0 atm to 2.0 atm at a constant volume (ΔV=0 and thus w=-PΔV=0). The total work for path 1 is thus the work resulting from the first segment: -(5.0 atm)x(3.0 L). (The answer is left in L atm for this problem so I assume we don't have to convert it into joules like we normally do.)

Path 2 is similar, but the first segment has a change in pressure at a constant volume (ΔV=0 and thus w=-PΔV=0) and the second segment goes from 1.0 L to 4.0 L at a constant pressure of 2.0 atm. Thus, the total work of path 2 is that of the second segment, since the first segment results in zero work: -(2.0 atm)x(3.0 L).

### Re: Quiz 1 preparation, winter 2013 #3

Posted: **Sat Jan 24, 2015 8:33 pm**

by **Justin Le 2I**

These two pathways are similar to when we use multiple reversible pathways to describe an irreversible pathway. We just take it in terms of one change at a time and depending on which you choose first, you will get a different answer.

### Re: Quiz 1 preparation, winter 2013 #3

Posted: **Mon Jan 26, 2015 8:53 pm**

by **Maria Davila 2I**

Definitely looking at your graph helps just like Niharika mentioned. By looking at the paths taken you can see when Pressure is constant in each pathway. The equation for work is -P(Delta V) , so you need Pressure when it is constant, in pathway 1 it is at 5atm, so you would use the Volume change when Pressure was constant (4.0L - 1.0L) . Same thing goes for Pathway 2.

Looking at the graph plotted helps a a lot to visualize what is happening, once you see this, I think it becomes obvious which ones to choose.

Thanks Niharika!