[confusedstud]

I just had an exam question that was basically this:

A gas is trapped in the piston which expands isothermally. The initial pressure is 200kPa and the initial volume is 2m^3 and the final pressure is 100kPa. Calculate the work done by the gas.

So my thought process was that this was a reversible expansion scenario since the gas expanded isothermally. So I calculated the final volume which was 4m^3. And from here we need to use the equation work done by gas=-nRTlnVf/Vi however, the question did not specify the temperature and number of moles of gas so I assumed that the temperature was 298K and calculated the number of moles of the gas using n=PV/RT which was (200000Pa)(2m3)/(8.314)(298)=161.44moles. With this I substituted it back into the -nRTlnVf/Vi equation such that -(161.44)(8.314)(298)ln(4/2)=-277258J

However, some of my classmates used the formula -PextΔV where ΔV=2m^3 and Pext was the final pressure 100000Pa. So they got -2x10^6 J.

Which method is the most appropriate? I feel like my concept was correct as they wanted the work done in an expansion. However, I was not given the T or n which i had to assume that the former was 298K. On the other hand, the methods my friends used did not have any variables that weren't specified.

[mjc123]

You can't answer which method is appropriate, as you aren't given enough information. You can't assume the expansion is reversible because it's isothermal; it can be either reversible or irreversible, and you aren't told which (unless there's a part of the question you haven't quoted - "basically this"?).
If it is expanding against a constant external pressure, it is irreversible and the work is -P_extΔV, as your friends used.
If it is reversible, that means the external pressure is always equal to the internal pressure, and the work is the integral of -PdV, so you get your result. You don't need to know n and T; you know nRT = PV = 400 kPa*m³.

[confusedstud]

You can't answer which method is appropriate, as you aren't given enough information. You can't assume the expansion is reversible because it's isothermal; it can be either reversible or irreversible, and you aren't told which (unless there's a part of the question you haven't quoted - "basically this"?).
If it is expanding against a constant external pressure, it is irreversible and the work is -P_extΔV, as your friends used.
If it is reversible, that means the external pressure is always equal to the internal pressure, and the work is the integral of -PdV, so you get your result. You don't need to know n and T; you know nRT = PV = 400 kPa*m³.

Oh right i just saw this link http://opencourseware.kfupm.edu.sa/colleges/cs/phys/phys102/files%5C4-_Summaries_Thermody102.PDF that say that nRTlnVf/Vi=P1V1lnVf/Vi = P2V2lnVf/Vi which makes so much sense right now.

But would it be correct to say that the pressure to use in the -PextΔV would be 100kPa?

[mjc123]

If you're told so, or if you assume that the "final" pressure is where it stops expanding because external and internal pressure are equal, but you're not told that. Apparently. Did the question give any extra information besides what you've quoted?

[confusedstud]

If you're told so, or if you assume that the "final" pressure is where it stops expanding because external and internal pressure are equal, but you're not told that. Apparently. Did the question give any extra information besides what you've quoted?

Hmm I cannot really remember because this was from an exam paper i just took. But those were the main parts. They didn't specify if the internal and external pressures remained the same or not. I used the nRTlnVf/Vi equation because I saw that they mentioned 'isothermal expansion' which seemed like a maximum work question to me.