Calculating Work Done by a Gas on a p-V Diagram

[jr662]

45. A gas sample expands from Vo to 4.0Vo while its pressure decreases from po to po/4.0. If Vo = 1.0m^3 and po = 40 Pa, how much work is done by the gas if its pressure changes with volume via (a) path A, (b) path B, and (c) path C?

The p-V diagram can be found at the following addrs on slide/pg #37: http://people.virginia.edu/~ral5q/classes/phys631/summer09/Lecture_pdf/FINAL_LECTURE13.pdf

I know that the relevant equation here is Delta E = Q - W, and based on the given data is adiabatic and isothermal --> meaning Q=0 and T=>constant..

 

[kuruman]

Please do not post the entire lecture notes in the future. It is confusing. Just select the piece that you need.

There is no adiabatic or isothermal process in the pV diagram that you mention. The work done by the gas is the area under the curve in the pV diagram. If you are looking for kust the work, that's all you need to use.

 

[jr662]

Kuruman,,,sorry for any confusion. Anyhow, could you please confirm if the equation i am suppose to use is => W = nRT ln (Vf/Vi) ? Where n =1 R = 8.31...im not sure how to plug in the different values with the given info..

 

[kuruman]

Not confirmed. The expression you have quoted is for an isothermal process in which the pressure and the volume are inversely proportional. Just because the final temperature and the initial temperature are the same, this does not mean that the process is isothermal. The pV diagram shows three different paths for going from the initial state to the final, all of which involve straight lines, none of which is isothermal.

The work done by the gas is given by

W=\int p dV

which is the area under the curve in a pV diagram. If the processes involve straight lines only, you don't really need integration. Just a little bit of geometry ...

 

[jr662]

Kuruman, I've read that W = integral pdV is true only if pressure is constant and the question states that the "pressure decreases" and "how much work is done by the gas if its pressure changes with volume." ?? I'm still confused, I guess this is what happens when your textbook doesn't hasnt come in yet and your researching all the assigned homework problems online...Anyhow, hopefully you can help.

 

[jr662]

Ok, so apparently W is always the area under the curve of a p-V graph. So would W for Path (a) be = 160 and (c) 10 and (b) = ??

a) delta p = 0 ... and v increased (4.0 x 1.0 m^3) = 4 x po = 4 x 40 Pa = 160

? I'll never figure this problem out!

 

[jr662]

Sorry, so i got path a and c:

a) delta P = 0...delta V=4.0Vo-Vo and Po = 40 so W=Po(deltaV) = 120 Joules (i don't know how joules is derived, besides the fact that we are looking for W and it is joules usually..
c) delta V= 0...delta p = (Po/4.0) - Po = 30 and W = p(delta v) so W = 30 x 1 = 30 joules
b) clueless...i think i found somewhere online where it = 75 joules...i have no idea how they got that number...

 

[kuruman]

Sorry, so i got path a and c:

a) delta P = 0...delta V=4.0Vo-Vo and Po = 40 so W=Po(deltaV) = 120 Joules (i don't know how joules is derived, besides the fact that we are looking for W and it is joules usually..

Do dimensional analysis on pV to see how you get Joules.

c) delta V= 0...delta p = (Po/4.0) - Po = 30 and W = p(delta v) so W = 30 x 1 = 30 joules

Looks OK.

b) clueless...i think i found somewhere online where it = 75 joules...i have no idea how they got that number...

You have a right triangle sitting on top of a rectangle. Can you figure out their combined area?

 

[jr662]

Awesome!