(GR0177 #6) need help with thermo problem

[MikeTwo]

ETS Practice Test: GR0177 (First problem involves a pendulum)

Problem 6: An ideal monatamic gas expands quasi-statically to twice its volume. If the process is isothermal, work done by the gas is Wi. If adiabatic, work done is Wa. Which is true?
(list of options, the correct answer being E)
E) 0 < Wa < Wi

Can someone explain to me why isothermal work is necessarily larger than adiabatic work?

[aranovB]

Hi its my attempt to explain that:

look at infamous PV = nRT equation!!
1. now lets see whats happen in a isothermal expansion of our problem:
1.1 The volume increases and temperature remains constant, so
P = nRT/V,
from the equation above we see that the pressure will be decreased.
2. Lets see now in an adiabatic expansion:
in an adiabatic expansion temperatures decreases, now looking our eq.
P = nRT/V,
now not just the volume increase making the pressure decrease bu the temperature also decrease that makes pressure decrease still more.
so the pressure in this case (adiabatic/) is less than that in the isothermal case, so remembering that
Wi = PiV
Wa = PaV
if Pa < Pi

Wa < Wi ence
0 < Wa < Wi

what abou it, Im right!!!?

[MikeTwo]

Yes, that's perfect.

Since I posted, I had found a P vs T graph with isotherms on them. It also had an adiabatic expansion line, and it was a straight line tangent to the isotherm running underneath it. I know work is the area under the curve, so by remembering the graph I should be able to solve this problem, but I like your method better. I'm an equation kind of guy. So thanks.

Sidenote: I wish this site would find a way to post the publically available ETS Practice Test questions, and allow people to fill out the answers, rationale for them, and ways to remember them. It's a lot of work to set up, but I think the community would pick up the torch if it was lit.

[liwenzao]

Hi, non-mathematically, I think it's easier to solve the problem with the P-V frame.

Just start to draw the isothermal line and the adiabatic line from the same point, since the isothermal line is always higher than the adiabatic line, and they both end at the same volume, the area under the isothermal line is surely bigger, therefore the work Wi > Wa.

[Zhomper]

hmm...here's how I got the answer. I don't know how proper this is.

The first law of thermodynamics dU=dQ-dW. For an isothermal expansion dU = 0 and so dQ=dWi. For an adiabatic dQ is 0 and so dU=-dWa

dU+dW=dQ so dU<dQ 0<Wa<Wi