# Adiabatic condition and equation of state

## Homework Statement

8.02 × 10−1 moles of nitrogen gas ( γ= 1.40) is contained in a volume of 2.00 × 10−2 m3 at a pressure of 1.00 × 105 Pa and temperature of 300 K. The sample is adiabatically compressed to half its original volume. IT behaves as an ideal gas.

(i) What is the change in entropy of the gas?

(ii) Show from the adiabatic condition and the equation of state that TV γ-1
remains constant, and hence determine the final temperature of the gas

## Homework Equations

PV=nRT
PVγ = A (constant)

## The Attempt at a Solution

For I, I think that because the compression is a reversible adiabatic process there will be no entropy change.

It is ii, that I am stuck on. I think that the equation of state will need to be rearranged to give P=nRT / V and that this will need to be substituted in to the adiabatic condition to give (nRT Vγ) / V = constant which feels close but I cannot think of the last step.

PVVγ-1 = constant
nRT *Vγ-1 = constant and ignore n and R seeing as they themselves are constants? giving TVγ-1 = constant
(I am not sure whether n and R can just be ignored or whether this method is even correct or valid).

or some other way entirely, any insight would be appreciated.

Chestermiller
Mentor
Both methods are correct, and give the correct result. In the first method, Vγ/V=Vγ-1

Chet

I think the answer might be along the lines:
PV=nRT
V=nRT/P
so fitting this into the equation:
P(nRT/P)gamma

then you get TgammaP1-gamma=A/(nR)gamma

Use the same principle to substitute P=nRT/V into the adiabatic condition
hope that helps

Sorry about the subscripts, my first post and new to forums. But since I always check here for help, is only fitting that I help when I can

Chestermiller
Mentor
I think the answer might be along the lines:
PV=nRT
V=nRT/P
so fitting this into the equation:
P(nRT/P)gamma

then you get TgammaP1-gamma=A/(nR)gamma

Use the same principle to substitute P=nRT/V into the adiabatic condition
hope that helps

Sorry about the subscripts, my first post and new to forums. But since I always check here for help, is only fitting that I help when I can
Hi Astoreth. Welcome to Physics Forums.

It looks like LivvyS had already solved the problem correctly in post #1. Do you feel that there was a problem with what he/she did?

Chet

Hello:

sorry about that. No, the answer was not incorrect, but given this:

which feels close but I cannot think of the last step.
I thought I'd help.

I am sorry if I overstepped. I was only trying to help

Also, i didn't noticed that it was marked as resolved.

Again, am sorry

Chestermiller
Mentor
Hello:

sorry about that. No, the answer was not incorrect, but given this:

I thought I'd help.

I am sorry if I overstepped. I was only trying to help

Also, i didn't noticed that it was marked as resolved.

Again, am sorry
No need to apologize. As you said, you were just trying to help.

The OP used one method to solve it in post #1, and got it correct. And he tried another method to solve it, but couldn't complete the final step. In post #2, I pointed out how to complete the final step, which, when applied, leads to the same correct solution.

Again, Welcome to Physics Forums. It's great to have you in our community.

Chet

• Greg Bernhardt
Thanks, Chet :)
Is good to be here

All help much appreciated, thanks guys!