Homework Statement

2.5 mol of an ideal gas which starts at 2.2 atm and 50 degrees Celsius does 2.4 kJ of work during an adiabatic expansion. What is the final volume of the gas? Express the result in the unit [m3].

Homework Equations

V(initial) x T(initial)^(3/2) = V(final) x T(final)^(3/2)

The Attempt at a Solution

I found the initial volume first using pV=nRT which I got to be 30.15L. But I'm not sure how to get the final temperature, nor the final volume for that matter.

haruspex
Homework Helper
Gold Member
What's the internal energy of the gas? How does a change in that relate to work done?

What I've posted is literally all the information I was given when asked to complete the question... my professor is known for his incomprehensible and ambiguous questions.

Andrew Mason
Homework Helper
What I've posted is literally all the information I was given when asked to complete the question... my professor is known for his incomprehensible and ambiguous questions.
It is not possible to answer this question without knowing the heat capacity of the gas. You appear to be using 1/(γ-1)=3/2 which would make Cv = 3/2 and γ = 5/3

Are you told that this is a monatomic gas?

AM

Last edited:
The question posted is all that I'm given. This professor is known for questions like these. It's a course called Biophysics.

Andrew Mason
Homework Helper
The question posted is all that I'm given. This professor is known for questions like these. It's a course called Biophysics.
So how do you know that 1/(γ-1) = 3/2? (you have put this in your equation). Only a monatomic gas would have Cv=3R/2; γ=5/3.

AM

Just a thought
ΔU=nR∫(T/V)dV
where T can be represent by V
(VixTi)^3/2)=(VxT)^3/2=constant <-- calculate by sub initial conditions.

Andrew Mason
Homework Helper
Just a thought
ΔU=nR∫(T/V)dV
Ok except for a - sign. If Q = 0 then ΔU = -∫PdV
where T can be represent by V
??How do you do that?

AM

Re AM:
(VixTi)^3/2=(VxT)^3/2=constant
isn't it, since the whole process is Adiabiatic Expansion, the above equation satisfy in any time during the process.

Andrew Mason