1. Nov 12, 2012

kineticwave

1. The problem statement, all variables and given/known data

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].

2. Relevant equations

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

3. 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.

2. Nov 12, 2012

haruspex

What's the internal energy of the gas? How does a change in that relate to work done?

3. Nov 13, 2012

kineticwave

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.

4. Nov 14, 2012

Andrew Mason

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: Nov 14, 2012
5. Nov 14, 2012

kineticwave

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

6. Nov 15, 2012

Andrew Mason

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

7. Nov 15, 2012

henry407

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.

8. Nov 15, 2012

Andrew Mason

Ok except for a - sign. If Q = 0 then ΔU = -∫PdV
??How do you do that?

AM

9. Nov 16, 2012

henry407

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.

10. Nov 16, 2012

Andrew Mason

Ok. I see what you mean. That will work.

AM