Van der Waals gas

1. Oct 11, 2015

lee403

1. For a van der Waals gas πT=a/Vm2. Calculate ΔUm for the isothermal expansion of nitrogen gas from an initial volume of 1.00 dm3 to 24.8 dm3at 298K. What are values for q and w?

2. Relevant equations
I looked up a for N2 gas to be 1.352 atm*dm6/mol2
dU(V,T)=(∂U/∂V)TdV+(∂U/∂T)VdT
(∂U/∂V)TT
(∂U/∂T)V=Cv

3. The attempt at a solution

which gives dU(V,T)=πTdV+CvdT

Since this is isothermal dt=0 so dU(V,T)=πTdV

I integrate both sides ∫dU=∫πTdV. πT= a/Vm2.
a is a constant so it comes out of the integral leaving
ΔU=a∫dV/Vm2.= a*n2 ∫ dV/V2

Evaluating the integral
ΔU= -((a*n2)/2)*V-3.
After plugging in values for a, Vi, and Vf I get:
ΔU=-((1.352 *n2)/2)*[(1/24.83)-(1/13)]

I get 0.676 *n 2 J/mol2. I just don't know what to do with the moles.

I know how to get work which is just w=-nrt∫dV/V and the solution is -7.95 kJ of work
but without ΔU I cannot calculate q.

2. Oct 11, 2015

lee403

I actually solved my own problem. I apparently don't know how to integrate.

3. Oct 11, 2015

ValyrianSteel

Could you post the solution, please? I'm having trouble with this one myself.

4. Oct 12, 2015

lee403

Everything is right up until I pulled out the n^2 from the integral, just keep it as Vm. I also integrated wrong so if you get the proper solution of the integral, which is -1/Vm then you can just plug everything in. With unit conversion you get U then you can calculate q.

5. Oct 12, 2015

Thank you!