Van der Waals and nitrogen gas

[lee403]

1. For a van der Waals gas π_T=a/V_m². Calculate ΔU_m 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?

Homework 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)_T=π_T
(∂U/∂T)_V=C_v

3. The Attempt at a Solution
I start with (∂U/∂V)_T=π_T and substitute in π_T and C_v
which gives dU(V,T)=π_TdV+C_vdT

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

I integrate both sides ∫dU=∫π_TdV. π_T= a/V_m².
a is a constant so it comes out of the integral leaving
ΔU=a∫dV/V_m².= a*n² ∫ dV/V²

Evaluating the integral
ΔU= -((a*n²)/2)*V^-3.
After plugging in values for a, V_i, and V_f 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.

 

[lee403]

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

 

[ValyrianSteel]

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

 

[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.

 

[ValyrianSteel]

Thank you!