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Van der Waals and nitrogen gas

  1. Oct 11, 2015 #1
    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

    3. The attempt at a solution

    I start with (∂U/∂V)TT and substitute in πT and Cv
    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. jcsd
  3. Oct 11, 2015 #2
    I actually solved my own problem. I apparently don't know how to integrate.
  4. Oct 11, 2015 #3
    Could you post the solution, please? I'm having trouble with this one myself.
  5. Oct 12, 2015 #4
    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.
  6. Oct 12, 2015 #5
    Thank you!
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