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Van der Waals and nitrogen gas

  • Thread starter lee403
  • Start date
  • #1
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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?


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)TT
(∂U/∂T)V=Cv

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.
 

Answers and Replies

  • #2
16
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I actually solved my own problem. I apparently don't know how to integrate.
 
  • #3
Could you post the solution, please? I'm having trouble with this one myself.
 
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  • #4
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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
Thank you!
 

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