Kinetic theory of gasses Integration problem

  • #1

Homework Statement


In the temperature range 310 K and 330 K, the pressure p of a certain nonideal gas is related to volume V and temperature T by
p = (24.9 J/K) T/V - (0.00662 J/K2)T^2/V
How much work is done by the gas if its temperature is raised from 314 K to 324 K while the pressure is held constant?


Homework Equations


Pv=NRT
W=integral(p dv)


The Attempt at a Solution



W=integral((24.9 J/K) T/V - (0.00662 J/K2)T^2/V dv)
I am not sure what to use for the limits of integration. the problem gives me a range of temperatures, is there any way that I can use those? I am very sure the limits of integration should be in units of m^3.
 

Answers and Replies

  • #2
Hootenanny
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HINT: Can you use the given equation to calculate the change in volume from the change in temperature?
 
  • #3
Then I get V1=7165.9J/P and V2=7414.6J/P. can I use those as the limits of Integration?
 
  • #4
Hootenanny
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Then I get V1=7165.9J/P and V2=7414.6J/P. can I use those as the limits of Integration?
I would say so :smile:
 
  • #5
what do I do with the T and T^2 parts of the pressure equation?
 
  • #6
Hootenanny
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what do I do with the T and T^2 parts of the pressure equation?
Notice that when you substitute the limits in, the expressions for pressure cancel.
 
  • #7
I think i am doing somthing wrong. I integrate the equation p = (24.9 J/K) T/V - (0.00662 J/K2)T^2/V right? so what do I do with the T and T^2? Would I treat them as a constant during integration?
 
  • #8
HallsofIvy
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No, T is not constant. You know that
p = (24.9 J/K) T/V - (0.00662 J/K2)T^2/V
with P constant. That is the same as
V = (24.9 J/K) T/P - (0.00662 J/K2)T^2/P

dV= [(24.9 J/K)/V - (0.00662 J/K2)(2T)/V]dT

and you can do the entire integral in terms of T.
 
  • #9
how did you get p to cancel out?
 
  • #10
Hootenanny
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how did you get p to cancel out?
'Cancel out' was a poor phrase to use. I should have said 'change of variable', it's been a long day :zzz:
 

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