# 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?

2. Homework Equations
Pv=NRT
W=integral(p dv)

3. 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.

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Hootenanny
Staff Emeritus
Gold Member
HINT: Can you use the given equation to calculate the change in volume from the change in temperature?

Then I get V1=7165.9J/P and V2=7414.6J/P. can I use those as the limits of Integration?

Hootenanny
Staff Emeritus
Gold Member
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 what do I do with the T and T^2 parts of the pressure equation?

Hootenanny
Staff Emeritus
Gold Member
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.

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?

HallsofIvy
Homework Helper
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.

how did you get p to cancel out?

Hootenanny
Staff Emeritus