Thermodynamics gas work done question

AI Thread Summary
The discussion revolves around calculating the work done by a non-ideal gas when its temperature increases from 310K to 320K at constant pressure. The pressure is defined by the equation P=(24.9 J/K) T/V− (0.00662 J/K^2)T^2/V, and the volume can be expressed as V = aT/P + bT^2/P. To find the work done, the integral of pressure with respect to volume, \int PdV, is used, simplifying to P \int dV when pressure remains constant. Participants express confusion about how to derive dV when pressure is constant, indicating a need for clarity on the relationship between volume and temperature in this context. The thread highlights the challenges of applying thermodynamic principles to non-ideal gases in homework problems.
Norton
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Homework Statement


In the temperature range 310K and 330K the pressure of a non ideal gas is related to volume V and temperature T by P=(24.9 J/K) T/V− (0.00662 J/K^2)T^2/V. How much work is done by the gas if it's temperature is raised from 310K to 320K whilst pressure held constant?


Homework Equations


P=(24.9 J/K) T/V− (0.00662 J/K^2)T^2/V


The Attempt at a Solution


I'm quite stumped on this one, initially I thought maybe something to do with W=NkT(ln)Vf/Vi, but I was wrong since the temperature has changed. I also tried using Q=n(Cp)(deltaT), but it is for an ideal gas and the question is talking about a non ideal gas. If someone could just give me a bit of a hand showing me the equations I should be using that would be great, since that is where I think I am falling down at the moment.

Thanks
 
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Norton said:

Homework Statement


In the temperature range 310K and 330K the pressure of a non ideal gas is related to volume V and temperature T by P=(24.9 J/K) T/V− (0.00662 J/K^2)T^2/V. How much work is done by the gas if it's temperature is raised from 310K to 320K whilst pressure held constant?
The expression for V is in the form:

V = aT/P + bT^2/P

You want to find \int PdV. To do that, you need to find dV.

What is dV/dT if P is constant?

AM
 
Sorry I'm still lost. I understand the expression for V, but what happens when P is constant?
 
Norton said:
Sorry I'm still lost. I understand the expression for V, but what happens when P is constant?
It enables you to calculate work easily: \int Pdv = P \int dV.

AM
 
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