Adiabatic expansion with temperature dependent heat capacity.

[michael]

Homework Statement

1 mole of an ideal gas initially at 100° C and 10 atm is expanded adiabatically against a constant pressure of 5 atm until equilibrium is re-established. Given that the temperature dependence of the heat capacity is CV = 18.83 + 0.0209T calculate deltaU, deltaH and deltaS for the change in the gas.

Homework Equations

The Attempt at a Solution

I'm trying to find delta U. I know that U is given by the integral of CvdT between the initial and final temperatures so I have that expression. My problem is in finding the final temperature. To get this value most expressions use either gamma or the final volume(or both), both of which I don't know. Since gamma=1+R/Cv and Cv is temperature dependant I'm not sure gamma is even useful anymore. So the temperature dependence of Cv is really throwing me off here. Any hints would be great.

 

[Chestermiller]

Hi Michael. Welcome to Physics Forums!

What is the initial volume of the gas? Let T be the final temperature and V be the final volume. In terms of V, how much work is done by the system on the surroundings if the expansion takes place against a constant pressure of 5 atm.? How is this work related to the change in internal energy?

Chet

 

[michael]

I know I can get the initial volume from the ideal gas law and since dQ=0 dU=dW and dW can be expressed as dW=-p(deltaV) or dW=(Cv/R)*(p1V1-p2V2) so either of these will give me dU. But since Cv is dependent on temperature and I don't know the final temperature since I don't know the final volume I'm just getting a bit lost in the maths.

 

[Chestermiller]

I know I can get the initial volume from the ideal gas law and since dQ=0 dU=dW and dW can be expressed as dW=-p(deltaV) or dW=(Cv/R)*(p1V1-p2V2) so either of these will give me dU. But since Cv is dependent on temperature and I don't know the final temperature since I don't know the final volume I'm just getting a bit lost in the maths.

Try this:
W=-P₂(V₂-V₁)

The reason I'm using P₂ here is that the work done on the surroundings is ∫P_IdV, where P_I is the pressure at the interface with the surroundings. In this problem, the pressure at the interface is constant at 5 atm. (For more details about this, see my Blog at my PF page).

The change in internal energy is ΔU=∫C_VdT, which you can obtain by integration from the initial temperature to T2. P2, V2, and T2 are related by the ideal gas law. And the change in internal energy is equal to the work on the system (as you said). So you have enough information to solve for V2 and T2.

Chet

 

[michael]

Sorry for being dense, but every expression I come up with using W=-p2(V2-V1) ends up having both T2 and V2 in it. Such as -((Cvdt)/p2)+V1=V2=RT2/P2 or ln(T2/T1)=-(R/Cv)*ln(V2/V1). Thus the root of my problem. Any hints on eliminating T2?

 

[Chestermiller]

Sorry for being dense, but every expression I come up with using W=-p2(V2-V1) ends up having both T2 and V2 in it. Such as -((Cvdt)/p2)+V1=V2=RT2/P2 or ln(T2/T1)=-(R/Cv)*ln(V2/V1). Thus the root of my problem. Any hints on eliminating T2?

Yes. Substitute RT₂/P2 for V2 to eliminate V2. Also integrate CvdT for the left hand side with respect to T. Then you'll have an equation with only one unknown: T2.

 

[Trevor D]

This is an interesting problem. I would go about the same way to solve this. You should be able to eliminate T2 by solving this way. I have had experience with this type of question in the past.

Best Wishes

T