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Thermodynamics question (Adiabatic expansion)

  • Thread starter ArmanG
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Hello,
I am taking a course on thermodynamics and there was a question on our textbook( "Thermodynamics of Materials" by David V. Ragone) and I can't understand the solution given to me by my prof and I think it is wrong.

Homework Statement


An evacuated (P=0), insulated tank is surrounded by a very large volume (assume infinite volume) of an ideal gas at a temperature To. The valve on the tank is opened and the surrounding gas is allowed to flow quickly into the tank until the pressure inside the tank equals the pressure outside. Assume that no heat flow takes place. What is the final temperature of the gas in the tank?
The heat capacities of the gas, C p and C v, each may be assumed to be constant over the temperature range spanned by the experiment. Your answer may be left in terms of C p and C v.
Hint: One way to approach the problem is to define the system as the gas that ends up in the tank.
(Ragone: problem 1.5)

Homework Equations



The Attempt at a Solution


Ok First of all, the question is asking us the relation between T1 (To) and T2. I think they should be the same given that the volume gained by the expansion(Vg) would be so minimal (read: negligible) compared to the infinite volume we have that P*(V(infinite)+Vg)=nRT would be the same as P*V(infinite)=nRT meaning that T1=T2 temperature would not change

and I think what we should do is apply the adiabatic expansion formula for temperature change (as this is exactly that, I think) : T2/T1=(P2/P1)^(R/Cp)
this solution is given in the following link under the problem 1.5 and the result overlaps with my line of thinking (T2=T1)
http://wenku.baidu.com/view/c19452bb960590c69ec376c5.html

But the solution we were given and another one I've found on the internet on the OCW of MIT is (the solution numbered 3.4 given in the first solution set as pdf in the following link) http://ocw.mit.edu/courses/materials-science-and-engineering/3-20-materials-at-equilibrium-sma-5111-fall-2003/assignments/
I will write the solution here with my comments in paranthesis:
Closed system solution
System is gas flowing into the tank
U2-U1= Q+w = w (since Q=0, adiabatic)
1)U2-U1=P1*V1 (How?? I thought w=-P(external)*ΔV , NOT P(of the gas)*V(of the gas) )
For an ideal gas,
2)U2-U1=N*Cv*(T2-T1)
3)P1*V1=nR*T1
From 1), 2) and 3) --> n*Cv*(T2-T1)=nR*T1
and since R=Cp-Cv
n*Cv*T2-n*Cv*T1=n*Cp*T1-n*Cv*T1
so T2=(Cp/Cv)*T1

I think this solution is wrong and I don't really understand it. Can any of you help me with understanding it or confirm that it is wrong?

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
256bits
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Ok First of all, the question is asking us the relation between T1 (To) and T2. I think they should be the same given that the volume gained by the expansion(Vg) would be so minimal (read: negligible) compared to the infinite volume we have that P*(V(infinite)+Vg)=nRT would be the same as P*V(infinite)=nRT meaning that T1=T2 temperature would not change
You are asked about the temperature change of the gas in the tank, not that of the infinite volume !!

and I think what we should do is apply the adiabatic expansion formula for temperature change (as this is exactly that, I think) : T2/T1=(P2/P1)^(R/Cp)
this solution is given in the following link under the problem 1.5 and the result overlaps with my line of thinking (T2=T1)
No, it is not exactly that if you are referring to a free expansion of an amount of gas of definite volume into a vacuum chamber.

The MIT solution is correct.
the solution from MIT finallizes into the following:
The enthalpy of the gas entering the tank is equal to the final internal energy of the gas in the tank. Or H1 = U2.

1)U2-U1=P1*V1 (How?? I thought w=-P(external)*ΔV , NOT P(of the gas)*V(of the gas) )
You should consider what is pushing the amount of gas into the tank.

The rest of the solution is manipulation of the basic equations.

The other solution listed at the MIT site has the tank as the control volume.
Write down your basic fisrt law and eliminate the valuse that are zero. Remember that initially there is no gas in the tank, and finally, after the process is complete, all the gas is in the tank. ( ie the amount of gas that could fit into the tank is is the tank and not outside )
 
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