Homework Help: Thermodynamics problem: Adiabatic free expansion

1. Sep 8, 2011

nakamura25

1. The problem statement, all variables and given/known data
A rigid ( ie. Constant volume), non-conducting ( ie. Perfectly insulated, no heat losses or gain) tank with a volume of 4.6 m^3. The tank is divided into two unequal parts by a thin membrane. One side of the membrane representing 1/3 of the tank is filled with nitrogen ( may be assumed to be an ideal gas) at 3.1 bar and 333.5 K. The other side of the membrane is evacuated. The membrane ruptures and the gas fills the entire tank. Heat capacity of nitrogen, Cp=3.5 R.
What is the final pressure of the gas in the tank?

3. The attempt at a solution
My thought was:
Non-conducting means adiabatic, hence Q = 0.
Free expansion, no external work, W = 0.
delta u = Q + W = Cp*(deltaT) = 0. So the temperature stays the same.
Since nitrogen can be assumed to be ideal gas, I used ideal gas law to calculated final pressure.
(P1)*(V1) = (P2)*(V2)
(3.1 bar)*(1/3)V = (P2)*(V)
P2 = 3.1/3 = 1.03 bar

Later on I saw this video lecture by UC Boulder:
http://www.learncheme.com/page/ideal-gas-expansion-closed [Broken]
I think the question and result are the same for my case. So I summited my answer. But it turned out that the temperature does change, since the gas will do work to expand. Also, the final pressure should be 1.76 bar. I tried to use the adiabatic process equation p2/p1 = (V1/V2)^(Cp/Cv). Still could not get the right answer.

My questions are:
What's wrong with my reasoning? Does that mean the solution in the video is also wrong?
How to solve this problem for final pressure and temperature?

I'd really appreciate your effort. Any hint or solution would be helpful. Thank you.

Best Regards,
Naka

Last edited by a moderator: May 5, 2017
2. Sep 29, 2015

SN94

The adiabatic process relationship you used is strictly for reversible process, which in this case the system is not. That's why you're getting the wrong answer

3. Sep 29, 2015

SN94

Also, it depends on what you choose as your system. If the whole entity inside the tank is the system then V is constant. Which means Work is zero...

4. Sep 29, 2015

Staff: Mentor

Your original analysis was flawless. I like SN94's approach in post #3, taking the system as the entire contents of the tank. That certainly confirms that W = 0. So ΔU=0, and there is no change in temperature. And your calculation of the final pressure is correct. My only criticism if that you wrote down the equation for ΔU incorrectly. It should be ΔU=mCvdT, not Cp.

If your answer was marked wrong, whoever marked it wrong was incorrect.

Chet