Thermodynamics problem: Adiabatic free expansion

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Discussion Overview

The discussion revolves around a thermodynamics problem involving adiabatic free expansion of an ideal gas in a rigid, insulated tank. Participants explore the implications of the adiabatic process on temperature and pressure changes following the rupture of a membrane separating two gas volumes.

Discussion Character

  • Homework-related
  • Debate/contested
  • Technical explanation

Main Points Raised

  • One participant states that since the system is non-conducting and experiences free expansion, the heat transfer (Q) is zero and no work is done (W = 0), leading to the conclusion that the internal energy change (ΔU) is also zero, implying constant temperature.
  • Another participant points out that the adiabatic process relationship used by the original poster is only valid for reversible processes, suggesting that this is a source of error in their reasoning.
  • A different participant emphasizes the importance of defining the system correctly, noting that if the entire contents of the tank are considered the system, then the volume remains constant, confirming that work done is zero.
  • One participant supports the original poster's analysis, agreeing that if ΔU is zero, then there is no change in temperature, and they affirm the calculation of final pressure as correct, while also correcting the equation for ΔU to use Cv instead of Cp.
  • There is a suggestion that the original poster's answer was marked incorrect, but one participant argues that the marking was erroneous based on their analysis.

Areas of Agreement / Disagreement

Participants express differing views on the correctness of the original poster's reasoning and calculations. Some agree with the original poster's conclusion about constant temperature and pressure, while others challenge the validity of the assumptions made regarding the adiabatic process and the nature of the expansion.

Contextual Notes

Participants highlight the need for careful consideration of the definitions of systems and processes in thermodynamics, as well as the distinction between reversible and irreversible processes. There is also a correction regarding the use of heat capacities in the context of internal energy changes.

nakamura25
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Homework Statement


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?

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
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
 
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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
 
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...
 
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
 
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