Well this got me stumbled, because I've been wondering what the "question" is. One of the example examns got the following question (strangely there are no supplied solution books):

A $0.2 {\rm m^3}$ thermally insulated rigid container is divided into two equal volumes by only a thin membrane. Initially, one of these chambers is filled with air at a pressure of $700\, {\rm kPa}$ and $37 \, {\rm C}$ while the other chamber is evacuated.

$C_p = 1.005 \frac{{\rm kJ}}{{\rm kg \cdot K}}\,$ and $\, C_v =0.721 \frac{{\rm kJ}}{{\rm kg \cdot K}}$

Now the questions:

a) Determine the change in internal energy of the air when the membrane is ruptured.
b) Determine the final air pressure in the container

Is this now a really silly question or am I missing something important? Cause isn't the internal energy an intrinsic property that has to be looked up/ experimentally determined?

And for the second problem, as there can be no energy transfer the internal energy also has to stay the same - so the temperature doesn't change and the pressure simply halves. ($PV = {\rm constant}$). Or am I missing something?

  • $\begingroup$ In Joule expansion, (irreversible process) internal energy is conserved. For a reversible process, it is an isentropic process. For a Joule-Thomson expansion, enthalpy is conserved. $\endgroup$
    – Trimok
    Jul 1, 2013 at 19:08
  • $\begingroup$ @Trimok Well how could I determine what the actual process is, I can rule out an isentropic process I guess. But the other two seem very similar though. $\endgroup$
    – paul23
    Jul 1, 2013 at 21:15
  • $\begingroup$ It looks like a Joule Expansion, so, for the first question, you would be right ; the internal energy is conserved. For the second question, if we consider air at 700kPa as an ideal gas, then the internal energy does depends only on temperature, so the temperature is conserved, and the relation $pV$= Constant is correct. However, if you do not consider air as an ideal gas, then the temperature is not conserved, then the relation $pV$= Constant is not correct. However, the questions seem very odd (at least the first) if it is a Joule expansion !! $\endgroup$
    – Trimok
    Jul 2, 2013 at 6:50
  • $\begingroup$ What you can do is to answer the questions for both cases -Joule expansion - and Joule-Thomson expansion $\endgroup$
    – Trimok
    Jul 2, 2013 at 6:54

3 Answers 3


The internal energy of an ideal gas can only change if heat is added or removed from the system, or if the system does some work. Neither is the case in this example so the change in internal energy is zero.

The final pressure is, as you say, just half the initial pressure.

I'm not sure why they give you the specific heats, but you can use them to work out how ideal your gas is. The specific heats of an ideal diatomic gas should be 2.5 and 3.5 joules/mole/R, and converting the specific heats you're given to J/mole/R (assuming an average $M_w$ of 28.8 for air) gives 2.50 and 3.48 to two SF. So approximating the air as an ideal gas seems entirely justified.

Response to comment:

You're given $C_v$ = 721 J/kg/K. The average molecular weight of air is 28.8 (assuming 20% oxygen and 80% nitrogen) so one mole is 0.0288kg. Multiplying by 0.0288 to convert to moles gives $C_v$ = 20.76 J/mole/K. Now divide by the molar gas constant, $R$ = 8.314 J/mole/K, to get $C_v$ = 2.50 (to 2 significant figures). Exactly the same calculation gives 3.48 for $C_p$.

  • $\begingroup$ How would you do this conversion? That seems the missing link I'm missing? $C_p = C_v + R$? But then R is also a dependent variable? $\endgroup$
    – paul23
    Jul 2, 2013 at 12:58

The way the question is worded "when the membrane is ruptured" sounds like a Joule expansion. The answer depends (sort of) on what class this is for. Based on the title of your question, it sounds to me that this exam is for just a basic thermodynamics course in which case this is likely just a "trick" question.

So it's just a classic Joule expansion, no change in internal energy, and the pressure is halved (to 350 kPA). Neither heat capacity is relevant because the process is not reversible, not at equilibrium, and neither the volume nor pressure is constant. All that added information is just to try to throw you off, to see if you really understand what's going on. That's my take on this (unless I'm missing something). HTH.


This is a case of adiabatic expansion. So change in internal energy is work done by the gas or system.

$dU=-dW=-\dfrac{P_iV_i-P_fV_f}{\gamma -1}\tag*{}$

where $\gamma =C_p/C_v$

Also, change in internal energy is given by,

$\Delta U=mC_v(T_f-T_i)\tag*{}$

where, $T_f=T_i\left(\dfrac{V_i}{V_f}\right)^{\gamma -1}$ and, $m=\dfrac{PV}{RT}M$

$dU=-42,650$ joules

Now final air pressure can be given by,

$P_iV_i^\gamma =P_fV_f^\gamma \implies P_f=266.4$ kPa

Above many answers and comments considered it as irreversible adaibatic case, but this is not. Instead of membrane if piston is placed and which moves quickly upto double the volume of the container. Now piston can be pushed back to same volume and if heat is not exchanged then it can gain its initial state.

The work done on the system to move back to its initial position is equal to work done by the system or gas during expansion. Thus by calculating external work, we can find the change in internal energy during expansion.

Edit: This is to show that there is change in internal energy in OP's question. Calculation from online calculator is pasted. https://www.omnicalculator.com/physics/thermodynamic-processes


  • $\begingroup$ This is (irreversible) expansion into a vacuum. The gas doesn't do any work as it expands to fill the remaining part of the container. $\endgroup$
    – J. Murray
    Oct 18, 2022 at 15:27
  • $\begingroup$ Volume of gas changes, thus this is equivalent to work done. One can't utilize it's another question. Also, change in internal energy was not derived by work done in case. $\endgroup$ Oct 18, 2022 at 15:32
  • $\begingroup$ See any introductory resource which describes free expansion (in particular, the last equation). There is no work done by the gas and no heat transferred through the container, so the internal energy does not change (unless you model air as a non-ideal gas, which is clearly beyond the scope of the OP). $\endgroup$
    – J. Murray
    Oct 18, 2022 at 15:39
  • 2
    $\begingroup$ $P dV$ is the work done when the gas expands against an external pressure $P$. In this case, the gas is not expanding against any external pressure - it is expanding into vacuum. No heat is added and no work is done. The pressure decreases and the volume increases, but the temperature and internal energy remain constant, while the entropy increases. My previous comment contains a link to an online chemistry textbook which you may find useful. $\endgroup$
    – J. Murray
    Oct 18, 2022 at 16:45
  • 2
    $\begingroup$ Neil, this will be my last response. The expression $PV^\gamma=const$ holds for a reversible, adiabatic process. Joule expansion is not reversible. If you are a student, I suggest you ask an instructor or somebody you are willing to listen to to explain this, since obviously you think that none of us know what we're talking about. If you are not a student, please search any introductory textbook (such as the one I linked) for Joule expansion or "free expansion of a gas". $\endgroup$
    – J. Murray
    Oct 22, 2022 at 1:33

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