3
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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 '13 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 '13 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 '13 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 '13 at 6:54
1
$\begingroup$

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 $C_p$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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 '13 at 12:58
0
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.