# Expansion of gas through valve

1. Aug 21, 2013

### CAF123

1. The problem statement, all variables and given/known data
A thick walled insulating chamber contains $n_1$ moles of helium gas at a high pressure $P_1$ and temperature $T_1$. It is allowed to leak out slowly to the atmosphere at a pressure $P_o$ through a small valve. Show that the final temperature of the $n_2$ moles of helium left in the chamber is $$T_2 = T_1\left(\frac{P_0}{P_1}\right)^{1-1/\gamma}\,\,\,\text{with}\,\,\,n_2 = n_1 \left(\frac{P_0}{P_1}\right)^{1/\gamma}$$

2. Relevant equations

For a reversible adiabatic process, $P_i V_i^{\gamma} = P_f V_f^{\gamma}$

3. The attempt at a solution

System is insulated from environment => process is adiabatic. Slow release means process reversible and that final pressure of gas in chamber = P0. The eqn is: $$P_1 \left(\frac{n_1 R T_1}{P_1}\right)^{\gamma} = P_0 \left(\frac{n_2 R T_2}{P_0}\right)^{\gamma}$$ Simplifying, $$T_2 = \frac{n_1}{n_2} T_1 \left(\frac{P_0}{P_1}\right)^{1-1/\gamma}$$
This means that, to obtain the given expression, $n_1 = n_2$ but this a) does not make sense and b) is in contradiction to the eqn for $n_2$ in the question.

Many thanks.

2. Aug 21, 2013

### voko

This formula assumes that the amount of gas is constant.

3. Aug 21, 2013

### Stealth95

I think that your equation must be:
$$\displaystyle{P_1 \left(\frac{n_2 R T_1}{P_1}\right)^{\gamma} = P_0 \left(\frac{n_2 R T_2}{P_0}\right)^{\gamma}}$$
Although we have $\displaystyle{n_1}$ moles, only $\displaystyle{n_2}$ expand adiabaticaly. The rest $\displaystyle{n_1 -n_2 }$ moles leave the chamber without participating in the process. So we can apply the first law of thermodynamics with $\displaystyle{Q=0}$ only for $\displaystyle{n_2}$. Hence, the equation LHS must have the volume corresponding to $\displaystyle{n_2}$.

Now using the ideal gas equation it's easy to find the ratio $\displaystyle{\frac{n_2 }{n_1 }}$.

4. Aug 21, 2013

### CAF123

It makes sense, thanks voko and Stealth95.
For the ratio, before the expansion, $P_1 V_1 = n_1 R T_1$ and afterwards, $P_0 V_2 = n_2 R T_2$. No work is done on the chamber (the chamber does not expand) so $V_1 = V_2$.
So, $n_2 = n_1 \frac{P_0 T_1}{P_1 T_2}$
Reexpress using $T_1/P_1^{1-1/\gamma} = T_2/P_0^{1-1/\gamma}$ gives the result.

5. Aug 22, 2013

### Staff: Mentor

Just to add my two cents, the statement
is incorrect. A high-pressure gas leaking out of a container to a lower pressure is definitely not reversible (there is no parameter you could change by an infinitesimal amount that would get the gas to go back into the container). Considering a slow release allows you to assume that the gas in the container is always in internal equilibirum.

6. Aug 22, 2013

### CAF123

Thank you for pointing this out - what then is it that allows me to apply the equations which I have used- they are only applicable for a reversible process.

7. Aug 22, 2013

### Staff: Mentor

The formula $PV^\gamma = \textrm{const.}$ is valid for all quasi-static processes, not just reversible ones.

8. Aug 22, 2013

### CAF123

Ok, so reversible => quasistatic but the right to left implication is not true?

The two books I have and the link from Wikipedia mention excessively that the formula PVγ=const. is valid only for a reversible process.

From the first book: 'We define a change to be adiabatic if it is both adiathermal and reversible;
From the other: 'There is a very simple relation between P and V if the expansion is performed both adiabatically and reversibly...(derivation starting from dQ = dU + PdV - which is only true for reversible processes)...we have finally PVγ=const. which is our eqn for a reversible adiabatic'

I think it must apply only to a reversible process otherwise how could you sketch the adiabat in the first place on a P-V diagram?

9. Aug 22, 2013

### Staff: Mentor

Yes. A reversible process has to be quasistatic. If a process is not quasistatic, i.e., sudden, then you create additional entropy, and of course this means that it cannot be reversible. But you can come up with quasistatic processes that are not reversible, like in the OP.

:yuck:

Looking on my bookshelf, I find that D. V. Schroeder, An Introduction to Thermal Physics (Addison Wesley Longman, 2000) derives the formula referring only to a quasistatic criteria. Same with P.M. Morse, Thermal Physics (Benjamin, 1964). M.W. Zemansky and R.H. Dittman, Heat and Thermodynamics (McGraw-Hill, 1981) derive it in section 5-5 Quasi-Static Adiabatic Processes.

I only find Atkins' Physical Chemistry to impose reversibility, but then again the entire book does not mention quasistatic processes. My guess is that authors who say that the formula can only be applied to reversible processes are overzealous.

I don't see how the fact that it applies to all quasistatic processes affects how you draw the adiabat for a reversible process.

10. Aug 22, 2013

### Staff: Mentor

For the n2 moles that remain in the chamber, their expansion is adibatic and reversible. Just imagine an invisible boundary between the n1-n2 moles which leave the chamber, and the n2 moles which remain in the chamber. If the seepage is slow, those n2 moles expand adiabatically and reversibly from a volume of n2V/n1 to a volume of V.

Chet

11. Aug 22, 2013

### CAF123

Oh. I am using Thermal Physics Finn and Concepts in Thermal Physics Blundell and Blundell. I will hopefully check out some of those other books when I go to the library.

As far as I know, irreversible processes may not be represented on a PV diagram as a smooth continuous curve.

What could you do to make the $n_2$ moles reoccupy their original volume $(n_2/n_1)V$, i.e what is the source of the reversibility here?

Thanks.

12. Aug 22, 2013

### Staff: Mentor

Just add a piston to the chamber and compress the gas to its original volume. The reversible compression work you do with the piston would be equal to the reversible expansion work done by the n2 moles in expelling the n1-n2 moles from the chamber.