# Homework Help: Gas released from container

1. Jul 11, 2013

### CAF123

1. The problem statement, all variables and given/known data
A metal container of volume $V$ and with diathermal walls contains $n$ moles of an ideal gas at high pressure. The gas is allowed to leak out slowly from the container through a small valve to the atmosphere at a pressure $P_0$. The process occurs isothermally at the temperature of the surroundings. Show that the work done by the gas against the surrounding atmosphere is $W = P_0 (nv_o - V)$, where $v_o$ is the molar volume of the gas at atmospheric pressure and temperature.

2. Relevant equations

Work done by gas: $W = \int_{V_i}^{V_f} P\,dV$

3. The attempt at a solution
I think I have the answer, but I have one question and I don't get a physical implication that the equation suggests.
Reasoning for answer: Initially volume occupied by gas: V. After all gas released, the volume occupied by n moles is nvo. The gas does work against surroundings so $W = \int_V^{nv_o} P_0 dV$ gives the answer.

However, one small point: Why can I say the volume occupied by the gas at the end when it is all released is $nv_o$? This is only true under STP conditions and it is given that the pressure of the atmosphere is P0 which need not be atmospheric pressure. We don't know anything about the temperature either.

Physical implication: The equation appears to be independent of $P$ inside the container. But I think, via the ideal gas law, I can write $V = \frac{nRT}{P}$ and so the equation becomes $$W = nP_0\left(v_o - \frac{RT}{P}\right)$$ but I don't think this makes sense:
As P gets larger, RT/P tends to zero and so this tends to make W bigger. I would expect that W get smaller.

Many thanks.

2. Jul 11, 2013

### voko

Does this not follow from the definition of $v_0$ given above?

A higher pressure gas, everything else being equal, does more work when expanding - is this not obvious?

3. Jul 11, 2013

### CAF123

Hi voko,
Assuming atmospheric pressure here refers to $P_0$ and temperature refers to the temperature of the surroundings, then I understand. Is this the case?

I was taking the approach of saying if the pressure inside the container is so much less than that of the atmosphere, then more work will be needed to expand against the atmosphere.
Force provided by atmosphere: P_0 = F/A so F proportional to P and W proportional to F.

4. Jul 11, 2013

### voko

This is how I read the definition.

The gas is said to be "at high pressure". If the gas is at low pressure, then the work is negative, because the gas will not expand at all.

5. Jul 11, 2013

### CAF123

So since the pressure inside the container is greater than that of the surroundings, the gas will expand. If the pressure inside the container is less than the pressure of the surroundings, then the gas will not expand. Is this correct?

Last edited: Jul 11, 2013
6. Jul 11, 2013

### voko

If the pressure inside is lower, then the atmosphere will get into the container and equalize the pressures. I do not think you can technically call that "compression", but this is definitely not expansion.

7. Jul 12, 2013

### CAF123

Would the common pressure be $P_0$ after the pressures have been equalised?

Similarly, for the case in the question, when all the gas escapes the container, the pressure inside is then zero, but this causes an increase in the pressure of the atmosphere because of the expansion?

8. Jul 12, 2013

### voko

I do not understand the question.

Why will the gas escape if its pressure becomes equal to or lower than atmospheric?

If the amount of gas in the box is not much smaller than in the atmosphere, yes.

9. Jul 12, 2013

### CAF123

In #6, you said that if the pressure inside the container was lower than that of the atmosphere, then (I am assuming provided there is an entrance) the atmosphere will tend to equilibriate the pressure. My question was: when the pressure is equalised, this means pressure inside container = pressure of surroundings so is this pressure $P_0$?

In the question the final volume of gas is nvo so after some of the gas is released, (not all of it), the pressure inside the container is $P_0$. At this point, the gas no longer expands and so the pressure inside container = pressure of atmosphere and there is a total of n moles of gas in the container and surroundings environment, hence a final volume nv_o Is this correct?

10. Jul 12, 2013

### voko

Both are correct.

One should keep in mind, though, that this all correct when the gas exchange happens slowly. If the gas (or the atmosphere) is allowed to rush out (in), then it is entirely possible that the pressure inside can become both lower and greater than atmospheric, and it will oscillate up and down before settling at the atmospheric equilibrium. In the problem the slowness of the release is implicit because it is isothermal.

11. Jul 12, 2013

### CAF123

Is the mathematical argument simply: The gas is at a high pressure $P$ inside the container. So to expand it must exert a force infinitesimally greater than $PA$. Force here is proportional to pressure and, in turn, work is proportional to force. So a higher pressure gas needs to do more work on expanding than a gas at a pressure $p < P$.

12. Jul 12, 2013

### voko

Your use of "must" and "need" is somewhat perplexing. It is not that a gas 'must' exert a greater force; it simply does.

There is a more intuitive (and more verbose) argument. Let's say we have two equal volumes of some gas at different pressures and equal temperature. The ideal gas law implies that we actually have more of the higher pressure gas, as their their masses are proportional to their pressures. We can expand the higher pressure gas, isothermally, so that the final pressure is equal to that of the lower pressure gas. Clearly that does some positive work. Now we have proportionally more volume of the formerly higher pressure gas than the lower pressure gas. Clearly a greater volume of otherwise identical gas will do greater work when expanding.

13. Jul 12, 2013

### CAF123

Okay, thanks. But returning to the mathematical argument for a moment: If what I said in my last post is correct, then when I derived $W = P_0 (nv_o - V)$ I didn't seem to worry about the pressure of the gas inside the container at all, only $P_0$.

I think it is then correct to say that given that the pressure of the gas $P$ is greater than the pressure of the atmosphere $P_0$ (as given in Q), for it to expand, the force it exerts on the surroundings is only (infinitesimally) greater than $P_0A$. That is what $W = \int_{V_i}^{V_f} P_0 dV$ suggests.

However, this neglects the pressure of the gas in the container, P, that the gas also has to do work against.

Last edited: Jul 12, 2013
14. Jul 12, 2013

### voko

In fact, $P$ and $T$ are completely determined by $n$, $P_0$, $v_0$, and $V$. I am sure you can convince yourself of this.

15. Jul 12, 2013

### CAF123

Yes, the problem I have is one of those cases where despite it being an irreversible process, $W = \int P dV$ is valid.

At each stage, PV = nRT holds so in particular,$$T = PV/nR = P_0 nv_o/nR \Rightarrow P = nP_ov_o/V$$

So this means $P_o = PV/nv_o$ and so $$W = \frac{PV}{nv_o} \int_{v_i}^{v_f} dV,$$ where $v_i = V$ and $v_f = nv_o$. (Sloppy notation but V and P denote pressure and volume before the expansion and are fixed quantities so they can be taken outside the integral)

So this argument relies on T being constant. Yes?

This is essentially the last eqn in the OP rearranged.

Last edited: Jul 12, 2013
16. Jul 13, 2013

### voko

I think you are making this more complex than it needs to be. The gas expand against a constant external pressure. So the work integral has this constant external pressure and dV, which makes it trivial. The dependence on the internal pressure is implicit via the constants of the problem.

17. Jul 13, 2013

### CAF123

Do you mean that $P = nP_0v_o/V$ here? (I.e the internal pressure P is related to the external pressure $P_0$ in this functional form)
This is only true provided the process is isothermal, right?

18. Jul 13, 2013

### voko

What I mean is that the constants given fully define the initial state. This is not specific to the isothermal process - adiabatic expansion, for example, would be fully defined just as well.

19. Jul 13, 2013

### CAF123

How would you express $P_0$ in terms of $P, n, v_o, V$(and T) if the process was not isothermal? In my method in #15, this assumes the process was isothermal.

Would it be simply $$P_0 = \frac{PV}{nv_o} \left(\frac{T_2}{T_1}\right)?$$

20. Jul 13, 2013

### voko

I regard $P_0$ as given, and, as I said, you can determine $P$ and $T$ before any process occurs. Then, of course, the internal pressure and the internal temperature will depend on the nature of the process.

21. Jul 13, 2013

### CAF123

The reason I want to express $P_0$ in terms of $P$ is so that in the work integral $W = \int P_0 dV$ I can see explicity how the work done in the gas expanding is related to the internal pressure (P) of the gas.

In #15, for an isothermal case, I got $W = \frac{PV}{nv_o} \int dV$, i.e W proportional to P, which is what we expected in the end. I think this is fine?

Now for the non-isothermal case, (so assuming the walls are not diathermal and the process does not occur at some fixed temperature T) the pressure at the beginning is $P = nRT_1/V$ and the pressure at the end is (when the pressure of gas = pressure of atmosphere) $P_0 = nRT_2/nv_o$. Solving for $P_0$ gives $P_0 = \frac{PV}{nv_o} \left(\frac{T_2}{T_1}\right)$?

22. Jul 13, 2013

### voko

Yes, that looks correct.