Gas temperature in a constant volume

[Ebi Rogha]

An insulated container (constant volume, adiabatic) contains an Ideal gas with pressure P1 and temperature T1.

We open the container's hatch for a few seconds and let some particles escape from the container, then we close the hatch again. We know container's pressure has reduced by exiting particles P2<P1. What can we say about gas temperature T2?

Can I say when we open the hatch, that particles with more kinetic energy escape and particles with less kinetic energy remain in the container. This lowers the average kinetic energy of the particles, therefore lower temperature T2<T1?

 

[anuttarasammyak]

From the equation of state for ideal gas
[tex]T_1=\frac{P_1 V}{N_1 k}[/tex]
[tex]T_2=\frac{P_2 V}{N_2 k}[/tex]
We may get the answer if we know changes of not only P but N.

 

[Chestermiller]

I can help answer this question if we treat the gas as a continuum, but not from the molecular standpoint.

 

[anuttarasammyak]

1
Say a wall in the container moves outward for volume v. The gas expands adiabatically
[tex]\frac{P_2}{P_1}=\frac{V^\gamma}{(V+v)^\gamma}<1[/tex]
where
[tex]\gamma = \frac{C_p}{C_v}>1[/tex]
Then we insert a new partion where there was the wall. The number of gas molecules in the container is
[tex]N_2=N_1\frac{V}{V+v}[/tex]
From the equation in my previous post
[tex]\frac{T_2}{T_1}=(1+\frac{v}{V})^{1-\gamma}<1[/tex]

2
Say there is a vacuum adiabatic chamber of volume v next to the container and we open the door between them. There is no work to be done in such a adiabatic free expansion so
[tex]T_1=T_2[/tex]
Then we close the door which has nothing to do with temperature.

3
It seems that we need two conditions for temperature of container gas to change.:
- Escape molecules do or undertake work, and
- (a part of )These molecules come back into container.

I hope these considerations will help you to deal with your case.

 

[Chestermiller]

1
Say a wall in the container moves outward for volume v. The gas expands adiabatically
[tex]\frac{P_2}{P_1}=\frac{V^\gamma}{(V+v)^\gamma}<1[/tex]
where
[tex]\gamma = \frac{C_p}{C_v}>1[/tex]
Then we insert a new partion where there was the wall. The number of gas molecules in the container is
[tex]N_2=N_1\frac{V}{V+v}[/tex]
From the equation in my previous post
[tex]\frac{T_2}{T_1}=(1+\frac{v}{V})^{1-\gamma}<1[/tex]

This is incorrect because the problem statement says the container volume is constant. Furthermore, the gas the leaves the container does not experience an adiabatic reversible expansion; it experiences an irreversible Joule Thomson expansion. So different parts of the initial container contents experience different processes, and can't be treated homogeneously.

This problem can be done in two different ways, and both methods give the same answer.

METHOD 1

Let ##n_1## be the number of moles of gas present in the tank to start with and ##n_2## be the number of moles of gas remaining in the tank at the ned. The initial volume occupied by the ##n_1## moles of gas was the tank volume V and the initial volume occupied the ##n_2## moles of gas remaining in the end was ##\frac{n_2}{n_1}V##. So the gas remaining in the tank in the end expanded in volume from volume ##\frac{n_2}{n_1}V## to volume V, and it did so adiabatically and reversibly. So we have $$P_1\left(\frac{n_2}{n_1}V\right)^{\gamma}=P_2V^{\gamma}$$or$$\frac{P_2}{P_1}=\left(\frac{n_2}{n_1}\right)^{\gamma}$$Also, from the ideal gas law, $$\frac{P_2}{P_1}=\frac{n_2}{n_1}\frac{T_2}{T_1}$$From these equations, it follows that $$\frac{T_2}{T_1}=\left(\frac{n_2}{n_1}\right)^{\gamma-1}$$

 

[Chestermiller]

METHOD 2

From the open system version of the 1st law of thermodynamics, we can write $$d(nu)=hdn=(u+RT)dn$$where u is the internal energy per mole of gas in the tank and n is the number of moles of gas in the tank. This equation reduces to $$ndu=nC_vdT=RTdn$$ Dividing both sides of this equation by nT gives $$d\ln{T}=(\gamma-1)d\ln{n}$$Integrating this equation then yields the same result as from Method 1, $$\frac{T_2}{T_1}=\left(\frac{n_2}{n_1}\right)^{\gamma-1}$$

 

[Ebi Rogha]

This is incorrect because the problem statement says the container volume is constant. Furthermore, the gas the leaves the container does not experience an adiabatic reversible expansion; it experiences an irreversible Joule Thomson expansion. So different parts of the initial container contents experience different processes, and can't be treated homogeneously.

This problem can be done in two different ways, and both methods give the same answer.

METHOD 1

Let ##n_1## be the number of moles of gas remaining in the tank to start with and ##n_2## be the number of moles of gas remaining in the tank at the ned. The initial volume occupied by the ##n_1## moles of gas was the tank volume V and the initial volume occupied the ##n_2## moles of gas remaining int the end was ##\frac{n_2}{n_1}V##. So the gas remaining in the tank iii the end expanded in volume from volume ##\frac{n_2}{n_1}V## to volume V, and it did so adiabatically and reversibly. So we have $$P_1\left(\frac{n_2}{n_1}V\right)^{\gamma}=P_2V^{\gamma}$$or$$\frac{P_2}{P_1}=\left(\frac{n_2}{n_1}\right)^{\gamma}$$Also, from the ideal gas law, $$\frac{P_2}{P_1}=\frac{n_2}{n_1}\frac{T_2}{T_1}$$From these equations, it follows that $$\frac{T_2}{T_1}=\left(\frac{n_2}{n_1}\right)^{\gamma-1}$$

That's a great approach! thanks, Chestermiller.
I will share this and ask for comments, I will get back to you if I get comments worth discussing.
Thanks again

 

[haruspex]

I'm puzzled as to what is outside the tank. If any gas is outside, some of that will have come into the tank. If a vacuum, it is now likely empty.

There is some ambiguity in "We know container's pressure has reduced by exiting particles P2<P1." On the surface, this seems to be a logical deduction from the implied fact that no particles entered. However, it could be intended as additional information, i.e. P2<P1 despite the possibility that some particles entered, and we are being asked to determine whether P2<P1 is enough to infer that the temperature has dropped.

 

[Chestermiller]

I'm puzzled as to what is outside the tank. If any gas is outside, some of that will have come into the tank. If a vacuum, it is now likely empty.

There is some ambiguity in "We know container's pressure has reduced by exiting particles P2<P1." On the surface, this seems to be a logical deduction from the implied fact that no particles entered. However, it could be intended as additional information, i.e. P2<P1 despite the possibility that some particles entered, and we are being asked to determine whether P2<P1 is enough to infer that the temperature has dropped.

You are correct that this problem is poorly stated. The usual way this problem is expressed is that the pressure inside is higher than the pressure outside, and the process consists of opening a valve a small amount, and allowing some of the gas in the tank to slowly escape, after which the valve is closed.