In summary, the gas temperature in the end of the container is lower than the gas temperature in the beginning because the container's pressure has been lowered by the exiting particles.
  • #1
Ebi Rogha
24
6
Thread moved from the technical forums to the schoolwork forums
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?
 
Physics news on Phys.org
  • #2
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.
 
  • #3
I can help answer this question if we treat the gas as a continuum, but not from the molecular standpoint.
 
  • #4
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.
 
Last edited:
  • Like
Likes andrewkirk
  • #5
anuttarasammyak said:
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 $$\frac{T_2}{T_1}=\left(\frac{n_2}{n_1}\right)^{\gamma-1}$$
 
Last edited:
  • Love
Likes Ebi Rogha
  • #6
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, $$\frac{T_2}{T_1}=\left(\frac{n_2}{n_1}\right)^{\gamma-1}$$
 
  • Like
Likes Ebi Rogha
  • #7
Chestermiller said:
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
 
  • #8
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.
 
  • #9
haruspex said:
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.
 

What is gas temperature in a constant volume?

Gas temperature in a constant volume refers to the temperature of a gas when it is held at a constant volume, meaning its volume does not change despite any changes in temperature or pressure.

How is gas temperature affected in a constant volume?

In a constant volume, the temperature of a gas is directly proportional to its pressure. This means that as the pressure increases, the temperature also increases, and vice versa.

What is the relationship between gas temperature and volume in a constant volume?

In a constant volume, the temperature of a gas is inversely proportional to its volume. This means that as the volume decreases, the temperature increases, and vice versa.

What is the ideal gas law and how does it relate to gas temperature in a constant volume?

The ideal gas law, PV=nRT, states that in a closed system, the pressure and volume of a gas are directly proportional to its temperature and the number of moles present. This means that in a constant volume, an increase in temperature will result in an increase in pressure, while a decrease in temperature will result in a decrease in pressure.

How is gas temperature measured in a constant volume?

Gas temperature in a constant volume can be measured using a thermometer or a temperature sensor placed inside the container holding the gas. The temperature can also be calculated using the ideal gas law, if the other variables (pressure, volume, and number of moles) are known.

Similar threads

  • Introductory Physics Homework Help
Replies
2
Views
533
  • Introductory Physics Homework Help
Replies
8
Views
742
  • Introductory Physics Homework Help
Replies
1
Views
967
  • Introductory Physics Homework Help
Replies
16
Views
2K
  • Introductory Physics Homework Help
Replies
8
Views
1K
  • Introductory Physics Homework Help
2
Replies
39
Views
3K
  • Introductory Physics Homework Help
Replies
12
Views
1K
Replies
2
Views
4K
  • Introductory Physics Homework Help
Replies
8
Views
1K
  • Introductory Physics Homework Help
Replies
21
Views
2K
Back
Top