Why is temperature constant after gas has expanded?

In summary, the answer given for part (c) in the back is that temperature doesn't change as the gas in cylinder A expands to fill cylinder B.
  • #1
vcsharp2003
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Homework Statement
The problem from the chapter on Thermodynamics is as given in the screenshot below. I am interested in part (c) of the question.
Relevant Equations
##PV=nRT## (Ideal Gas Law)
##Q = \Delta U + W## (First law of Thermodynamics)
CamScanner 01-28-2023 20.39.jpg

The answer given for part (c) in the back is that temperature doesn't change as the gas in cylinder A expands to fill cylinder B.

The thermodynamic system here is composed of the two cylinders A and B joined by some pipe.

Screenshot_20230128-210554.jpg


But, I cannot find a satisfactory explanation for temperature remaining constant as the ideal gas expands to fill cylinder B.

I know the system is completely thermally insulated from its surroundings so no heat can flow in or out of the system, but that doesn't mean temperature of the ideal gas cannot change as it expands as for an ideal gas ## \dfrac {PV} {T} =~ constant## which means as V increases, so can T. Also, the process of expansion of gas is not given to be isothermal in the question. So, I am at a loss to explain that temperature is remaining constant in the process.
 
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  • #3
Lnewqban said:
Temperature is related to the kinetic energy of the molecules, which in this case has no reason to change.
Lower pressure (higher area of walls against the molecules can colide) fully compensates for higher volume.

Please, see:
http://hyperphysics.phy-astr.gsu.edu/hbase/Kinetic/kintem.html#c1
I see. But how can one conclude that ##P## and ##V## exactly compensate each other so that ##T## remains constant? May be ##P## and ##V## do not exactly compensate and their product ##PV## changes.
 
  • #4
Because “the entire system is enterely insulated”.
Same amount of energy remains inside the system because heat transfer is zero.
In real life, there is no perfect insulation and what you properly describe happens.
 
  • #5
Lnewqban said:
Because “the entire system is enterely insulated”.
Same amount of energy remains inside the system because heat transfer is zero.
In real life, there is no perfect insulation and what you properly describe happens.
But temperature can change in an adiabatic process of an ideal gas. Atleast, that's the idea I have and so this process being an adiabatic process can also have temperature change.
 
  • #6
vcsharp2003 said:
But temperature can change in an adiabatic process of an ideal gas. Atleast, that's the idea I have and so this process being an adiabatic process can also have temperature change.
Only if the gas does work. Here the gas expands freely which means no work done. Since we have already established that no heat transfer takes place, the right hand side of the first law is zero which makes the change in internal energy also zero.
 
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  • #7
vcsharp2003 said:
But temperature can change in an adiabatic process of an ideal gas. Atleast, that's the idea I have and so this process being an adiabatic process can also have temperature change.
No work is done because the walls of the system are rigid.
No heat transfer occurs.
Therefore, it seems to me that the process shown in our problem should be simultaneously adiabatic and isothermal.

Please, see:
https://en.m.wikipedia.org/wiki/Adiabatic_process

I may be wrong, so let’s page @Chestermiller
 
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  • #8
The gas does work on itself while it is expanding so its temperature drops as it’s internal energy is converted to kinetic energy of the flow. However, at the final equilibrium state, the KE has been thermalized back into internal energy.
 
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  • #9
kuruman said:
Only if the gas does work. Here the gas expands freely which means no work done. Since we have already established that no heat transfer takes place, the right hand side of the first law is zero which makes the change in internal energy also zero.
I see. That makes sense.
So for the thermodynamic system, we can say that ##Q = 0## and ##W = 0## (since the gas expansion is against vacuum or maybe because the the thermodynamic system doesn't expand i.e. cylinders A,B do not expand).

Thus, by Thermodynamics First Law we have ##0= \Delta U + 0 \implies \Delta U= 0##.

Since internal energy of an ideal gas depends only on it's temperature, so we can say the process occurs at constant temperature.
 
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  • #10
vcsharp2003 said:
W=0 (since the gas expansion is against vacuum or maybe because the the thermodynamic system doesn't expand i.e. cylinders A,B do not expand)
I am not too sure if ##W=0## because of the first reason or the second reason, that are mentioned in above extract from my post#9.
 
  • #11
vcsharp2003 said:
so we can say the process occurs at constant temperature.
No, we can say that the initial and final states are at constant temperature. The stopcock is suddenly opened so my comment in post 8 applies.
 
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  • #12
Frabjous said:
No, we can say that the initial and final states are at constant temperature. The stopcock is suddenly opened so my comment in post 8 applies.
I think what you have said is true because thermodynamic processes in real life, which this one is, are always happening irreversibly i.e. without thermodynamic equilibrium in the intermediate states between the initial and final states. Is that what you meant?
 
  • #13
vcsharp2003 said:
I think what you have said is true because thermodynamic processes in real life, which this one is, are always happening irreversibly i.e. without thermodynamic equilibrium in the intermediate states between the initial and final states. Is that what you meant?
We can only talk about equilibrium states when there are equilibrium states. While the gas is flowing into B, it has a spatially varying kinetic energy and consequently a spatially varying internal energy.
 
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  • #14
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  • #15
Suppose we say that the boundary of our system with its surroundings is comprised of the inner surfaces of the two rigid cylinders, and our system consists of the gas within the two cylinders. How much work W does the gas do on the inner surfaces of the two rigid cylinders?

This is a tricky problem. If there is heat exchange allowed between the two cylinders (say through the valve), then their answer that the temperature does not change is correct. But, if no heat exchange is allowed between the two cylinders, then the air remaining in the left cylinder will finally be colder than initially, and the gas transferred to the right cylinder will finally be hotter than initially. Any interest in analyzing this version of the problem?
 
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  • #16
Chestermiller said:
But, if no heat exchange is allowed between the two cylinders, then the air remaining in the left cylinder will finally be colder than initially, and the gas transferred to the right cylinder will finally be hotter than initially. Any interest in analyzing this version of the problem?
OK, I'll bite. With the valve open and both chambers having reached equilibrium conditions, why would the molecules in the initially empty chamber end up having a higher average kinetic energy?
 
  • #17
kuruman said:
OK, I'll bite. With the valve open and both chambers having reached equilibrium conditions, why would the molecules in the initially empty chamber end up having a higher average kinetic energy?
Viscous heating in the stopcock valve and recompression of the gas in the cylinder initially containing vacuum.
 
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  • #18
kuruman said:
For those interested, this Wikipedia entry treats exactly the case in this thread.
A great find. Thankyou.

As per this article on Wikipedia, it clearly states that "Also, since the system's total volume is kept constant, the system cannot perform work on its surroundings.". So, in my post#9, it's the second mentioned reason that makes ##W=0## in this question. I had asked in post #9 the following:
W=0 (since the gas expansion is against vacuum or maybe because the the thermodynamic system doesn't expand i.e. cylinders A,B do not expand).
 
  • #19
Frabjous said:
While the gas is flowing into B, it has a spatially varying kinetic energy and consequently a spatially varying internal energy.
That's a beautiful description of what's happening. Very precise and accurate. I love the phrase "spatially varying". It makes so much sense.
 
  • #20
Chestermiller said:
How much work W does the gas do on the inner surfaces of the two rigid cylinders?
Zero, since the walls do not expand or contract.
 
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  • #21
Frabjous said:
The gas does work on itself
Could you please explain how the gas does work on itself while expanding? I'm confused. I was thinking the molecules of the gas get further apart and so gas is doing work against the intermolecular attractive forces; but an ideal gas has no intermolecular forces according to what I know. Maybe I'm wrong about the absence of intermolecular forces in an ideal gas.
 
  • #22
Chestermiller said:
Viscous heating in the stopcock valve and recompression of the gas in the cylinder initially containing vacuum.
I can understand the viscous heating in the stopcock part but can you explain recompression a bit more? Won't the two chambers eventually reach the same equilibrium pressure? Also, let's say that the temperature of the valve rises initially, will it not equilibrate eventually at the same temperature of the gas? My naive thinking is that any temperature difference will drive heat flow. Unless of course there is a Maxwell's demon at the valve.
 
  • #23
kuruman said:
I can understand the viscous heating in the stopcock part but can you explain recompression a bit more?
The gas comes out of the stopcock at a pressure less than the final pressure (in the right cylinder) and ends up being compressed by subsequent gas discharge to eventually reach the final pressure.
kuruman said:
Won't the two chambers eventually reach the same equilibrium pressure?
Yes.
kuruman said:
Also, let's say that the temperature of the valve rises initially, will it not equilibrate eventually at the same temperature of the gas? My naive thinking is that any temperature difference will drive heat flow. Unless of course there is a Maxwell's demon at the valve.
Like the walls of the cylinders, the valve is assumed to have negligible thermal inertia, so that its internal energy change is always negligible. And it is assumed to have negligible thermal conductivity,, so that no heat can be conducted between the lower pressure chamber and the higher pressure chamber during the process.
 
  • #24
vcsharp2003 said:
Could you please explain how the gas does work on itself while expanding? I'm confused. I was thinking the molecules of the gas get further apart and so gas is doing work against the intermolecular attractive forces; but an ideal gas has no intermolecular forces according to what I know. Maybe I'm wrong about the absence of intermolecular forces in an ideal gas.
It’s just PdV work.
 
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  • #25
Chestermiller said:
But, if no heat exchange is allowed between the two cylinders, then the air remaining in the left cylinder will finally be colder than initially, and the gas transferred to the right cylinder will finally be hotter than initially. Any interest in analyzing this version of the problem?
I’d be interested in seeing the numbers quantified.
 
  • #26
Frabjous said:
I’d be interested in seeing the numbers quantified.
Are you referring to the solution to the 2nd version of the problem (the version I posed)
 
  • #27
Yes.
 
  • #28
Frabjous said:
Yes.
I would like to see what other members come up with
 
  • #29
I’m coming into the thread rather late, but a few thoughts...

For an ideal gas, under perfect adiabatic conditions, only purely elastic particle collisions (with the walls and with other particles) are possible.

[Edi: I should add that the walls need to be stationary because rebound from a moving wall will change a particle's kinetic energy.]

Therefore the total kinetic energy of the gas particles can’t change. And so the average kinetic energy per particle can’t change.

The temperature of an ideal gas depends only on the average kinetic energy per particle. Therefore the final (equilibrium) temperature must equal the initial temperature.

Particle speeds are determined by the Maxwell-Boltzmann distribution. Soon after the valve is opened, the first particles to reach the far end of B will be the fastest ones. This mean initially the gas at the far end of B is hottest (but low density and pressure).

There will be a time and position-dependant variation of temperature during the period from opening the valve to reaching equilibrium. There is also the possibility of some sort of longitudinal wave being produced to complicate matters!
 
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  • #30
Steve4Physics said:
Therefore the total kinetic energy of the gas particles can’t change. And so the average kinetic energy per particle can’t change.
That's new to me but I can see how it makes complete sense in this situation. Thankyou.
 
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  • #31
Steve4Physics said:
.... There is also the possibility of some sort of longitudinal wave being produced to complicate matters!
With the proper timing and velocity through the valve, an oscillation of pressure values could occur between both tanks, before pressure equalizes.

For the first cycle at least, low pressure inside the left A cylinder could lead to lower temperature in there, while simultaneously, the opposite happens inside the right B cylinder.

Please, see:
https://en.wikipedia.org/wiki/Kadenacy_effect

https://en.wikipedia.org/wiki/Expansion_chamber
 
  • #32
vcsharp2003 said:
That's new to me but I can see how it makes complete sense in this situation. Thankyou.
@vcsharp2003, please note I edited Post #29 as I thought something I wrote wasn’t clear.

In a little more detail...

For an adiabatic compression or expansion of an ideal gas, the total kinetic energy of the particles changes if particles bounce off a moving wall. However that’s not the case in your Post #1 problem, so the total kinetic energy is constant in your problem.

(It's a useful exercise to think about how, say compressing a gas by pushing-in a piston, increases the kinetic energy of the gas particles. (You consider the behaviour of particles bouncing off a moving wall.).)
 
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  • #33
Mass Balance: $$n_1+n_2=n_0$$where ##n_0## is the number of moles of gas initially in the left chamber, ##n_1## is the final number of moles of gas in the left chamber, and ##n_2## is the final number of moles in the right chamber. From the ideal gas law, this gives: $$\frac{PV}{RT_1}+\frac{PV}{RT_2}=\frac{P_0V}{RT_0}$$where P is the final pressure in both chambers, V is the volume of each of the two chambers, ##P_0## is the initial pressure in the left chamber, ##T_1## is the final temperature in the left chamber, ##T_2## is the final temperature in the right chamber, and ##T_0## is the initial temperature of the gas in the left chamber. So, $$\frac{T_0}{T_1}+\frac{T_0}{T_2}=\frac{P_0}{P}\tag{1}$$Energy Balance: $$\Delta U=n_1C_V(T-T_0)+n_2C_v(T_2-T_0)=0$$, or, combining with the ideal gas law:$$1-\frac{T_0}{T_1}+1-\frac{T_0}{T_2}=0$$or$$\frac{T_0}{T_1}+\frac{T_0}{T_2}=2\tag{2}$$So, combining Eqns. 1 and 2, we have $$P=\frac{P_0}{2}\tag{3}$$Thus, the final pressure in both chambers is half the initial pressure in the left chamber.

The gas finally remaining in the left chamber has experienced an adiabatic reversible expansion in gradually pushing the gas ahead of it into the stopcock, and into the right chamber. Thus, $$\frac{T_1}{T_0}=\left(\frac{P}{P_0}\right)^{\frac{\gamma-1}{\gamma}}=\left(\frac{1}{2}\right)^{\frac{\gamma-1}{\gamma}}\tag{4}$$Next, combining Eqns. 2 and 4 yields:$$\frac{T_2}{T_0}=\frac{1}{2-2^{\frac{\gamma-1}{\gamma}}}>1$$
 
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  • #34
Given these results, what is the entropy change for this version of the problem and how does it compare quantitatively with the entropy change for the other version?
 

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