# There is a thermodynamics problem I cannot solve

1.
Consider a balloon which has thick rigid walls and
from which all the air has been pumped out. Now, the valve of
the balloon is slightly opened, and the balloon is slowly filled
with the air from outside. Find the temperature of the air
inside the balloon once the air flow has stopped (since a mech-
anical equilibrium has been reached). The room temperature
is T, the balloon walls have low heat conductance and heat ca-
pacitance so that heat flux through the walls can be neglected.

2. P*V=const
ΔQ=PΔV+ΔU

3. I cannot figure out how is the work done by external air converted to heat and how is the external volume change related to the volume of cavity.

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• Klaus von Faust This one I don't think is very simple because of things like the Joule-Thomson effect: https://en.wikipedia.org/wiki/Joule–Thomson_effect
I will try to solve it again, with fresh mind, but it is indeed complicated. The only thing that I can get from Joule-thomson effect is that the entropy is constant. As I understand, this means that all work done by external air is incrementing the internal energy. Intuitively, I found out that the temperature must have changed by the adiabatic constant of air. But I am pretty unsure whether my answer is correct or not.

•  This one I don't think is very simple because of things like the Joule-Thomson effect: https://en.wikipedia.org/wiki/Joule–Thomson_effect
Could you give me another hint, please?

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Could you give me another hint, please?
This one I really don't know how to do. @Chestermiller is the Physics Forums Thermodynamics expert. I don't know that the parameters of the problem, such as the hole size are sufficiently specified. In addition, with this type of problem, I have seen effusion calculations to determine what goes through the aperture, (both the number of particles, as well as their velocity distribution), as well as an alternative calculation of laminar flow through the aperture, where I think it is Bernoulli's equation that gets employed. With the Bernoulli approach, I have very limited experience. The effusion case I can readily compute, but I don't know that it is applicable.

• Klaus von Faust
This one I really don't know how to do. @Chestermiller is the Physics Forums Thermodynamics expert. I don't know that the parameters of the problem, such as the hole size are sufficiently specified. In addition, with this type of problem, I have seen effusion calculations to determine what goes through the aperture, (both the number of particles, as well as their velocity distribution), as well as an alternative calculation of laminar flow through the aperture, where I think it is Bernoulli's equation that gets employed. With the Bernoulli approach, I have very limited experience. The effusion case I can readily compute, but I don't know that it is applicable.
Thank you very much!

• Homework Helper
Gold Member
Just an additional item: In the effusion case, the distribution of speeds of molecules that enter the cavity will be higher than the distribution of speeds that exists outside the cavity, because the effusion rate formula is ## R=\frac{n \bar{v}}{4} ##. Thereby the temperature of the gas in the box will be higher than the temperature outside the box, but again, I don't think the effusion calculation is applicable here.

• Klaus von Faust
Chestermiller
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I will try to solve it again, with fresh mind, but it is indeed complicated. The only thing that I can get from Joule-thomson effect is that the entropy is constant. As I understand, this means that all work done by external air is incrementing the internal energy. Intuitively, I found out that the temperature must have changed by the adiabatic constant of air. But I am pretty unsure whether my answer is correct or not.
The entropy is not constant in the Joule-Thompson effect.

• Klaus von Faust and Charles Link
Chestermiller
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There are two equivalent ways of doing this problem. I'm going to assume that you haven't had any experience with the open system version of the first law of thermodynamics, so I'm going to help you using the closed system version of the first law. Imagine that your system consists of all the gas that will eventually be within the balloon in the final state of the system, and imagine that this gas is surrounded by an invisible membrane. So all the gas inside this membrane starts out outside the balloon, and ends up inside the balloon. If V is the volume of this gas before being pushed by the surrounding atmosphere into the balloon, how much work is done by the surrounding atmosphere in pushing the gas into the balloon? Is this the total work done by the surroundings on the system?

• Klaus von Faust
The entropy is not constant in the Joule-Thompson effect.
Excuse me, I meant enthalpy

There are two equivalent ways of doing this problem. I'm going to assume that you haven't had any experience with the open system version of the first law of thermodynamics, so I'm going to help you using the closed system version of the first law. Imagine that your system consists of all the gas that will eventually be within the balloon in the final state of the system, and imagine that this gas is surrounded by an invisible membrane. So all the gas inside this membrane starts out outside the balloon, and ends up inside the balloon. If V is the volume of this gas before being pushed by the surrounding atmosphere into the balloon, how much work is done by the surrounding atmosphere in pushing the gas into the balloon? Is this the total work done by the surroundings on the system?
I think that the total work done by system equals the external pressure(which is P) multiplied by change in volume. The system changes it's volume from V(volume of external gas located in the invisible membrane) to the volume of balloon.

Chestermiller
Mentor
I think that the total work done by system equals the external pressure(which is P) multiplied by change in volume. The system changes it's volume from V(volume of external gas located in the invisible membrane) to the volume of balloon.
This is not exactly what I had in mind, but close. If the membrane also passes through the value and surrounds the volume of vacuum initially present within the rigid balloon, then the initial volume of our system is ##(V_B+V)##, where ##V_B## is the volume of the balloon and V is the volume of outside gas that eventually enters the balloon. In the final state of the system, the membrane containing the gas entirely fills the balloon, so the final volume of our system is ##V_B##. So, the change in volume of this system is just -V, and the work done by the surrounding air on this system is just ##P_0V##, where ##P_0## is the outside pressure. Note that no work takes place at the rigid surface of the balloon, since the displacements there are zero. If the work done on the system is ##P_0V##, how is this related to the number of moles of gas that eventually enters n and the outside temperature of the gas ##T_0##? In terms of the n, the heat capacity of the gas ##C_v##, the initial temperature of the gas ##T_0##, and the final temperature of the gas within the balloon T, what is the change in internal energy ##\Delta U## of the gas that has entered the balloon? From the closed system version of the first law of thermodynamics, how is ##\Delta U## related to the work done by the surroundings on the system?

• Charles Link and Klaus von Faust
This is not exactly what I had in mind, but close. If the membrane also passes through the value and surrounds the volume of vacuum initially present within the rigid balloon, then the initial volume of our system is ##(V_B+V)##, where ##V_B## is the volume of the balloon and V is the volume of outside gas that eventually enters the balloon. In the final state of the system, the membrane containing the gas entirely fills the balloon, so the final volume of our system is ##V_B##. So, the change in volume of this system is just -V, and the work done by the surrounding air on this system is just ##P_0V##, where ##P_0## is the outside pressure. Note that no work takes place at the rigid surface of the balloon, since the displacements there are zero. If the work done on the system is ##P_0V##, how is this related to the number of moles of gas that eventually enters n and the outside temperature of the gas ##T_0##? In terms of the n, the heat capacity of the gas ##C_v##, the initial temperature of the gas ##T_0##, and the final temperature of the gas within the balloon T, what is the change in internal energy ##\Delta U## of the gas that has entered the balloon? From the closed system version of the first law of thermodynamics, how is ##\Delta U## related to the work done by the surroundings on the system?
Well, the problem states that heat capacity of the rigid walls of balloon is to be neglected, this means that Q is zero and all work done is used in internal energy change.
So P*V=ΔU and ΔU=n*Cv*(T-##T_0##)

Well, the problem states that heat capacity of the rigid walls of balloon is to be neglected, this means that Q is zero and all work done is used in internal energy change.
So P*V=ΔU and ΔU=n*Cv*(T-##T_0##)
Thank you very much. I solved it, and the answer is 1.4

• Chestermiller
Chestermiller
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Thank you very much. I solved it, and the answer is 1.4
Very nicely done. Excellent.