What Happens to Kinetic Energy of the Piston in an Isothermal System?

In summary, the conversation discusses a system with two chambers of gas, where one chamber increases in volume while the other decreases, and a reservoir that remains at constant volume and is not worked on. The entropy changes of the gas and reservoir are calculated using thermodynamic equations, and it is found that the entropy change of the universe is zero. The conversation also touches on the work and heat exchanged between the chambers and the surroundings. Overall, the final state of the system is that the piston has been moved and the system has returned to equilibrium, with no net heat exchanged between the system and surroundings.
  • #1
Pendleton
20
3
Homework Statement
A clamped piston partitions a container of ideal gas into halves of volume V, the left one at pressure ##P_1## and right at pressure ##P_2 < P_1##, and is kept at temperature T by a reservoir. Were the piston unclamped, moved by δ, and reclamped, how would the entropy of each gas, the reservoir, and universe change?
Relevant Equations
$$dU = dQ + dW$$
$$dU = TdS - PdV$$
Attempted Solution:

Gas Entropy
This system is isothermal: the energy of each gas remains constant.
$$dU = 0$$

By the combined statement of the first and second laws,
$$dU = TdS - PdV$$

Therefore,
$$0 = TdS - PdV$$
$$dS = \frac {PdV}{T}$$

Therefore,
$$dS_1 = \frac {P_1 dV_1}{T} = \frac {P_1 Aδ}{T}$$
$$dS_2 = \frac {P_2 dV_2}{T} = \frac {P_2 A(-δ)}{T} = \frac {-P_2 Aδ}{T}$$


Reservoir Entropy Change
By the fundamental thermodynamic relation,
$$TdS - PdV = dQ + dW$$

The reservoir has constant volume and is not worked on.
$$TdS_R - 0 = dQ_R + 0$$
$$dS_R = \frac{dQ_R}{T}$$

The heat absorbed by the reservoir is emitted by the gases.
$$dQ_R = dQ_1 + dQ_2$$

By the first law,
$$dU = dQ + dW$$
$$0 = dQ + dW$$
$$dQ = -dW$$

The work done on each gas is the pressure-volume work done by the other.
$$dW_1 = P_2 dV_2 = P_2 A(-δ) = -P_2 Aδ$$
$$dW_2 = P_1 dV_1 = P_1 Aδ$$

Therefore,
$$dQ_1 = -(-P_2 Aδ) = P_2 Aδ$$
$$dQ_2 = -P_1 Aδ$$
$$dQ_R = P_2 Aδ + (-P_1 Aδ) = (P_2 - P_1)Aδ$$
$$dS_R = \frac {P_2 - P_1}{T} Aδ$$

Universe Entropy Change
The change of the entropy of the universe is the sum of the changes of the entropy of the gasses and reservoir.

$$dS_U = dS_1 + dS_2 + dS_R$$
$$ = \frac {P_1 Aδ}{T} + \frac {-P_2 Aδ}{T} + \frac {P_2 - P_1}{T} Aδ$$
$$ = \frac{P_1 - P_2}{T} Aδ + \frac {P_2 - P_1}{T} Aδ$$
$$dS_U = 0$$
 
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  • #2
what is the final state of the system?
 
  • #3
Chestermiller said:
what is the final state of the system?

My goodness, it’s you! Thanks for being here. The final state of the system is that the piston has been reclamped, having been moved δ rightward, and the system returned to equilibrium.
 
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  • #4
Pendleton said:
My goodness, it’s you! Thanks for being here. The final state of the system is that the piston has been reclamped, having been moved δ rightward, and the system returned to equilibrium.
So what are the final pressures, temperatures, and volumes of the two chambers?
 
  • #5
Chestermiller said:
So what are the final pressures, temperatures, and volumes of the two chambers?

The volume of chamber 1 increases by Aδ and volume of chamber 2 decreases by Aδ. The temperature of both chambers is T, the same as the initial temperature. Pressure changes only with the second order of δ.
 
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  • #6
The final volume of the p2 chamber is ##(V+A\delta)## and the final pressure of the p1 chamber is ##(V-A\delta)##. So, for these ideal gases, the changes in entropy are $$\Delta S_2=n_2R\ln{(V+A\delta)/V}=\frac{P_2V}{T}\ln{\left(1+\frac{A\delta}{V}\right)}$$
$$\Delta S_1=n_1R\ln{(V-A\delta)/V}=\frac{P_1V}{T}\ln{\left(1-\frac{A\delta}{V}\right)}$$

How much work does the combination of the two chambers do on the external surroundings? If the change in internal energy of the combination of the two chambers is zero, how much heat has the combination of the two chambers (i.e., the system) exchanged with the surroundings?
 
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  • #7
Chestermiller said:
The final volume of the p2 chamber is ##(V+A\delta)## and the final pressure of the p1 chamber is ##(V-A\delta)##. So, for these ideal gases, the changes in entropy are $$\Delta S_2=nR\ln{(V+A\delta)/V}=\frac{P_2V}{T}\ln{\left(1+\frac{A\delta}{V}\right)}$$
$$\Delta S_1=n_1R\ln{(V-A\delta)/V}=\frac{P_1V}{T}\ln{\left(1-\frac{A\delta}{V}\right)}$$

Hold on. The entropy change formula

$$dS = nRln{\frac{V_f}{V_o}}$$

implies that

$$dS_1 = nRln{\frac{V+Aδ}{V}}$$
$$= \frac{P_1}{T}ln(1+\frac{Aδ}{V})$$$$dS_2 = nRln{\frac{V-Aδ}{V}}$$
$$= \frac{P_2}{T}ln(1-\frac{Aδ}{V})$$

because, just to clarify, the volume of chamber 1, which is the one on the left and which has the higher pressure, increases by Aδ. The volume of the other chamber decreases by same.

Also, why not just say that

$$TdS - PdV = dU = 0$$
$$dS = \frac{PdV}{T}$$

Chestermiller said:
How much work does the combination of the two chambers do on the external surroundings? If the change in internal energy of the combination of the two chambers is zero, how much heat has the combination of the two chambers (i.e., the system) exchanged with the surroundings?

The combination of the two chambers does not work on the surroundings because the container they are in is of fixed volume. The heat absorbed by each chamber from the reservoir is

$$dQ = -dW$$

where ##dW## is the work done on the chamber by the other chamber. That work is

$$dW_1 = P_2 dV_2 = P_2 A(-δ) = -P_2 Aδ$$
$$dW_2 = P_1 dV_1 = P_1 A-δ = P_1 Aδ$$

Therefore,
$$dQ_1 = -dW_1 = P_2 Aδ$$
$$dQ_2 = -dW_2 = -P_1 Aδ$$
 
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  • #8
Pendleton said:
Hold on. Isn’t the entropy change formula

$$dS = nRln{\frac{V_f}{V_o}}$$
Sure. Isn't that what I used for each chamber?
Also, just to clarify, the volume of chamber 1, which is the one on the left and which has the higher pressure, increases by Aδ. The volume of the other chamber decreases by same.
Correct. That's what I used in my equations.
The combination of the two chambers does not work on the surroundings because the container they are in is of fixed volume.
If the combination of the two chamber does no work and the change in internal energy of the combination of the two chambers is zero, then if ##\Delta U=Q-W##, this means that Q = 0. That is, there is no net heat exchanged between the combination of the two chambers and the surroundings. What does that tell you about the change in entropy of the reservoir?
 
  • #9
Chestermiller said:
Sure. Isn't that what I used for each chamber?
Correct. That's what I used in my equations.

First point, yes. I see now. The formula was right. Second point, I think you had it backward. Also, I edited in a question about why we don’t just say that

$$dS = \frac{PdV}{T}$$

If the combination of the two chamber does no work and the change in internal energy of the combination of the two chambers is zero, then if ##\Delta U=Q-W##, this means that Q = 0. That is, there is no net heat exchanged between the combination of the two chambers and the surroundings. What does that tell you about the change in entropy of the reservoir?
If the combination of the two chamber does no work and the change in internal energy of the combination of the two chambers is zero, then if ##\Delta U=Q-W##, this means that Q = 0. That is, there is no net heat exchanged between the combination of the two chambers and the surroundings. What does that tell you about the change in entropy of the reservoir?

That means the overall process is both adiabatic and isothermal. How is this possible?[/QUOTE]
 
  • #10
Pendleton said:
Hold on. The entropy change formula

$$dS = nRln{\frac{V_f}{V_o}}$$

implies that

$$dS_1 = nRln{\frac{V+Aδ}{V}}$$
$$= \frac{P_1}{T}ln(1+\frac{Aδ}{V})$$$$dS_2 = nRln{\frac{V-Aδ}{V}}$$
$$= \frac{P_2}{T}ln(1-\frac{Aδ}{V})$$

because, just to clarify, the volume of chamber 1, which is the one on the left and which has the higher pressure, increases by Aδ. The volume of the other chamber decreases by same.

Also, why not just say that

$$TdS - PdV = dU = 0$$
$$dS = \frac{PdV}{T}$$
The combination of the two chambers does not work on the surroundings because the container they are in is of fixed volume. The heat absorbed by each chamber from the reservoir is

$$dQ = -dW$$

where ##dW## is the work done on the chamber by the other chamber. That work is

$$dW_1 = P_2 dV_2 = P_2 A(-δ) = -P_2 Aδ$$
$$dW_2 = P_1 dV_1 = P_1 A-δ = P_1 Aδ$$

Therefore,
$$dQ_1 = -dW_1 = P_2 Aδ$$
$$dQ_2 = -dW_2 = -P_1 Aδ$$
The work that one gas does on the other gas has to be equal and opposite in sign to the work that work that the other gas does on the first gas. So their sum is zero.
 
  • #11
Chestermiller said:
The work that one gas does on the other gas has to be equal and opposite in sign to the work that work that the other gas does on the first gas. So their sum is zero.

That is not true because the pressures are different. The expanding gas in the left chamber, where the pressure is higher, does more work on the gas it is compressing than the latter does on the former.
 
  • #12
Pendleton said:
That is not true because the pressures are different. The expanding gas in the left chamber, where the pressure is higher, does more work on the gas it is compressing than the latter does on the former.
This is an irreversible process, and there are viscous stresses that modify the force per unit area that each of the gases exert on each of the faces of the partition. So the force per unit area exerted by the gases on the piston faces during this irreversible change are not P1 and P2. If the partition is massless and frictionless, the force per unit area exerted by each of the gases on its own face of the piston must match the force per unit area exerted the other gas. on its face of the piston (Newton's 2nd law).
 
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  • #13
It is not valid to use the equation ##dS=\frac{PdV}{T}## because this equation applies to a tiny change between two differentially neighboring thermodynamic equilibrium states (i.e., a reversible path), and this process is irreversible. However, for a small ##\delta##, it provides a good approximation.
 
  • #14
Chestermiller said:
This is an irreversible process, and there are viscous stresses that modify the force per unit area that each of the gases exert on each of the faces of the partition. If the partition is massless and frictionless, the force per unit area exerted by each of the gases on its own face of the piston must match the force per unit area exerted the other gas. on its face of the piston (Newton's 2nd law).

Woaaaaah—viscous stresses in an ideal gas? I thought their particles never interacted. Tell me more.

It is not valid to use the equation ##dS=\frac{PdV}{T}## because this equation applies to a change between differentially neighboring thermodynamic equilibrium states (i.e., a reversible path), and this process is irreversible.

My lecture notes state that because T, S, and V are all state variables, ##dU = TdS - PdV## is true for all processes, reversible and irreversible.
 
  • #15
Pendleton said:
Woaaaaah—viscous stresses in an ideal gas? I thought their particles never interacted. Tell me more.
I know that they say this sometimes, but they really shouldn't. What is the mean free path of air molecules at 1 atm. and room temperature (i.e.,conditions under which air exhibits virtually ideal gas behavior)? In an irreversible problem like this, it is necessary to acknowledge the viscous behavior of the gases, even in the ideal gas limit. Bird, Stewart, and Lightfoot, Transport Phenomena actually have a corresponding states graph for gas viscosity, which shows the limiting line in the limit of low pressures.
My lecture notes state that because T, S, and V are all state variables, ##dU = TdS - PdV## is true for all processes, reversible and irreversible.
If you have a process which has a highly convoluted irreversible path with large excursion from the starting thermodynamic equilibrium state, only to arrive at the closely neighboring thermodynamic equilibrium state in the end, then this equation will describe the final differences in the parameters between the two end states only. But it can't be used at each increment of the irreversible path which involves a continuous sequence of non-equilibrium states.
 
  • #16
Chestermiller said:
I know that they say this sometimes, but they really shouldn't. What is the mean free path of air molecules at 1 atm. and room temperature (i.e.,conditions under which air exhibits virtually ideal gas behavior)? In an irreversible problem like this, it is necessary to acknowledge the viscous behavior of the gases, even in the ideal gas limit. Bird, Stewart, and Lightfoot, Transport Phenomena actually have a corresponding states graph for gas viscosity, which shows the limiting line in the limit of low pressures.

Yet this seems to contradict our previous agreement that the change of entropy of an ideal gas isothermally compressed a small distance by a piston of area A was

$$dS = \frac{-PA}{T} δ$$

If you have a process which has a highly convoluted irreversible path with large excursion from the starting thermodynamic equilibrium state, only to arrive at the closely neighboring thermodynamic equilibrium state in the end, then this equation will describe the final differences in the parameters between the two end states only. But it can't be used at each increment of the irreversible path which involves a continuous sequence of non-equilibrium states.

This problem, despite describing an infinitesimal change, mercifully does not require a path integral over a finite change. We are free to consider the difference between the initial and final states without regard for this issue.
 
  • #17
Pendleton said:
Yet this seems to contradict our previous agreement that the change of entropy of an ideal gas isothermally compressed a small distance by a piston of area A was

$$dS = \frac{-PA}{T} δ$$
This is just the limit of my equations for small ##\delta##

This problem, despite describing an infinitesimal change, mercifully does not require a path integral over a finite change. We are free to consider the difference between the initial and final states without regard for this issue.
In my judgment and experience, for whatever it's worth, d's should never be used for irreversible paths, except when evaluating an alternate reversible path between the same two end points. For irreversible differences between the end points, one should always use ##\Delta 's##. This will help to avoid a lot of confusion on the part of the analyst. For a short path, one can always take the limit of the result as some change parameter like ##\delta##, in our case, becomes small. It hurts my eyes to see differentials used for irreversible paths.
 
  • #18
Chestermiller said:
This is just the limit of my equations for small ##\delta##

Yes, and here ##\delta## is small.

In my judgment and experience, for whatever it's worth, d's should never be used for irreversible paths, except when evaluating an alternate reversible path between the same two end points. For irreversible differences between the end points, one should always use ##\Delta 's##. This will help to avoid a lot of confusion on the part of the analyst. For a short path, one can always take the limit of the result as some change parameter like ##\delta##, in our case, becomes small. It hurts my eyes to see differentials used for irreversible paths.

Oh, ok! It is very inconvenient to type out

By the above formula,
$$\Delta S_1 = \frac{P_1 A\delta}{T}$$
$$\Delta S_2 = \frac{P_2 A (- \delta)}{T} = - \frac{P_2 A \delta}{T}$$

By the fundamental relation,
$$\Delta S_R = \frac{dQ_R}{T}$$

The heat absorbed by the reservoir is the heat emitted by the chambers.
$$dQ_R = -(dQ_1 + dQ_2)$$

By the first law,
$$dU = dQ + dW, ~ dU = 0$$
$$dQ = -dW$$

The work done on each chamber is the pressure-volume work done by the other.
$$dW_1 = P_2 dV_2 = P_2 A (- \delta) = -P_2 A \delta$$
$$dW_2 = P_1 dV_1 = P_1 A \delta$$

Therefore,
$$dQ_R = -(-P_2 A \delta + P_1 A \delta)$$
$$= P_2 A \delta - P_1 A \delta$$
$$=(P_2 - P_1)A \delta$$

Therefore,
$$\Delta S_R = \frac{(P_2 - P_1)A \delta}{T}$$

Therefore,
$$\Delta S_U = \Delta S_1 + \Delta S_2 + \Delta S_R$$
$$= \frac{P_1 A\delta}{T} - \frac{P_2 A\delta}{T} + \frac{(P_2 - P_1)A \delta}{T}$$
$$ = 0$$

This seems strange for an irreversible process.
 
  • #19
Pendleton said:
Yes, and here ##\delta## is small.
Oh, ok! It is very inconvenient to type out

By the above formula,
$$\Delta S_1 = \frac{P_1 A\delta}{T}$$
$$\Delta S_2 = \frac{P_2 A (- \delta)}{T} = - \frac{P_2 A \delta}{T}$$

By the fundamental relation,
$$\Delta S_R = \frac{dQ_R}{T}$$

The heat absorbed by the reservoir is the heat emitted by the chambers.
$$dQ_R = -(dQ_1 + dQ_2)$$

By the first law,
$$dU = dQ + dW, ~ dU = 0$$
$$dQ = -dW$$

The work done on each chamber is the pressure-volume work done by the other.
$$dW_1 = P_2 dV_2 = P_2 A (- \delta) = -P_2 A \delta$$
$$dW_2 = P_1 dV_1 = P_1 A \delta$$

Therefore,
$$dQ_R = -(-P_2 A \delta + P_1 A \delta)$$
$$= P_2 A \delta - P_1 A \delta$$
$$=(P_2 - P_1)A \delta$$

Therefore,
$$\Delta S_R = \frac{(P_2 - P_1)A \delta}{T}$$

Therefore,
$$\Delta S_U = \Delta S_1 + \Delta S_2 + \Delta S_R$$
$$= \frac{P_1 A\delta}{T} - \frac{P_2 A\delta}{T} + \frac{(P_2 - P_1)A \delta}{T}$$
$$ = 0$$

This seems strange for an irreversible process.
That's because it is incorrect. I keep telling you that, during the partition motion, the forces on the partition are not P1A and P2A. If F1 and F2 are the forces exerted by the gases on the partition during the irreversible process and m is the mass of the partition, what is Newton's 2nd law applied to the partition (assuming frictionless partition)?

Doesn't it bother you that the 1st law applied to the combination of the two chambers indicates no work on the combination and no heat transfer from the reservoir to the combination, while your analysis does not?

The heat transferred from the reservoir to the combination is zero, and the change in entropy of the reservoir is zero. The change in entropy of the universe is thus just equal to the change in entropy of the system. Yes, the system is both isothermal and adiabatic. This same thing happens in a free expansion with a partitioned chamber.
 
  • #20
Chestermiller said:
That's because it is incorrect. I keep telling you that, during the partition motion, the forces on the partition are not P1A and P2A. If F1 and F2 are the forces exerted by the gases on the partition during the irreversible process and m is the mass of the partition, what is Newton's 2nd law applied to the partition (assuming frictionless partition)?

The partition is unclamped, allowed to move to the right by the force of the higher-pressure gas against it, reclamped, and then allowed to equilibrate. Therefore, if F1 and F2 are the forces of the gases,

$$F_{NET} = F_1 - F_2 = ma$$

where a is the rightward acceleration of the partition. If

$$F_1 = P_1 A = P_2 A = F_2$$

then the chambers would be in mechanical equilibrium, and the partition would not move unclamped. Furthermore,

$$P_1 = P_2$$

which directly contradicts our given that ##P_1 > P_2$$

Doesn't it bother you that the 1st law applied to the combination of the two chambers indicates no work on the combination and no heat transfer from the reservoir to the combination, while your analysis does not?

Is the internal energy of the system equal to merely the sum of the internal energy of the gases, or could useful work could be extracted by unclamping the partition and letting it reach its equilibrium position? Otherwise,

By the first law,
$$dU = dQ + dW$$
$$dU_1 = dQ_1 + dW_1$$
$$dU_2 = dQ_2 + dW_2$$

The system being isothermal,
$$dU_S = dU_1 + dU_2 = 0 + 0 = 0$$
$$dQ_1 = -dW_1$$
$$dQ_2 = -dW_2$$

The system is of fixed volume
$$dW_S = dW_1 + dW_2 = 0$$

Returning to the first law,
$$dU_S = dQ_S + dW_S, ~ dU_S = 0, ~ dW_S = 0$$

$$0 = dQ_S + 0$$
$$dQ_S = 0$$

The heat absorbed by the reservoir is the heat emitted by the system.
$$dQ_R = -dQ_S = 0$$

By the fundamental relation,
$$dS_R = \frac{dQ_R}{T} = 0$$

This seems bizarre, but perhaps it’s true.

The heat transferred from the reservoir to the combination is zero, and the change in entropy of the reservoir is zero. The change in entropy of the universe is thus just equal to the change in entropy of the system. Yes, the system is both isothermal and adiabatic. This same thing happens in a free expansion with a partitioned chamber.

In a free expansion with a partitioned chamber, but no reservoir,

$$dU = dQ + dW = 0 + 0 = 0$$

The entropy of an ideal gas is
$$S = S_0 + nRln(\frac{V}{V_0}) + nC_V ln(\frac{T}{T_0})$$

Free expansion being isothermal,
$$S - S_0 = nRln(\frac{V}{V_0})$$

Where the difference of entropies is a change due to infinitesimal expansion,
$$S - S_0 = \frac{\partial S}{\partial V} = \frac{\partial}{\partial V} nRln(\frac{V}{V_0})$$
$$=nR \frac{\partial}{\partial V} (ln(V) - ln(V_0))$$
$$= \frac{PV}{T} \frac{1}{V}$$
$$= \frac{P}{T}$$
 
  • #21
You forgot to account for what happens to the kinetic energy of the piston once it is reclamped. Let ##F_1(x)## be the force exerted by gas 1 on the piston when it has moved distance x to the right of center, and left ##F_2(x)## be the force exerted by gas 2. Then, a force balance on the piston becomes:
$$F_2-F_1=m\frac{dv}{dt}$$where v is the velocity of the piston dx/dt. If we multiply this equation by v = dx/dt and integrate with respect to time, from t = 0 to time ##t_c## just before the instant that the piston is clamped, we obtain $$W_2+W_1=\frac{1}{2}mv_c^2=KE$$where ##v_c## is the piston velocity just as the piston reaches ##x=\delta##, and where $$W_2=\int_0^{\delta}{F_2(x)dx}$$and $$W_1=-\int_0^{\delta}{F_1(x)dx}$$with ##W_1## and ##W_2## representing the work done by gases 1 and 2 on the piston. Once the piston is reclamped, no more work occurs, but heat can continue to flow between the compartments, and the internal energies of the gases in the compartments can change until the system re-equilibrates. In addition, the kinetic energy of the piston is converted to heat, and a fraction of this heat f is released into compartment 2, and the remaining fraction (1-f) is released into compartment 1. So the final 1st law energy balances on the two chambers become: $$\Delta U_2=Q_{1,2}+(KE)f-W_2$$and $$\Delta U_1=-Q_{1,2}+(KE)(1-f)-W_1$$where ##Q_{12}## is the heat flow from compartment 1 to compartment 2 (this is over and above the heat flow derived from the piston KE to the two compartments). Note that no heat flow from the reservoir to the compartments is included in these equations. If we add the two equations together, we obtain:
$$\Delta U=\Delta U_1+\Delta U_2=KE-(W_1+W_2)=0$$So the condition for no change of internal energy of the combination is satisfied without having to invoke heat flow from the surroundings.

Like I said, the error you made in your analysis was failing to account for the kinetic energy of the piston.
 
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  • #22
Chestermiller said:
You forgot to account for what happens to the kinetic energy of the piston once it is reclamped. Let ##F_1(x)## be the force exerted by gas 1 on the piston when it has moved distance x to the right of center, and left ##F_2(x)## be the force exerted by gas 2. Then, a force balance on the piston becomes:
$$F_2-F_1=m\frac{dv}{dt}$$where v is the velocity of the piston dx/dt. If we multiply this equation by v = dx/dt and integrate with respect to time, from t = 0 to time ##t_c## just before the instant that the piston is clamped, we obtain $$W_2+W_1=\frac{1}{2}mv_c^2=KE$$where ##v_c## is the piston velocity just as the piston reaches ##x=\delta##, and where $$W_2=\int_0^{\delta}{F_2(x)dx}$$and $$W_1=-\int_0^{\delta}{F_1(x)dx}$$with ##W_1## and ##W_2## representing the work done by gases 1 and 2 on the piston. Once the piston is reclamped, no more work occurs, but heat can continue to flow between the compartments, and the internal energies of the gases in the compartments can change until the system re-equilibrates. In addition, the kinetic energy of the piston is converted to heat, and a fraction of this heat f is released into compartment 2, and the remaining fraction (1-f) is released into compartment 1. So the final 1st law energy balances on the two chambers become: $$\Delta U_2=Q_{1,2}+(KE)f-W_2$$and $$\Delta U_1=-Q_{1,2}+(KE)(1-f)-W_1$$where ##Q_{12}## is the heat flow from compartment 1 to compartment 2 (this is over and above the heat flow derived from the piston KE to the two compartments). Note that no heat flow from the reservoir to the compartments is included in these equations. If we add the two equations together, we obtain:
$$\Delta U=\Delta U_1+\Delta U_2=KE-(W_1+W_2)=0$$So the condition for no change of internal energy of the combination is satisfied without having to invoke heat flow from the surroundings.

Like I said, the error you made in your analysis was failing to account for the kinetic energy of the piston.

Sorry for taking so long to reply! You are right—the chambers are in thermal contact with each other. Therefore, here is my analysis.

For the system,
$$dU_S = Q_S + W_S$$
$$dU_S = 0, ~ dW_S = 0$$
$$0 = dQ_S + 0$$
$$dQ_S = 0$$

For the reservoir,
$$dQ_R = -dQ_S$$
$$dS_R = \frac{dQ_R}{T}$$
$$dS_R = 0$$

For each chamber,
$$dU = TdS - PdV$$
$$dU = 0$$
$$dS = \frac{PdV}{T}$$
$$dV_1 = A\delta, ~ dV_2 = A(-\delta)$$
$$dS_1 = \frac{P_1 A\delta}{T}$$
$$dS_2 = \frac{P_2 A(-\delta)}{T}$$

For the universe,
$$dS_U = dS_1 + dS_2 + dS_R$$
$$= \frac{P_1 A\delta}{T} +\frac{P_2 A(-\delta)}{T}$$
$$dS_U = \frac{P_1 - P_2}{T} A\delta$$
 

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A piston between two ideal gases is used in scientific experiments to study the behavior of gases under different conditions. It allows for the manipulation of pressure, volume, and temperature, which are important variables in understanding the properties of gases.

4. What factors affect the behavior of a piston between two ideal gases?

The behavior of a piston between two ideal gases is affected by several factors, including the temperature, volume, and number of moles of gas present. Changes in these variables can cause the gases to expand or contract, resulting in changes in pressure.

5. How is the ideal gas law applied to a piston between two ideal gases?

The ideal gas law, PV = nRT, is applied to a piston between two ideal gases by using the pressure and volume of the gases to calculate the number of moles (n) present. This allows for the determination of other properties, such as temperature, within the system.

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