# Impossible refrigerator (thermodynamics and PV diagrams)

DrClaude
Mentor
Well, what else can I use...?
I aim aiming at something simpler. What can we say about the required conditions on Tc, Th, and Tws in the different legs of the cycle?

Just to add some more..The problem cannot lie in one of the legs A, B, C or D alone since all those parts also exist in a rectangular heat engine (which supposedly does not have this problem). So, it must be a combination of the legs causing the problem.
In a sense, yes. To give an additional hint: there is a fundamental difference in the behavior of Tws for heat engines and refrigerators.

Avatrin
I aim aiming at something simpler. What can we say about the required conditions on Tc, Th, and Tws in the different legs of the cycle?

In a sense, yes. To give an additional hint: there is a fundamental difference in the behavior of Tws for heat engines and refrigerators.
Okay, well, since I am still working with ideal gasses: dT ∝ dU, and dU = (f/2)Δ(PV) (I've used the ideal gas law and the equipartition theorem)
So, dT ∝Δ(PV) since proportionality is an equivalence relation.

This is true for both heat engines and refrigerators. I do not see the fundamentally different behaviour of the temperature of the working substance.

I'll try something more fundamental: Tc < Th..

For net heat to flow from the cold reservoir to the working substance, the working substance has to be colder than the reservoir. Then work is done to heat the working substance, and when Tws>Th, the system is brought into contact with the hot reservoir and heat flows to the hot reservoir. Then work is done again to cool the system down so that the cycle can start anew.
At least, that's how I understood refrigerators before. Now, I am no longer certain, but lets roll with it. Assuming my intuition is/was correct:
D is when pressure falls. So, it is in contact with the hot reservoir. B, when pressure rises, it is in contact with the cold reservoir. The work that is done in A lowers its volume and thus its temperature goes up. At this point the temperature is at its highest. C is when work is done to expand the system.
So, again, I do not see the problem.

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DrClaude
Mentor
I think you're almost there

I'll try something more fundamental: Tc < Th..

For net heat to flow from the cold reservoir to the working substance, the working substance has to be colder than the reservoir. Then work is done to heat the working substance, and when Tws>Th, the system is brought into contact with the hot reservoir and heat flows to the hot reservoir. Then work is done again to cool the system down so that the cycle can start anew.
At least, that's how I understood refrigerators before. Now, I am no longer certain, but lets roll with it. Assuming my intuition is/was correct:
D is when pressure falls. So, it is in contact with the hot reservoir. B, when pressure rises, it is in contact with the cold reservoir. The work that is done in A lowers its volume and thus its temperature goes up. At this point the temperature is at its highest. C is when work is done to expand the system.
And what about heat in leg C? And how does that fit in with what you said (correctly) happens in leg D?

Avatrin
I think you're almost there

And what about heat in leg C? And how does that fit in with what you said (correctly) happens in leg D?
I can use dT ∝Δ(PV)... So, since the system is expanding in volume while pressure remains the same, temperature goes up. So, in legs A and C the opposite is happening of what I wrote should be happening. After coming into contact with the hot reservoir, the temperature of the working substance keep rising.

Andrew Mason
Homework Helper
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Well, the website is not alone in claiming it is not possible. The textbook I am using is specifically asking me to explain why it is impossible to use the cycle in a refrigerator.
It is probably impossible to make this cycle reversible. But the question says it is a reversible heat engine. That means that the working substance is in equilibrium with the surroundings at all times. If so, the cycle can be reversed with an infinitesimal change in conditions. If it is reversed, work will be applied and the heat flow will go in the reverse direction. How is that not a refrigerator or heat pump?

AM

It is probably impossible to make this cycle reversible. But the question says it is a reversible heat engine. That means that the working substance is in equilibrium with the surroundings at all times. If so, the cycle can be reversed with an infinitesimal change in conditions. If it is reversed, work will be applied and the heat flow will go in the reverse direction. How is that not a refrigerator or heat pump?

AM
The question doesn't say it is a reversible heat engine. All it says is that we have a rectangular cycle on a PV diagram, and we are trying run it counterclockwise to make a refrigerator (unlike the image where it's running in clockwise direction).

Andrew Mason
Homework Helper
The question doesn't say it is a reversible heat engine. All it says is that we have a rectangular cycle on a PV diagram, and we are trying run it counterclockwise to make a refrigerator (unlike the image where it's running in clockwise direction).
Sorry. I was confusing it with this thread that I am also posting on: https://www.physicsforums.com/threads/carnot-cycle.784322/

The question does not say that the surroundings are at constant temperature. A stirling cycle has constant volume expansion and compression parts and can be made, at least in theory, reversible. It uses a 'regenerator' to change the temperature of the surroundings that the working substance is in thermal contact with. I don't see a material difference between the Stirling cycle and this rectangular one. So, if it could be a reversible heat engine cycle, it has to be able to operate in the reverse direction and use work to cause heat flow in the opposite direction.

AM

DrClaude
Mentor
I can use dT ∝Δ(PV)... So, since the system is expanding in volume while pressure remains the same, temperature goes up. So, in legs A and C the opposite is happening of what I wrote should be happening. After coming into contact with the hot reservoir, the temperature of the working substance keep rising.
We're getting confused here. Lets take a step back. Could you please summarize the results you have by answering:

What is the sign of Q in each leg?

What is the relation between the sign of Q and whether it is in contact with the hot or cold reservoir?

What is the relation between temperature of the working substance and the hot or cold reservoir during these exchanges?

We're getting confused here. Lets take a step back. Could you please summarize the results you have by answering:

What is the sign of Q in each leg?

What is the relation between the sign of Q and whether it is in contact with the hot or cold reservoir?

What is the relation between temperature of the working substance and the hot or cold reservoir during these exchanges?
Well, I do stand by my calculations in post #20 (I am quite sure my calculations for A and C in post #11 were wrong). So,
$$Q_A = dU_A + P_A(V_D - V_B)<0$$
$$Q_B = dU_B = \frac{f}{2} V_B (P_A - P_C)>0$$
$$Q_C = dU_C + P_C(V_B - V_D)>0$$
$$Q_D = dU_D = \frac{f}{2} V_D (P_C - P_A)<0$$

If Q is positive, the internal energy of the working substance is rising. So, it must be in contact with the cold reservoir when Q is positive since this is supposed to be a refrigerator. The system is in contact with the hot reservoir when it Q is negative.

If heat is going from the cold reservoir to the system, the system must be colder than the cold reservoir... .... Not necessarily...? But, I did assume that in my previous answer.. C = Q/dT.. So Q = CdT. Since we are not working with something like black holes, C must be positive. A positive Q for the working substance means a negative Q for the reservoir; Temperature must fall there (although, since it is a reservoir, it is negligible). Since they are approaching equilibrium by coming into contact, the first sentence in this paragraph must be correct.

DrClaude
Mentor
Well, I do stand by my calculations in post #20 (I am quite sure my calculations for A and C in post #11 were wrong). So,
$$Q_A = dU_A + P_A(V_D - V_B)<0$$
$$Q_B = dU_B = \frac{f}{2} V_B (P_A - P_C)>0$$
$$Q_C = dU_C + P_C(V_B - V_D)>0$$
$$Q_D = dU_D = \frac{f}{2} V_D (P_C - P_A)<0$$

If Q is positive, the internal energy of the working substance is rising. So, it must be in contact with the cold reservoir when Q is positive since this is supposed to be a refrigerator. The system is in contact with the hot reservoir when it Q is negative.

If heat is going from the cold reservoir to the system, the system must be colder than the cold reservoir... .... Not necessarily...? But, I did assume that in my previous answer.. C = Q/dT.. So Q = CdT. Since we are not working with something like black holes, C must be positive. A positive Q for the working substance means a negative Q for the reservoir; Temperature must fall there (although, since it is a reservoir, it is negligible). Since they are approaching equilibrium by coming into contact, the first sentence in this paragraph must be correct.
That's pretty much correct. Then take for instance the junction between legs D and C. In leg D, as you calculated, Q < 0, so heat is going from the working substance to the hot reservoir, so Tws > Th. Then in leg C Q > 0, so Tws < Tc. What happens then at point 4 on the diagram?

That's pretty much correct. Then take for instance the junction between legs D and C. In leg D, as you calculated, Q < 0, so heat is going from the working substance to the hot reservoir, so Tws > Th. Then in leg C Q > 0, so Tws < Tc. What happens then at point 4 on the diagram?
Well, if we graph Tws, there will either be an impossible discontinuity, or Th<Tws<Tc, which is a contradiction.

DrClaude
Mentor
Well, if we graph Tws, there will either be an impossible discontinuity,
But Tws is a state variable, so it has a defined value at each point in the cycle.

or Th<Tws<Tc, which is a contradiction.
And that's the solution.

To summarize, any cycle can be made into a heat engine, as we only need to use a hot reservoir with Th > max(Tws) and a cold reservoir with Tc < min(Tws). The problem with a refrigerator is that it has to cycle between Tws < Tc and Tws > Th, which is not possible for all cycles.