How Do Thermodynamic Potentials Apply in a Two-Cylinder System with a Heat Pump?

[Feynmanfan]

Dear friends,

I’m having trouble with this thermodynamic problem. I apologise if you don’t understand my poor English (I’m writing to you from Spain!)

We’ve got two cylinders (1 is adiabatically isolated from 2 but not from the outside, where Pressure P and Tº are constant) and cylinder 1 can move inside cylinder 2. Cylinder 2 is isolated from the outside but it is connected to a heat pump
Suppose both cylinders contain IDEAL GAS.
Everything is in equilibrium but now we lower Tº and make it (T-Gamma).Therefore Q1 heat will be delivered from cylinder 1. As said, Cylinder 2 is connected to a Carnot HEAT PUMP (that is maximum efficiency) and we use Q1 to heat up cylinder 2.

I need to calculate the final volume of the gas in cylinder 2, entropy change in both systems and many other things I don’t want to bore you with.

Does this problem have to do with thermodynamic potentials? I’d be very grateful if you could help me.

 

[wais]

Dear friend,

Thank you for reaching out for help with your thermodynamic problem. I will do my best to assist you, despite any language barriers.

From the given information, it seems like you are dealing with a complex system involving two cylinders, an ideal gas, and a heat pump. This type of problem can indeed involve thermodynamic potentials, specifically the Helmholtz free energy and the Gibbs free energy, which are useful for analyzing systems that are not at constant temperature and pressure.

To start, I suggest setting up a diagram of the system, including the two cylinders and their connections, as well as the heat pump. From there, you can apply the first and second laws of thermodynamics to determine the final volume of gas in cylinder 2 and the entropy changes in both systems. It may also be helpful to consider the properties of an ideal gas, such as the ideal gas law and the specific heat capacity.

Additionally, since you mentioned that cylinder 2 is connected to a Carnot heat pump, it may be useful to consider the Carnot cycle and its efficiency in your calculations. Remember that the Carnot cycle is a reversible process, and the efficiency is equal to the ratio of the temperature difference between the hot and cold reservoirs to the temperature of the hot reservoir.

I hope this information helps you in solving your thermodynamic problem. If you need any further assistance, please do not hesitate to reach out. Best of luck!


Your friend in thermodynamics