Solve Thermodynamics Problem: Q12, W12, W21 & Entropy Change

[ashkash]

Homework Statement

A closed system contains 3 kg of R-134a at 10 °C and 90 kPa (state 1), and undergoes an internally reversible, isothermal process until the volume of the R-134a reaches 0.056 m3 (state 2). The system returns to state 1 by an irreversible process and the heat transfer for the return process is +3.2 kJ. Determine Q12, W12, W21 and the entropy change for each process.

Homework Equations

The Attempt at a Solution

I do not know how to go about doing this. Any help will be appreciated. thanks.

 

[teknodude]

Start out by drawing an T-s, P-V, and T-V diagram, to better see the cycle

 

[piercas]

I would begin by first understanding the given information and identifying any relevant equations or concepts from thermodynamics. From the problem statement, we are given the initial state (state 1) of the closed system, which contains 3 kg of R-134a at 10 °C and 90 kPa. We are also told that the system undergoes an internally reversible, isothermal process until it reaches state 2, with a volume of 0.056 m3. This process is followed by an irreversible process that returns the system to state 1, with a heat transfer of +3.2 kJ.

To solve this problem, we can use the first law of thermodynamics, which states that the change in internal energy of a closed system is equal to the heat transfer into the system minus the work done by the system. Mathematically, this can be represented as ΔU = Q - W.

For the first process, we know that it is isothermal (meaning the temperature remains constant) and internally reversible. Therefore, we can use the equation for isothermal processes, which is Q = W. This means that the heat transfer (Q12) and the work done (W12) are equal in magnitude but opposite in sign. We can also use the ideal gas law to calculate the final pressure of the system at state 2, since the temperature and volume are given. Once we have the final pressure, we can calculate the work done by the system using the equation W = -∫PdV, where the negative sign indicates work done by the system. This will give us the value for W12.

For the second process, we are told that it is irreversible and that the heat transfer is +3.2 kJ. Therefore, we know that Q21 = +3.2 kJ. To calculate the work done (W21), we can use the same equation as before, but this time we need to know the final pressure at state 1. This can be found using the ideal gas law again, but with the initial temperature and volume given.

To calculate the entropy change for each process, we can use the equation ΔS = Q/T, where Q is the heat transfer and T is the temperature of the process. For the first process, since it is isothermal, the temperature remains constant and we can use the value given in the problem statement. For the second process, we need to calculate