Question

The entropy of the working fluid of the ideal Carnot cycle (increases, decreases, remains the same) during the isothermal heat rejection process. (Explain the reasoning behind your answer selection!)

Answer 1

Decreases

Step-by-step explanation:

The Entropy change for a reversible process is;

ΔS = nRIn(V_b/V_a)

Now, applying this formula to an isothermal expansion in an irreversible process, it will have same initial and final states as we got for the reversible expansion process. Now, due to the fact that entropy is a state function, the change in entropy of the system will be the same as we got for the equation in reversible process and is given by the same formula.

In both the reversible and irreversible process, the surroundings of which they differ by the entropy there are usually at a constant temperature, T in a way that we have the change jn entropy relationship as;

ΔS_sur = −Q/T

Where;

- the negative sign is used since the heat transferred to the surroundings is equal in magnitude and opposite in sign to the heat, Q, transferred to the system.

Thus, it is clear that the change in the entropy decreases due to the negative sign.