Entropy change in a Carnot cycle

Question

The Carnot cycle is a reversible cycle, which means that by the end of the cycle the system is in it's initial state, which is characterized by the same entropy value. Therefore we can say that for a reversible cycle the total entropy change is zero.

The Carnot cycle itself is comprised by 4 rev. processes:

1. Reversible Isotherm Expansion
2. Reversible adiabatic expansion
3. Reversible isotherm compression.
4. Reversible adiabatic compression.

Let's consider one of these 4 processes i.e Reversible Isotherm Expansion. In this process we have 3 different entities which represent the total system : The hot reservoir, the machine and the cold reservoir.

Because the process is reversible that means that the total entropy change for this simple step should zero. The entities that do take part in this step are the hot reservoir and the machine. Therefore the total entropy should be zero, because the process is reversible.

The hot reservoir loses heat $-Q_H$ and the entropy change of it is $\Delta S_H=\frac{-Q_H}{T_H}$

Now regarding the machine/gas/motor, because we have isotherm expansion $\Delta S_{motor} = nR ln (\frac{V_2}{V_1})>0$.

Does it mean that the total entropy $\Delta S_{total}=\Delta S_{motor} + \Delta S_H=0$

Answer 1

Yes.

Using the first law, and assuming that the working substance is an ideal gas, so that isothermal implies no change in internal energy,

Heat input to gas = Work done by gas

So

$$Q_{in} = nRT_H \ln\left(\frac{V_2}{V_1}\right)$$

So for the gas, $\Delta S = \frac{Q_{in}}{T_H}=nR \ln\left(\frac{V_2}{V_1}\right)$

Answer 2

Let's consider one of these 4 processes i.e Reversible Isotherm Expansion. In this process we have 3 different entities which represent the total system : The hot reservoir, the machine and the cold reservoir.

The reversible isothermal expansion only involves the heat engine and the hot reservoir, not the cold reservoir.

Therefore the total entropy should be zero, because the process is reversible.

Correct.

The hot reservoir loses heat $-Q_H$ and the entropy change of it is $\Delta S_H=\frac{-Q_H}{T_H}$

Correct. And the entropy change of the motor is $\Delta S_{motor}=\frac {+Q_H}{T_H}$

Now regarding the machine/gas/motor, because we have isotherm expansion $\Delta S_{motor} = nR ln (\frac{V_2}{V_1})>0$.

Yes, assuming we are dealing with an ideal case because only then does internal energy depend only on temperature change. That means from the first laws $\Delta U=0$, $Q=W$, and then since

$$W=nRT_{H}\ln\frac{V_2}{V_1}=+Q_H$$

and

$$\Delta S_{motor}=\frac{+Q_H}{T_H}$$

then

$$\Delta S_{motor}=nR\ln\frac{V_2}{V_1}$$

Hope this helps.