**Problem Statement:**

A mole of a monatomic ideal gas performs the Carnot cycle illustrated in the figure below. The gas volume in state A is V_{A} = 2 10^{-3} m^{3}. The gas volume in state B is V_{B} = 5 10^{-3} m^{3}. After an adiabatic expansion, the gas volume is V_{C} = 20 10^{-3} m^{3}. The hot thermal reservoir is at a temperature of 500K. Calculate:

- The cold reservoir temperature.
- The cycle efficiency.
- The change in entropy of the gas for each step as well as the total entropy change over the entire cycle.
- The entropy change of both heat reservoirs.
- The total entropy change of the universe over each cycle.
- If a heat engine had 20% of the efficiency of a Carnot heat engine operating between the same thermal reservoirs, determine the change in entropy of its working fluid, that of both thermal reservoirs as well as the change in entropy of the universe over each cycle for the same amount of heat absorbed from the hot thermal reservoir.

Express the results in SI units.

__Givens__: R = 8.31 J/mol K; C_{V} = (3/2)R; γ = 1.67

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**Solution:**

The Carnot cycle consists of these four reversible processes:

**A-B** Isothermal expansion

**B-C** Adiabatic expansion

**C-D** Isothermal compression

**D-A** Adiabatic compression

We will make use of the adiabatic equation for the B-C process to determine the temperature of the cold thermal reservoir temperature (T_{2} in the figure):

The Carnot heat engine efficiency is given by:

The entropy change of the gas during the adiabatic processes of the Carnot cycle is zero (because Q_{R} = 0).

The entropy change of the gas during the isothermal expansion AB is given by:

We now calculate the volume in state D using the adiabatic equation applied to states D and A:

The entropy change of the gas during the isothermal compression CD is given by:

The total entropy change of the gas over the entire cycle is equal to the sum of the entropy changes for the different processes of the cycle:

And it is equal to zero because **entropy is a state function **and in a cycle, the initial and final states are the same.

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In a Carnot cycle, the ideal gas **absorbs heat **(Q_{1})** from the hot thermal reservoir** during the isothermal expansion. So the **hot thermal reservoir discharges the same amount of heat**, -Q_{1}, at constant temperature T_{1}. Remember that a **thermal reservoir is a system that can exchange heat indefinitely while keeping its temperature constant**.

The following figure represents the heat flow of a Carnot heat engine:

The isothermal compression of the Carnot cycle takes place while in contact with the cold thermal reservoir at temperature T_{2}. During this process, the heat Q_{2} is removed from gas and absorbed by the cold thermal reservoir (thus Q_{2} is positive from the *point of view* of the cold bath).

The entropy change between any two states A and B is defined:

And, as the temperature of the thermal reservoir does not change, we can pull T out of the integral. As a consequence, the **entropy changes of the hot and cold thermal reservoirs** are respectively given by:

As you can see, the entropy change of the hot thermal reservoir is minus the entropy change of the gas during the isothermal expansion; and the entropy change of the cold thermal reservoir is minus the entropy change of the gas during the isothermal compression So:

The entropy change of the universe is **the sum of the entropy change of both the thermodynamic system** (in this case the Carnot heat engine) **and its surroundings** (in this case the thermal reservoirs).

Adding all these quantities together we can see that **the entropy change of the universe is zero because the Carnot heat engine operates reversibly**.

If a heat engine had 20% of the efficiency of a Carnot heat engine operating between the same two thermal reservoirs at temperatures T_{1} y T_{2}, it would operate irreversible, because **all reversible heat engines operating between the same two thermal reservoirs have the same thermal efficiency**.

The efficiency of this engine would be:

Where W is the net work output of the engine and Q_{1} the heat absorbed by its working fluid from the hot thermal reservoir.

The heat absorbed from the hot thermal reservoir is the same as in the case of the Carnot heat engine:

And, since the efficiency of this engine is 20% of that of Carnot’s engine we have:

We can now deduce the heat Q_{2} discharged to the cold thermal reservoir:

And after substituting we have:

**The entropy change of the working fluid of this heat engine is zero**, because entropy is a state function and therefore, its change depends only on the initial and final states and in a cycle, they are the same.

The entropy change of the thermal reservoirs is given by:

The entropy change of the universe is the sum of these three:

**The entropy change of the universe is positive because the heat engine operates irreversibly**.