Defination of Carnot Engine : 

Carnot engine is a theoretical heat engine that sets the upper limit of efficiency for any engine operating between two thermal reservoirs. Its cycle consists of four reversible processes—two isothermal and two adiabatic—and its efficiency depends only on the temperatures of the hot and cold reservoirs. This post derives the Carnot efficiency while explicitly accounting for positive (expansion) and negative (compression) work.

Carnot cycle overview : 

The working substance (ideal gas) undergoes the following four reversible processes in sequence:

Work and heat in each process (with signs) : 

1. Isothermal expansion (A → B at ($T_1$))

For an ideal gas undergoing isothermal expansion, internal energy does not change, so heat absorbed equals work done by the gas.

- Work done (positive):  

$W_{AB} = n R T_1 \ln\!\left(\frac{V_2}{V_1}\right)$

$Q_{1} = +n R T_1 \ln\!\left(\frac{V_2}{V_1}\right)$

2. Adiabatic expansion (B → C)

No heat exchange; temperature drops from ($T_2$) to ($T_1$). For an ideal gas: Q =0

- Work done (positive):  

$W_{BC} = +\frac{n R}{\gamma - 1}\,\left(T_1 - T_2\right)$

3. Isothermal compression (C → D at ($T_{2}$))

Gas is compressed at constant ($T_2$); heat is rejected to the cold reservoir.

$W_{CD} = n R T_2 \ln\!\left(\frac{V_4}{V_3}\right)$

$Q_{2} = -n R T_2 \ln\!\left(\frac{V_4}{V_3}\right)$

4. Adiabatic compression (D → A)

No heat exchange; temperature rises from ($T_{2}$) back to ($T_{1}$). Work is done on the gas (negative).

- Heat exchange:  

$Q = 0$

Work done : 

$W_{DA} = \frac{n R}{\gamma - 1}\,\left(T_2 - T_1\right)$

Net work of the cycle (including negative work)

Sum the work over all four processes:</p>

- Total work:  

$W{\text{net}} = W_{AB} + W_{BC} + W_{CD} + W_{DA}$

Using the expressions above and noting that adiabatic works cancel in magnitude:

- Cancellation of adiabatic works:  

$W_{BC} + W_{DA} = 0$

Therefore, the net work reduces to the difference between isothermal works:

- Net work from isothermals:  

$W{\text{net}} = n R T_{1} \ln\!\left(\frac{V_2}{V_1}\right)+n R T_{2} \ln\!\left(\frac{V_4}{V_3}\right)$

$/ln\!\left(\frac{2}{1}\right) = - ln\!\left(\frac{1}{2}\right)$

$W{\text{net}} = n R T_{1} \ln\!\left(\frac{V_2}{V_1}\right)-n R T_{2} \ln\!\left(\frac{V_3}{V_4}\right)$

Using the adiabatic volume relation : 

Since points B and C lie on the same adiabatic curve

$T_1 V_2^{\gamma-1} = T_2 V_3^{\gamma-1}$

or

$\frac{T_1}{T_2} = \left(\frac{V_3}{V_2}\right)^{\gamma-1}$

Also points D and A lie on the same adiabatic curve

$T_1 V_1^{\gamma-1} = T_2 V_4^{\gamma-1}$

or

$\frac{T_1}{T_2} = \left(\frac{V_4}{V_1}\right)^{\gamma-1}$

From equations : 

$\left(\frac{V_3}{V_2}\right)^{\gamma-1}=\left(\frac{V_4}{V_1}\right)^{\gamma-1}$

or

$\frac{V_3}{V_2} = \frac{V_4}{V_1}$

Or

$\frac{V_3}{V_4} = \frac{V_2}{V_1}$

$\ln\!\left(\frac{V_3}{V_4}\right)=\ln\!\left(\frac{V_2}{V_1}\right)$

Hence equation becomes

$(\frac{V_2}{V_1} = \frac{V_3}{V_4})$:

- Compact form:  

$W{\text{net}} = n R \left(T_1 - T_2\right)\,\ln\!\left(\frac{V_2}{V_1}\right)$

Efficiency of the carnot's Engine : Efficiency of carnot's Engine is defined as the ratio of the work done per cycle by the engine to the amount of heat absorbed per cycle by the working substance of the engine from a hot reservoir (source).

$\eta = \frac{W{\text{net}}}{\text{Heat absorbed from source (input)}}$

$\eta = \frac{W_{net}}{Q_1}$

$\eta = \frac{n R \left(T_1 - T_2\right)\,\ln\!\left(\frac{V_2}{V_1}\right)}{n R T_1 \ln\!\left(\frac{V_2}{V_1}\right)}$

$\eta = \frac{T_1 - T_2}{T_1}$

$\eta = 1-\frac{T_2}{T_1}$

Discussion

1. The efficiency of Carnot engine depends upon the temperatures of the source and the sink.

2. The efficiency of Carnot engine is independent of the nature of the working substance.

3. The efficiency of Carnot engine becomes $100\%$ if the temperature of the sink $T_2 = 0\,\text{K}$. Since absolute zero cannot be achieved, a heat engine with $100\%$ efficiency is impossible in practice.