Welcome to Lecture 98 of our comprehensive JEE Physics Master Series. In this lecture, we examine the theoretical benchmark of thermodynamic efficiency: the Carnot Cycle. Conceived by Nicolas Léonard Sadi Carnot in 1824, this reversible quasi-static cycle establishes the fundamental upper limit on the efficiency of any heat engine operating between two thermal reservoirs.
1. Anatomy of the Carnot Cycle
A Carnot engine consists of an ideal working substance (typically $n$ moles of an ideal gas) subjected to a four-stage cyclic process executed between a hot reservoir at absolute temperature $T_H$ and a cold reservoir at absolute temperature $T_C$ ($T_H > T_C$).
The four sequential reversible processes represented on a $P$-$V$ indicator diagram are:
Step 1: Reversible Isothermal Expansion ($A \to B$)
The working gas is in thermal contact with the source reservoir at $T_H$. It expands isothermally and reversibly from $(P_A, V_A)$ to $(P_B, V_B)$ by absorbing heat $Q_H$:
$$Q_H = W_{AB} = n R T_H \ln\left(\frac{V_B}{V_A}\right)$$
Since temperature remains constant ($\Delta U_{AB} = 0$), all absorbed thermal energy is converted directly into quasi-static boundary work.
Step 2: Reversible Adiabatic Expansion ($B \to C$)
The cylinder is placed onto a perfectly insulating stand ($dQ = 0$). The gas continues to expand reversibly from $V_B$ to $V_C$ while cooling from $T_H$ to $T_C$:
$$W_{BC} = -\Delta U_{BC} = \frac{n R (T_H – T_C)}{\gamma – 1}$$
Governing Poisson relation: $T_H V_B^{\gamma-1} = T_C V_C^{\gamma-1}$.
Step 3: Reversible Isothermal Compression ($C \to D$)
The cylinder is placed in thermal contact with the cold sink at $T_C$. Compressive work is performed on the gas from $V_C$ to $V_D$, releasing heat $Q_C$ into the sink:
$$Q_C = -W_{CD} = n R T_C \ln\left(\frac{V_C}{V_D}\right)$$
Step 4: Reversible Adiabatic Compression ($D \to A$)
The cylinder is returned to the insulating stand ($dQ = 0$). The gas is compressed from $V_D$ back to its initial state $(P_A, V_A)$, raising its temperature from $T_C$ back to $T_H$:
$$W_{DA} = -\Delta U_{DA} = -\frac{n R (T_H – T_C)}{\gamma – 1}$$
Governing Poisson relation: $T_C V_D^{\gamma-1} = T_H V_A^{\gamma-1}$.
2. Rigorous Efficiency Derivation
The net work performed per complete thermodynamic cycle is given by the area enclosed by the cycle on the $P$-$V$ indicator diagram:
$$W_{\text{net}} = W_{AB} + W_{BC} + W_{CD} + W_{DA}$$
Observing that the two adiabatic expansion and compression steps involve identical magnitude work terms with opposite signs ($W_{BC} + W_{DA} = 0$), the net work simplifies strictly to the difference in heat exchanges:
$$W_{\text{net}} = Q_H – Q_C = n R T_H \ln\left(\frac{V_B}{V_A}\right) – n R T_C \ln\left(\frac{V_C}{V_D}\right)$$
Dividing the two adiabatic relations:
$$\frac{T_H V_B^{\gamma-1}}{T_H V_A^{\gamma-1}} = \frac{T_C V_C^{\gamma-1}}{T_C V_D^{\gamma-1}} \implies \left(\frac{V_B}{V_A}\right)^{\gamma-1} = \left(\frac{V_C}{V_D}\right)^{\gamma-1} \implies \frac{V_B}{V_A} = \frac{V_C}{V_D}$$
Consequently, the thermal efficiency $\eta$ is derived as:
$$\eta = \frac{W_{\text{net}}}{Q_H} = \frac{Q_H – Q_C}{Q_H} = 1 – \frac{Q_C}{Q_H} = 1 – \frac{T_C}{T_H}$$
Carnot’s Fundamental Theorem:
1. No engine operating between two given heat reservoirs can be more efficient than a reversible Carnot engine operating between the same reservoirs.
2. All reversible engines operating between the same two temperatures have the identical thermal efficiency, independent of the nature of the working substance.
3. The Reversed Carnot Engine (Refrigerator & Heat Pump)
When run in reverse (counter-clockwise on $P$-$V$ space), external work $W$ extracts heat $Q_C$ from the cold reservoir and discharges $Q_H$ into the hot reservoir.
Coefficient of Performance (COP) for Refrigerator ($\beta$):
$$\beta = \frac{\text{Desired Cooling Output}}{\text{Work Input}} = \frac{Q_C}{W} = \frac{T_C}{T_H – T_C} = \frac{1 – \eta}{\eta}$$
Coefficient of Performance for Heat Pump ($\beta_{\text{HP}}$):
$$\beta_{\text{HP}} = \frac{Q_H}{W} = \frac{T_H}{T_H – T_C} = \beta + 1 = \frac{1}{\eta}$$
4. High-Yield JEE Benchmark Problems
Problem 1 (Intermediate – JEE Main):
A Carnot engine has an efficiency of $40\%$ when its sink is at $300\text{ K}$. By how much must the temperature of the source be increased so that its efficiency becomes $50\%$?
Solution:
Initially, $\eta_1 = 1 – \frac{T_C}{T_{H1}} = 0.40 \implies \frac{300}{T_{H1}} = 0.60 \implies T_{H1} = 500\text{ K}$.
For the new efficiency $\eta_2 = 0.50$:
$1 – \frac{300}{T_{H2}} = 0.50 \implies \frac{300}{T_{H2}} = 0.50 \implies T_{H2} = 600\text{ K}$.
Therefore, the required temperature increase $\Delta T_H = 600 – 500 = \mathbf{100\text{ K}}$.
Problem 2 (Advanced – JEE Advanced):
Two Carnot engines $A$ and $B$ are operated in series. Engine $A$ absorbs heat at $T_1 = 900\text{ K}$ and rejects heat to an intermediate reservoir at temperature $T_2$. Engine $B$ absorbs this heat rejected by $A$ and rejects to a cold reservoir at $T_3 = 400\text{ K}$. Determine $T_2$ if both engines perform equal amounts of work ($W_A = W_B$).
Solution:
For engine $A$: $W_A = Q_1 – Q_2 = Q_1 \left(1 – \frac{T_2}{T_1}\right)$.
Since $Q_2 = Q_1 \left(\frac{T_2}{T_1}\right)$, for engine $B$: $W_B = Q_2 – Q_3 = Q_2 \left(1 – \frac{T_3}{T_2}\right) = Q_1 \left(\frac{T_2}{T_1}\right)\left(1 – \frac{T_3}{T_2}\right) = Q_1 \left(\frac{T_2 – T_3}{T_1}\right)$.
Equating $W_A = W_B$:
$$1 – \frac{T_2}{T_1} = \frac{T_2 – T_3}{T_1} \implies T_1 – T_2 = T_2 – T_3 \implies T_2 = \frac{T_1 + T_3}{2}$$
Substituting numerical values:
$$T_2 = \frac{900 + 400}{2} = \mathbf{650\text{ K}}$$