Welcome to Lecture 99 of our comprehensive JEE Physics Master Series. In this lecture, we formulate the mathematical foundation of thermodynamic irreversibility: the concept of Entropy ($S$) and the Clausius Inequality. While the First Law of Thermodynamics guarantees energy conservation, it imposes zero constraint on the directional flow of natural processes. The Second Law, formalized through the state function of entropy, dictates the arrow of time in physical systems.
1. Mathematical Formulation: Clausius Inequality
Consider an arbitrary cyclic process executed by a thermodynamic system exchanging thermal energy with any number of thermal reservoirs. The Clausius Theorem states that the cyclic integral of infinitesimal heat exchange $dQ$ divided by the absolute boundary temperature $T$ satisfies:
$$\oint \frac{dQ}{T} \le 0$$
where:
- $\oint \frac{dQ}{T} = 0$ for a strictly reversible cycle.
- $\oint \frac{dQ}{T} < 0$ for any irreversible (real) cycle.
- $\oint \frac{dQ}{T} > 0$ represents a physically impossible process (violating the Second Law).
2. Entropy as a State Function
Since $\oint_{\text{rev}} \frac{dQ_{\text{rev}}}{T} = 0$, the quantity $\frac{dQ_{\text{rev}}}{T}$ is an exact differential ($dS$). Entropy $S$ is therefore a rigorous state variable independent of the path taken:
$$\Delta S = S_B – S_A = \int_A^B \frac{dQ_{\text{rev}}}{T}$$
For an irreversible process transitioning between identical end states $A$ and $B$, $\Delta S$ remains identical, but the actual heat transferred satisfies:
$$\Delta S \ge \int_A^B \frac{dQ_{\text{irrev}}}{T} \implies dS \ge \frac{dQ}{T}$$
Principle of Increase of Entropy: For an isolated thermodynamic system ($dQ = 0$), $\Delta S_{\text{isolated}} \ge 0$. Every spontaneous natural process proceeds in the direction of increasing total entropy of the universe ($\Delta S_{\text{universe}} = \Delta S_{\text{system}} + \Delta S_{\text{surroundings}} \ge 0$).
3. Entropy Change of an Ideal Gas
From the First Law for a reversible process, $dQ = dU + P dV = n C_v dT + \frac{n R T}{V} dV$. Dividing by $T$ and integrating from $(T_1, V_1)$ to $(T_2, V_2)$:
$$\Delta S = n C_v \ln\left(\frac{T_2}{T_1}\right) + n R \ln\left(\frac{V_2}{V_1}\right)$$
Alternatively, expressing in terms of pressure and temperature ($V = \frac{n R T}{P}$):
$$\Delta S = n C_p \ln\left(\frac{T_2}{T_1}\right) – n R \ln\left(\frac{P_2}{P_1}\right)$$
4. Solved Examples
Example 1 (Foundational – JEE Main):
Calculate the change in entropy when $1\text{ kg}$ of ice at $0^\circ\text{C}$ is completely melted into water at $0^\circ\text{C}$. Latent heat of fusion of ice $L_f = 3.36 \times 10^5\text{ J/kg}$.
Solution:
The phase transition occurs isothermally and reversibly at $T = 273.15\text{ K}$.
Heat absorbed: $Q = m L_f = 1 \times 3.36 \times 10^5\text{ J} = 3.36 \times 10^5\text{ J}$.
$$\Delta S = \frac{Q}{T} = \frac{3.36 \times 10^5}{273.15} \approx \mathbf{1230.1\text{ J/K}}$$
Since heat is absorbed by the system, $\Delta S > 0$.
Example 2 (Multi-Step Thermal Mixing – JEE Main/Advanced):
An insulated vessel contains $2\text{ kg}$ of liquid water at $300\text{ K}$, which is mixed with $2\text{ kg}$ of liquid water at $360\text{ K}$. Neglect the heat capacity of the vessel. Specific heat of water $c = 4200\text{ J/(kg}\cdot\text{K)}$. Find $\Delta S_{\text{universe}}$.
Solution:
By energy conservation, the final equilibrium temperature $T_f$ is:
$$m c (T_f – 300) = m c (360 – T_f) \implies T_f = \frac{300 + 360}{2} = 330\text{ K}$$
For water heating from $300\text{ K}$ to $330\text{ K}$:
$$\Delta S_1 = m c \ln\left(\frac{330}{300}\right) = 2 \times 4200 \times \ln(1.10) \approx 8400 \times 0.09531 = +800.6\text{ J/K}$$
For water cooling from $360\text{ K}$ to $330\text{ K}$:
$$\Delta S_2 = m c \ln\left(\frac{330}{360}\right) = 2 \times 4200 \times \ln(0.9167) \approx 8400 \times (-0.08701) = -730.9\text{ J/K}$$
$$\Delta S_{\text{universe}} = \Delta S_1 + \Delta S_2 = 800.6 – 730.9 = \mathbf{+69.7\text{ J/K}} > 0$$
The positive entropy production confirms the fundamental irreversibility of heat conduction across finite temperature gradients.
Example 3 (Ideal Gas Processes – JEE Advanced):
One mole of a monoatomic ideal gas ($\gamma = 5/3$) undergoes a free expansion from volume $V_0$ to $3V_0$ inside an insulated container. Find the change in entropy of (a) the gas, (b) the surroundings, and (c) the universe.
Solution:
(a) In free expansion into vacuum, $W = 0$ and $Q = 0$. By the First Law, $\Delta U = 0 \implies T_f = T_i = T_0$.
Although the process is irreversible, entropy is a state function. To calculate $\Delta S_{\text{gas}}$, choose a reversible isothermal expansion between the same initial and final states:
$$\Delta S_{\text{gas}} = n C_v \ln\left(\frac{T_0}{T_0}\right) + n R \ln\left(\frac{3V_0}{V_0}\right) = 1 \times R \ln(3) = R \ln 3 = \mathbf{1.099 R}$$
(b) For the surroundings, because the system is thermally and mechanically isolated, $dQ_{\text{surr}} = 0 \implies \mathbf{\Delta S_{\text{surr}} = 0}$.
(c) $\Delta S_{\text{universe}} = \Delta S_{\text{gas}} + \Delta S_{\text{surr}} = \mathbf{R \ln 3} > 0$.
Example 4 (Clausius Inequality Cyclic Validation – JEE Advanced):
An inventor claims to have developed a cyclic heat engine operating between $T_H = 600\text{ K}$ and $T_C = 300\text{ K}$. During each cycle, the engine absorbs $1200\text{ J}$ of heat from the hot reservoir, does $700\text{ J}$ of work, and exhausts $500\text{ J}$ to the cold reservoir. Using the Clausius Inequality, determine whether this engine is reversible, irreversible, or impossible.
Solution:
Evaluate the cyclic integral $\oint \frac{dQ}{T}$:
$$\oint \frac{dQ}{T} = \frac{Q_H}{T_H} – \frac{Q_C}{T_C} = \frac{+1200\text{ J}}{600\text{ K}} – \frac{500\text{ J}}{300\text{ K}} = 2.0 – 1.67 = +0.33\text{ J/K}$$
Since $\oint \frac{dQ}{T} = +0.33\text{ J/K} > 0$, the cycle violates the Clausius Inequality ($\oint \frac{dQ}{T} \le 0$).
Alternatively, efficiency $\eta = \frac{W}{Q_H} = \frac{700}{1200} \approx 58.3\%$, while the maximum possible Carnot efficiency is $\eta_{\text{Carnot}} = 1 – \frac{300}{600} = 50\%$. The engine is physically impossible.
5. 10-Question Practice Worksheet (DPP)
Q1 (Single Choice): For an irreversible cyclic process, the cyclic integral $\oint \frac{dQ}{T}$ is always:
(A) Zero
(B) Greater than zero
(C) Less than zero
(D) Dependent on the working substance
Q2 (Single Choice): Two moles of an ideal diatomic gas ($\gamma = 7/5$) are heated isobarically from $T_0$ to $2T_0$. The change in entropy of the gas is:
(A) $5 R \ln 2$
(B) $7 R \ln 2$
(C) $2 R \ln 2$
(D) $\frac{7}{2} R \ln 2$
Q3 (Single Choice): A solid metal sphere of heat capacity $C$ at temperature $T_1$ is dropped into a large lake at temperature $T_2$ ($T_1 > T_2$). The change in entropy of the lake is:
(A) $C \ln(T_2/T_1)$
(B) $C (T_1 – T_2)/T_2$
(C) $C (T_2 – T_1)/T_1$
(D) $-C \ln(T_1/T_2)$
Q4 (Single Choice): During a reversible adiabatic process of an ideal gas, the change in entropy of the gas is:
(A) Positive
(B) Negative
(C) Zero
(D) Dependent on the ratio of specific heats
Q5 (Single Choice): An ideal gas expands isothermally and reversibly from $(P_0, V_0)$ to $(P_0/2, 2V_0)$ at temperature $T_0$. The entropy change of the surroundings is:
(A) $+n R \ln 2$
(B) $-n R \ln 2$
(C) 0
(D) $+2 n R \ln 2$
Q6 (Single Choice): Equal masses of two liquids $A$ and $B$ at temperatures $300\text{ K}$ and $400\text{ K}$ are mixed in an adiabatic calorimeter. If their specific heats are in the ratio $c_A : c_B = 1 : 1$, the net entropy change of the universe during mixing is:
(A) Negative
(B) Strictly zero
(C) Strictly positive
(D) Undefined without latent heat
Q7 (One or More than One Correct): Which of the following statements are strictly TRUE according to the Second Law of Thermodynamics?
(A) The entropy of an isolated system can never decrease.
(B) For any reversible cycle, $\oint \frac{dQ}{T} = 0$.
(C) The entropy of the universe decreases during spontaneous crystallization.
(D) During a spontaneous heat transfer between two thermal bodies, total entropy increases.
Q8 (One or More than One Correct): One mole of an ideal monoatomic gas undergoes a reversible process $P = \alpha V$, where $\alpha$ is a positive constant. As volume increases from $V_0$ to $2V_0$:
(A) The temperature of the gas quadruples ($T_f = 4T_0$).
(B) The molar heat capacity for this process is $2R$.
(C) The change in entropy is $2R \ln 2$.
(D) The change in entropy is $4R \ln 2$.
Q9 (Numerical Value): One mole of an ideal gas at temperature $300\text{ K}$ expands reversibly and isothermally until its volume is tripled. If the universal gas constant is taken as $R = 8.314\text{ J/(mol}\cdot\text{K)}$, the change in entropy of the gas in $\text{J/K}$ (rounded to two decimal places) is ________. (Take $\ln 3 = 1.0986$)
Q10 (Numerical Value): A heat reservoir at $600\text{ K}$ transfers $3000\text{ J}$ of heat directly through a non-insulated conducting rod to a second reservoir at $300\text{ K}$. Assuming the rod remains in a steady state, the total entropy generated in the universe in $\text{J/K}$ is ________.
6. Answer Key & Detailed Step-by-Step Solutions
Answer Key:
Q1: (C) | Q2: (B) | Q3: (B) | Q4: (C) | Q5: (B) | Q6: (C) | Q7: (A, B, D) | Q8: (A, B, D) | Q9: 9.13 | Q10: 5
Detailed Solutions
Solution 1:
By Clausius’s Theorem, $\oint \frac{dQ}{T} < 0$ for all irreversible cycles due to internal entropy generation ($\oint \frac{dQ}{T} + S_{\text{gen}} = 0$ with $S_{\text{gen}} > 0$). Hence, (C) is correct.
Solution 2:
At constant pressure ($P = \text{const}$), $dQ_{\text{rev}} = n C_p dT$.
For a diatomic gas, $C_p = \frac{7}{2}R$.
$$\Delta S = \int_{T_0}^{2T_0} \frac{n C_p dT}{T} = n C_p \ln\left(\frac{2T_0}{T_0}\right) = (2) \times \left(\frac{7}{2}R\right) \times \ln 2 = 7 R \ln 2$$
Hence, (B) is correct.
Solution 3:
Heat lost by the metal sphere: $Q_{\text{lost}} = C(T_1 – T_2)$.
Because the lake is an enormous reservoir at constant temperature $T_2$, heat absorbed by the lake is $Q_{\text{absorbed}} = C(T_1 – T_2)$ at constant $T_2$.
$$\Delta S_{\text{lake}} = \frac{Q_{\text{absorbed}}}{T_2} = \frac{C(T_1 – T_2)}{T_2}$$
Hence, (B) is correct.
Solution 4:
By definition of an adiabatic reversible process, $dQ_{\text{rev}} = 0$ at every infinitesimal step along the trajectory. Therefore, $dS = \frac{dQ_{\text{rev}}}{T} = 0 \implies \Delta S = 0$. Reversible adiabatics are isentropic. Hence, (C) is correct.
Solution 5:
For the gas during reversible isothermal expansion, $\Delta S_{\text{gas}} = +n R \ln\left(\frac{V_f}{V_i}\right) = +n R \ln 2$.
Since the process is completely reversible, $\Delta S_{\text{universe}} = \Delta S_{\text{gas}} + \Delta S_{\text{surr}} = 0 \implies \Delta S_{\text{surr}} = -\Delta S_{\text{gas}} = -n R \ln 2$. Hence, (B) is correct.
Solution 6:
Thermal mixing between bodies at unequal temperatures involves irreversible heat conduction across a finite temperature difference. For any spontaneous irreversible process in an isolated calorimeter, $\Delta S_{\text{universe}} > 0$. Mathematically, $\ln\left(\frac{T_{\text{AM}}}{T_{\text{GM}}}\right) > 0$ because arithmetic mean exceeds geometric mean. Hence, (C) is correct.
Solution 7:
(A) True: $\Delta S \ge 0$ for isolated systems.
(B) True: Clausius equality holds for all reversible closed loops.
(C) False: While the system’s entropy decreases during crystallization due to ordering, heat released to surroundings increases surroundings’ entropy by a greater magnitude ($\Delta S_{\text{universe}} > 0$).
(D) True: Natural heat transfer is irreversible, yielding net entropy generation. Hence, (A, B, D) are correct.
Solution 8:
For $P = \alpha V$, substituting ideal gas equation $P = \frac{R T}{V}$ gives $\frac{R T}{V} = \alpha V \implies T = \frac{\alpha}{R} V^2$.
If $V \to 2V_0$, $T_f = \frac{\alpha}{R} (2V_0)^2 = 4 \left(\frac{\alpha V_0^2}{R}\right) = 4T_0$. Statement (A) is correct.
The process equation is $P V^{-1} = \text{const}$ (polytropic index $m = -1$).
Molar heat capacity $C = C_v + \frac{R}{1 – m} = \frac{3}{2}R + \frac{R}{1 – (-1)} = \frac{3}{2}R + \frac{R}{2} = 2R$. Statement (B) is correct.
Entropy change: $\Delta S = \int \frac{n C dT}{T} = 1 \times (2R) \times \ln\left(\frac{T_f}{T_i}\right) = 2R \ln(4) = 2R \times 2 \ln 2 = 4R \ln 2$. Statement (D) is correct.
Hence, (A, B, D) are correct.
Solution 9:
$$\Delta S = n R \ln\left(\frac{V_2}{V_1}\right) = 1 \times 8.314 \times \ln 3 = 8.314 \times 1.0986 \approx \mathbf{9.13\text{ J/K}}$$
Answer is 9.13.
Solution 10:
The conducting rod operates in steady state, so $\Delta S_{\text{rod}} = 0$.
Entropy change of hot reservoir: $\Delta S_H = -\frac{Q}{T_H} = -\frac{3000}{600} = -5\text{ J/K}$.
Entropy change of cold reservoir: $\Delta S_C = +\frac{Q}{T_C} = +\frac{3000}{300} = +10\text{ J/K}$.
Total entropy generated in universe: $\Delta S_{\text{universe}} = \Delta S_H + \Delta S_C = -5 + 10 = \mathbf{5\text{ J/K}}$.
Answer is 5.