Welcome to Lecture 98 of our comprehensive JEE Physics Master Series. In this lecture, we examine the fundamental distinction between Reversible and Irreversible Processes. While classical thermodynamics often models transformations as ideal reversible pathways, all natural and spontaneous physical processes are inherently irreversible. We analyze the exact criteria for reversibility, explore the primary internal and external causes of irreversibility, resolve the crucial distinction between quasi-static and reversible processes, formulate the Clausius Inequality and entropy generation principle ($S_{\text{gen}} \ge 0$), and solve standard JEE Advanced problem archetypes including free expansions, irreversible heat exchange between finite bodies, and the Gouy-Stodola lost work theorem.
Detailed Concept Card: Reversible vs. Irreversible Processes
1. Definition & Fundamental Criteria of a Reversible Process
A thermodynamic process is defined as reversible if, after the process has been completed, both the system and every portion of its surroundings can be restored to their exact initial states without leaving any trace or net change anywhere in the universe.
For a physical transformation to be strictly reversible, it must satisfy two indispensable, non-negotiable conditions simultaneously:
- Quasi-Static Execution (Thermodynamic Equilibrium at All Intermediate Stages): The process must be carried out infinitely slowly through a continuous sequence of infinitesimal steps. At every instant, the driving forces (such as pressure difference $dP$ or temperature difference $dT$) differ from the opposing forces by only an infinitesimal amount. Consequently, the working system remains in mechanical, thermal, and chemical equilibrium throughout the path.
- Total Absence of Dissipative Effects: There must be zero dissipation of mechanical or electrical energy into disorganized thermal energy. This requires the complete elimination of internal fluid viscosity, dry friction between sliding surfaces, electrical resistance, inelastic deformation, and magnetic/dielectric hysteresis.
Internal vs. External Reversibility:
– Internally Reversible: No irreversibilities occur within the boundaries of the system itself during the process (e.g., quasi-static expansion of a gas without internal turbulence or viscosity).
– Externally Reversible: No irreversibilities occur across the boundary between the system and its surroundings (e.g., heat transfer occurs across an infinitesimal temperature gradient $dT \to 0$, so $T_{\text{sys}} = T_{\text{boundary}}$).
A process is totally reversible only when it is both internally and externally reversible.
2. Irreversible Processes & Primary Causes of Irreversibility
An irreversible process is any process that cannot be reversed without leaving a permanent, net change in the surrounding environment. In nature, every real physical process proceeds with a definite direction in time and is irreversible. The causes of irreversibility are categorized into two fundamental groups:
| Irreversibility Mechanism | Physical Manifestation | Key Thermodynamics Consequence |
|---|---|---|
| Mechanical & Dissipative Friction | Sliding solid friction, fluid viscosity, turbulent eddy currents, inelastic strain | Organized mechanical work is degraded into unorganized thermal energy; cannot be completely recovered without external work. |
| Unrestrained Expansion / Free Expansion | Gas expanding into a vacuum ($P_{\text{ext}} = 0$), throttling through a porous plug | Process proceeds under finite, non-equilibrium pressure gradients; no boundary work is performed ($W = 0$). |
| Heat Transfer across Finite $\Delta T$ | Direct conduction or radiation between bodies at temperatures $T_1 > T_2$ | Heat spontaneously flows down the temperature gradient; reversing it requires an external refrigerator consuming work $W_{\text{ext}} > 0$. |
| Spontaneous Chemical & Mixing Processes | Diffusion of two different gases, dissolution of solute in solvent, combustion | Spontaneous increase in molecular disorder / mixing entropy ($\Delta S_{\text{mix}} > 0$); impossible to unmix spontaneously. |
3. Crucial Conceptual Trap: Quasi-Static vs. Reversible
One of the most frequent conceptual traps in JEE Advanced is confusing a quasi-static process with a reversible process:
- Quasi-Static Process: Means infinitely slow. The system passes through a continuum of equilibrium states so that thermodynamic coordinates ($P, V, T$) are well-defined at all intermediate points.
- Reversible Process: Means quasi-static AND strictly non-dissipative.
- Every reversible process is necessarily quasi-static, but a quasi-static process is NOT necessarily reversible!
- Classic Counterexample: Consider a piston moving inside a rough cylinder containing a gas. If we compress the gas infinitely slowly, the gas remains in thermodynamic equilibrium at each instant (quasi-static). However, because mechanical friction opposes the piston motion in both directions, extra work is converted into frictional heat during compression, and additional work is lost to friction during re-expansion. The surroundings cannot be restored to their original state. Hence, the process is quasi-static but irreversible.
4. Mathematical Formulation: Clausius Inequality & Entropy Generation
The Second Law provides the quantitative mathematical test for reversibility via the Clausius Inequality. For any closed thermodynamic cycle:
$$\mathbf{\oint \frac{\delta Q}{T} \le 0}$$
- $\oint \frac{\delta Q}{T} = 0 \iff$ The cycle is completely reversible.
- $\oint \frac{\delta Q}{T} < 0 \iff$ The cycle is irreversible.
- $\oint \frac{\delta Q}{T} > 0 \iff$ The cycle is physically impossible (violates the Second Law).
For any process taking a system from initial state $1$ to final state $2$, the change in entropy of the system is given by:
$$S_2 – S_1 = \int_1^2 \frac{\delta Q}{T} + S_{\text{gen}}$$
where $S_{\text{gen}}$ is the entropy generation caused by internal irreversibilities:
- $S_{\text{gen}} = 0 \iff$ Internally reversible process.
- $S_{\text{gen}} > 0 \iff$ Irreversible process.
- $S_{\text{gen}} < 0 \iff$ Impossible process.
For an isolated system (or the entire Universe consisting of system plus surroundings):
$$\mathbf{\Delta S_{\text{universe}} = \Delta S_{\text{system}} + \Delta S_{\text{surroundings}} = S_{\text{gen}} \ge 0}$$
5. Work Output & The Gouy-Stodola Lost Work Theorem
A fundamental consequence of the Second Law is that irreversibility always destroys the capacity to perform useful mechanical work:
- In Expansion between given end states: A reversible path delivers the maximum possible work output: $$W_{\text{rev, expansion}} > W_{\text{irrev, expansion}}$$
- In Compression between given end states: A reversible path requires the minimum work input: $$|W_{\text{rev, compression}}| < |W_{\text{irrev, compression}}|$$
- Gouy-Stodola Theorem: The work lost due to irreversibility (the difference between reversible work and actual irreversible work) is directly proportional to the total entropy generated in the universe: $$\mathbf{W_{\text{lost}} = T_0 \, S_{\text{gen}} = T_0 \, \Delta S_{\text{universe}}}$$ where $T_0$ is the ambient temperature of the surroundings.
Four Graded Solved Examples
Example 1 (JEE Main Level: Reversible vs. Sudden Irreversible Isothermal Expansion)
Problem Statement: One mole of an ideal gas at temperature $T_0 = 300\text{ K}$ and initial pressure $P_1 = 4\text{ atm}$ expands isothermally to a final pressure $P_2 = 1\text{ atm}$. Compare the process under two different conditions:
(a) The expansion is carried out reversibly and quasi-statically.
(b) The expansion is carried out suddenly against a constant external pressure $P_{\text{ext}} = P_2 = 1\text{ atm}$ in a single irreversible step.
Given $R = 8.314\text{ J/(mol}\cdot\text{K)}$ and $\ln 2 = 0.693$, determine for both cases: the work done by the gas, the heat absorbed from the thermal reservoir at $300\text{ K}$, and the net entropy change of the universe ($\Delta S_{\text{univ}}$).
Solution:
Case (a): Reversible Isothermal Expansion.
Since the process is isothermal and ideal, $\Delta U = 0 \implies Q_{\text{rev}} = W_{\text{rev}}$.
$$W_{\text{rev}} = n R T_0 \ln\left(\frac{P_1}{P_2}\right) = (1)(8.314)(300) \ln(4) = 2494.2 \times (2 \ln 2) = 2494.2 \times (2 \times 0.693) = \mathbf{3457.0\text{ J}}$$
$$Q_{\text{rev}} = W_{\text{rev}} = \mathbf{3457.0\text{ J}}$$
Entropy change of the system:
$$\Delta S_{\text{sys}} = \frac{Q_{\text{rev}}}{T_0} = \frac{3457.0\text{ J}}{300\text{ K}} = \mathbf{+11.523\text{ J/K}}$$
Entropy change of the reservoir (surroundings):
$$\Delta S_{\text{surr}} = -\frac{Q_{\text{rev}}}{T_0} = \mathbf{-11.523\text{ J/K}}$$
Net entropy change of the universe:
$$\Delta S_{\text{univ}} = \Delta S_{\text{sys}} + \Delta S_{\text{surr}} = +11.523 – 11.523 = \mathbf{0}$$
Because $\Delta S_{\text{univ}} = 0$, the process is confirmed to be completely reversible.
Case (b): Irreversible Expansion against Constant $P_{\text{ext}} = P_2 = 1\text{ atm}$.
Work done against constant external pressure:
$$W_{\text{irrev}} = P_{\text{ext}} (V_2 – V_1) = P_2 \left(\frac{n R T_0}{P_2} – \frac{n R T_0}{P_1}\right) = n R T_0 \left(1 – \frac{P_2}{P_1}\right)$$
Substituting values:
$$W_{\text{irrev}} = (1)(8.314)(300)\left(1 – \frac{1}{4}\right) = 2494.2 \times 0.75 = \mathbf{1870.65\text{ J}}$$
Since the initial and final temperatures are both $T_0$, for an ideal gas $\Delta U = 0 \implies Q_{\text{irrev}} = W_{\text{irrev}} = \mathbf{1870.65\text{ J}}$.
Because entropy is a state function, $\Delta S_{\text{sys}}$ depends solely on the initial and final equilibrium states, which are identical to Case (a):
$$\Delta S_{\text{sys}} = \mathbf{+11.523\text{ J/K}}$$
Entropy change of the surrounding reservoir at $T_0 = 300\text{ K}$:
$$\Delta S_{\text{surr}} = -\frac{Q_{\text{irrev}}}{T_0} = -\frac{1870.65\text{ J}}{300\text{ K}} = \mathbf{-6.236\text{ J/K}}$$
Net entropy change of the universe:
$$\Delta S_{\text{univ}} = \Delta S_{\text{sys}} + \Delta S_{\text{surr}} = +11.523 – 6.236 = \mathbf{+5.287\text{ J/K} > 0}$$
Verification via Gouy-Stodola Lost Work:
$$W_{\text{lost}} = W_{\text{rev}} – W_{\text{irrev}} = 3457.0\text{ J} – 1870.65\text{ J} = 1586.35\text{ J}$$
$$T_0 \, \Delta S_{\text{univ}} = 300\text{ K} \times 5.287\text{ J/K} = 1586.1\text{ J} \approx W_{\text{lost}}$$
The irreversible drop in pressure wastes $1586.35\text{ J}$ of potential work into irreversibility!
Example 2 (Adiabatic Free Expansion vs. Reversible Adiabatic Expansion)
Problem Statement: A thermally insulated rigid vessel of total volume $2V_0$ is divided into two equal compartments by a thin partition. One compartment contains $n = 1\text{ mole}$ of an ideal monoatomic gas ($\gamma = 5/3$) at temperature $T_0$, while the other compartment is completely evacuated. The partition is suddenly removed, and the gas freely expands to occupy the entire vessel volume $2V_0$.
1. Determine the work done $W$, heat exchanged $Q$, change in internal energy $\Delta U$, final temperature $T_f$, and total entropy change $\Delta S_{\text{univ}}$.
2. If the same gas had instead expanded reversibly and adiabatically from volume $V_0$ to $2V_0$, what would be its final temperature $T_f’$ and work done $W_{\text{rev}}$?
Solution:
Part 1: Adiabatic Free Expansion (Joule Expansion).
– Insulated Vessel: $Q = 0$.
– Rigid Boundaries & Expansion into Vacuum ($P_{\text{ext}} = 0$): $W = \int P_{\text{ext}} dV = 0$.
– First Law: $\Delta U = Q – W = 0 – 0 = \mathbf{0}$.
– Since $U = n C_v T$ for an ideal gas, $\Delta U = 0 \implies \mathbf{T_f = T_0}$. The temperature remains unchanged!
– Entropy Calculation: Even though $Q = 0$, the process is highly irreversible. To compute $\Delta S_{\text{sys}}$, we connect the initial state $(T_0, V_0)$ and final state $(T_0, 2V_0)$ by an imaginary reversible isothermal path:
$$\Delta S_{\text{sys}} = n R \ln\left(\frac{2V_0}{V_0}\right) = n R \ln 2$$
Since the vessel is thermally insulated, the surroundings exchange zero heat: $\Delta S_{\text{surr}} = 0$.
$$\Delta S_{\text{univ}} = \Delta S_{\text{sys}} + \Delta S_{\text{surr}} = \mathbf{n R \ln 2 > 0}$$
Free expansion is intrinsically irreversible because entropy is generated internally with zero work extracted.
Part 2: Reversible Adiabatic Expansion from $V_0$ to $2V_0$.
By Poisson’s relation ($T V^{\gamma – 1} = \text{constant}$):
$$T_f’ (2V_0)^{5/3 – 1} = T_0 V_0^{2/3} \implies T_f’ = T_0 \left(\frac{V_0}{2V_0}\right)^{2/3} = \frac{T_0}{2^{2/3}} = \frac{T_0}{\sqrt[3]{4}} \approx \mathbf{0.630 \, T_0}$$
Work done by the gas:
$$W_{\text{rev}} = -\Delta U = n C_v (T_0 – T_f’) = \frac{3}{2} n R T_0 \left(1 – \frac{1}{2^{2/3}}\right) \approx \mathbf{0.555 \, n R T_0 > 0}$$
Entropy change of the universe in the reversible adiabatic path:
$$\Delta S_{\text{univ}} = \mathbf{0}$$
Takeaway: In free expansion, zero work is delivered ($W = 0$) and temperature is unchanged ($T_f = T_0$). In reversible adiabatic expansion, the gas delivers positive mechanical work ($W > 0$) at the expense of its internal energy, cooling down to $0.630 T_0$ while conserving total entropy ($\Delta S_{\text{univ}} = 0$).
Example 3 (JEE Advanced Level: Irreversible vs. Reversible Thermal Equilibration of Finite Bodies)
Problem Statement: Two identical solid bodies, each of mass $m$ and constant specific heat capacity $c$, are initially at temperatures $T_1 = 400\text{ K}$ and $T_2 = 100\text{ K}$.
1. If the two bodies are placed in direct thermal contact inside a thermally insulated enclosure until they reach a common final temperature $T_f$, find $T_f$ and show that the net entropy change of the universe is $\Delta S_{\text{univ}} = m c \ln(1.5625) > 0$.
2. Suppose instead that the bodies are brought to thermal equilibrium reversibly by operating a series of infinitesimal Carnot heat engines between them. Determine the final equilibrium temperature $T_f’$ and the maximum mechanical work $W_{\text{max}}$ that can be extracted from the system.
Solution:
Part 1: Direct Thermal Contact (Irreversible Heat Exchange).
Applying the First Law to the isolated combined system:
$$\Delta Q_{\text{lost by 1}} = \Delta Q_{\text{gained by 2}} \implies m c (T_1 – T_f) = m c (T_f – T_2)$$
$$T_1 – T_f = T_f – T_2 \implies \mathbf{T_f = \frac{T_1 + T_2}{2} = \frac{400 + 100}{2} = 250\text{ K}}$$
The final temperature is the arithmetic mean ($\text{AM}$).
Entropy change of Body 1 (cooling from $400\text{ K}$ to $250\text{ K}$):
$$\Delta S_1 = \int_{T_1}^{T_f} \frac{m c \, dT}{T} = m c \ln\left(\frac{T_f}{T_1}\right) = m c \ln\left(\frac{250}{400}\right)$$
Entropy change of Body 2 (heating from $100\text{ K}$ to $250\text{ K}$):
$$\Delta S_2 = \int_{T_2}^{T_f} \frac{m c \, dT}{T} = m c \ln\left(\frac{T_f}{T_2}\right) = m c \ln\left(\frac{250}{100}\right)$$
Total entropy change of the universe (since the enclosure is insulated, $\Delta S_{\text{surr}} = 0$):
$$\Delta S_{\text{univ}} = \Delta S_1 + \Delta S_2 = m c \ln\left(\frac{T_f^2}{T_1 T_2}\right) = m c \ln\left[\frac{250^2}{400 \times 100}\right] = m c \ln\left(\frac{62500}{40000}\right) = \mathbf{m c \ln(1.5625) \approx +0.4463 \, m c > 0}$$
Because $\text{AM} > \text{GM}$ for distinct positive temperatures, $\frac{T_f^2}{T_1 T_2} = \frac{((T_1+T_2)/2)^2}{T_1 T_2} > 1$, proving mathematically that spontaneous heat conduction always increases the entropy of the universe.
Part 2: Reversible Equilibration via Infinitesimal Carnot Engines.
For the entire combined process to be completely reversible, the total entropy change must be zero:
$$\Delta S_{\text{total}} = \Delta S_1 + \Delta S_2 = 0 \implies m c \ln\left(\frac{T_f’}{T_1}\right) + m c \ln\left(\frac{T_f’}{T_2}\right) = 0$$
$$m c \ln\left(\frac{T_f’^2}{T_1 T_2}\right) = 0 \implies \frac{T_f’^2}{T_1 T_2} = 1 \implies \mathbf{T_f’ = \sqrt{T_1 T_2}}$$
The reversible equilibrium temperature is the geometric mean ($\text{GM}$)!
$$T_f’ = \sqrt{400 \times 100} = \mathbf{200\text{ K}}$$
Maximum mechanical work extracted:
By energy conservation, the work delivered equals the total decrease in internal energy of the two bodies:
$$W_{\text{max}} = -\Delta U = [m c (T_1 – T_f’)] + [m c (T_2 – T_f’)] = m c (T_1 + T_2 – 2 T_f’)$$
Substituting $T_1 = 400\text{ K}, T_2 = 100\text{ K}, T_f’ = 200\text{ K}$:
$$W_{\text{max}} = m c (400 + 100 – 2 \times 200) = m c (500 – 400) = \mathbf{100 \, m c\text{ J}}$$
Takeaway: When two bodies at different temperatures are placed in direct thermal contact, the potential to perform $100 m c\text{ J}$ of work is completely destroyed, degraded entirely into thermal entropy!
Example 4 (JEE Advanced Level: Clausius Inequality Applied to an Irreversible Cycle)
Problem Statement: A cyclic heat engine operates between a source at $T_H = 600\text{ K}$ and a sink at $T_C = 300\text{ K}$. In each cycle, the engine absorbs $Q_H = 1200\text{ J}$ of heat from the source and delivers $W = 400\text{ J}$ of mechanical work. Determine:
1. The heat rejected $Q_C$ to the sink and the actual thermal efficiency $\eta_{\text{actual}}$.
2. The maximum theoretical efficiency $\eta_{\text{Carnot}}$ between these reservoirs.
3. Evaluate the cyclic integral $\oint \frac{\delta Q}{T}$ and verify the Clausius Inequality.
Solution:
Step 1: Heat Rejected and Actual Efficiency.
By First Law for a cycle ($\Delta U = 0$):
$$Q_C = Q_H – W = 1200\text{ J} – 400\text{ J} = \mathbf{800\text{ J}}$$
Actual efficiency:
$$\eta_{\text{actual}} = \frac{W}{Q_H} = \frac{400\text{ J}}{1200\text{ J}} = \frac{1}{3} \approx \mathbf{33.33\%}$$
Step 2: Maximum Reversible (Carnot) Efficiency.
$$\eta_{\text{Carnot}} = 1 – \frac{T_C}{T_H} = 1 – \frac{300\text{ K}}{600\text{ K}} = 1 – 0.5 = \mathbf{50.0\%}$$
Notice that $\eta_{\text{actual}} (33.33\%) < \eta_{\text{Carnot}} (50.0\%)$, which proves that the engine is irreversible.
Step 3: Evaluate the Cyclic Integral $\oint \frac{\delta Q}{T}$.
Heat is absorbed at the source temperature $T_H = 600\text{ K}$ and rejected at the sink temperature $T_C = 300\text{ K}$:
$$\oint \frac{\delta Q}{T} = \frac{Q_H}{T_H} – \frac{Q_C}{T_C} = \frac{+1200\text{ J}}{600\text{ K}} – \frac{800\text{ J}}{300\text{ K}} = +2.0\text{ J/K} – 2.667\text{ J/K} = \mathbf{-0.667\text{ J/K} < 0}$$
Because $\oint \frac{\delta Q}{T} < 0$, the Clausius Inequality is strictly satisfied for an irreversible cycle!
Entropy Generated per Cycle:
$$S_{\text{gen}} = -\oint \frac{\delta Q}{T} = -(-0.667\text{ J/K}) = \mathbf{+0.667\text{ J/K} > 0}$$
Work lost per cycle at ambient temperature $T_C = 300\text{ K}$:
$$W_{\text{lost}} = T_C \cdot S_{\text{gen}} = 300\text{ K} \times 0.667\text{ J/K} = \mathbf{200\text{ J}}$$
Indeed, a reversible Carnot engine would have delivered $W_{\text{rev}} = \eta_{\text{Carnot}} Q_H = 0.5 \times 1200 = 600\text{ J}$. The actual engine delivers only $400\text{ J}$, confirming $W_{\text{lost}} = 600 – 400 = 200\text{ J}$.
Practice Worksheet: 10 JEE Problems
Questions 1 to 5: JEE Main Level (Single Correct)
Q1. A thermodynamic process carried out on an ideal gas is strictly reversible if and only if:
(A) The process is carried out infinitely fast with insulated walls.
(B) The process is quasi-static and completely free from all dissipative effects such as friction and viscosity.
(C) The process is isothermal regardless of the rate of expansion.
(D) The initial and final pressures are equal.
Q2. One mole of an ideal gas expands from volume $V_1$ to volume $V_2$ at constant temperature $T$. Path 1 is a reversible quasi-static isothermal path, while Path 2 is an irreversible expansion against a constant external pressure $P_{\text{ext}} = P_2$. If $W_1$ and $W_2$ denote the work done by the gas in Paths 1 and 2 respectively, then:
(A) $W_1 > W_2$
(B) $W_2 > W_1$
(C) $W_1 = W_2$
(D) $W_1 = 0$ and $W_2 > 0$
Q3. An insulated rigid vessel is partitioned into two halves: one half contains an ideal gas at temperature $T_0$, and the other half is an absolute vacuum. If the partition is suddenly removed, which of the following sets of parameters correctly describes the free expansion?
(A) $Q > 0, W > 0, \Delta U = 0, \Delta S_{\text{univ}} = 0$
(B) $Q = 0, W = 0, \Delta U = 0, \Delta S_{\text{univ}} > 0$
(C) $Q = 0, W > 0, \Delta U < 0, \Delta S_{\text{univ}} = 0$
(D) $Q < 0, W = 0, \Delta U < 0, \Delta S_{\text{univ}} < 0$
Q4. For an irreversible thermodynamic cycle operating between multiple thermal reservoirs, the cyclic integral of $\frac{\delta Q}{T}$ must satisfy:
(A) $\oint \frac{\delta Q}{T} > 0$
(B) $\oint \frac{\delta Q}{T} = 0$
(C) $\oint \frac{\delta Q}{T} < 0$
(D) $\oint \frac{\delta Q}{T} \to \infty$
Q5. Two identical bodies of equal heat capacity are initially at temperatures $T_1$ and $T_2$ ($T_1 \neq T_2$). If they are brought to thermal equilibrium by direct thermal contact in an insulated box, the final temperature is $T_{\text{irrev}}$. If they are brought to equilibrium reversibly via Carnot engines, the final temperature is $T_{\text{rev}}$. Then:
(A) $T_{\text{irrev}} = \frac{T_1 + T_2}{2}$ and $T_{\text{rev}} = \sqrt{T_1 T_2}$ with $T_{\text{irrev}} > $T_{\text{rev}}$
(B) $T_{\text{irrev}} = \sqrt{T_1 T_2}$ and $T_{\text{rev}} = \frac{T_1 + T_2}{2}$ with $T_{\text{rev}} > T_{\text{irrev}}$
(C) $T_{\text{irrev}} = T_{\text{rev}} = \frac{T_1 + T_2}{2}$
(D) $T_{\text{irrev}} = T_{\text{rev}} = \sqrt{T_1 T_2}$
Questions 6 to 8: JEE Advanced Level (One or More Than One Correct)
Q6. Which of the following natural phenomena or processes are inherently irreversible?
(A) Spontaneous conduction of heat across a finite temperature gradient.
(B) Free expansion of an ideal gas into an evacuated space.
(C) Motion of a piston against mechanical friction, even when carried out infinitely slowly.
(D) Spontaneous diffusion and mixing of two non-reacting ideal gases at constant temperature and pressure.
Q7. Which of the following statements correctly distinguish a quasi-static process from a reversible process?
(A) Every reversible process is necessarily quasi-static.
(B) A quasi-static process is always reversible, even if dry friction is present.
(C) In a quasi-static process, the system is in internal thermodynamic equilibrium at every intermediate instant.
(D) A process can be quasi-static yet irreversible if mechanical dissipation or finite external temperature gradients exist.
Q8. An ideal gas undergoes a change of state from $(P_1, V_1, T_1)$ to $(P_2, V_2, T_2)$ via two different paths: Path A is completely reversible, and Path B is irreversible. Which of the following statements is/are correct?
(A) The change in entropy of the gas is the same for both paths ($\Delta S_{\text{sys, A}} = \Delta S_{\text{sys, B}}$).
(B) The total entropy change of the universe for Path A is zero ($\Delta S_{\text{univ, A}} = 0$).
(C) The total entropy change of the universe for Path B is strictly positive ($\Delta S_{\text{univ, B}} > 0$).
(D) If both paths represent expansion between the same volumes, the mechanical work delivered along Path A is greater than along Path B ($W_{\text{A}} > W_{\text{B}}$).
Questions 9 to 10: Numerical / Integer Answer Type
Q9. One mole of an ideal gas at temperature $T_0$ (where $n R T_0 = 2500\text{ J}$) expands from initial volume $V_0$ to final volume $4V_0$. In Experiment 1, the expansion is reversible and isothermal. In Experiment 2, the expansion is carried out isothermally at $T_0$ against a constant external pressure equal to the final gas pressure ($P_{\text{ext}} = P_f$). Taking $\ln 2 = 0.693$, calculate the lost work $W_{\text{lost}} = W_{\text{rev}} – W_{\text{irrev}}$ (in $\text{J}$).
Q10. Two identical bodies each of constant heat capacity $C = 100\text{ J/K}$ are initially at temperatures $T_1 = 900\text{ K}$ and $T_2 = 100\text{ K}$. A series of reversible Carnot engines is operated between the two bodies until they reach a common final equilibrium temperature $T_f$. Calculate the maximum total mechanical work $W_{\text{max}}$ delivered by the engines (in $\text{kJ}$).
Detailed Solutions & Answer Key
Answer Key
| Question | Correct Option / Answer | Topic / Concept |
|---|---|---|
| Q1 | (B) Quasi-static & non-dissipative | Criteria for Thermodynamic Reversibility |
| Q2 | (A) $W_1 > W_2$ | Work Output in Reversible vs. Irreversible Expansion |
| Q3 | (B) $Q=0, W=0, \Delta U=0, \Delta S_{\text{univ}}>0$ | Adiabatic Free Expansion Characteristics |
| Q4 | (C) $\oint \frac{\delta Q}{T} < 0$ | Clausius Inequality for Irreversible Cycle |
| Q5 | (A) $T_{\text{irrev}} = \text{AM}$, $T_{\text{rev}} = \text{GM}$ | Thermal Equilibration of Finite Bodies |
| Q6 | (A), (B), (C), (D) | Causes and Mechanisms of Irreversibility |
| Q7 | (A), (C), (D) | Quasi-Static vs. Reversible Conceptual Traps |
| Q8 | (A), (B), (C), (D) | Entropy and Work in Reversible vs. Irreversible Paths |
| Q9 | 1590 | Lost Work $W_{\text{rev}} – W_{\text{irrev}}$ ($3465 – 1875\text{ J}$) |
| Q10 | 40 | Maximum Work from Finite Bodies ($40\text{ kJ}$) |
Complete Step-by-Step Solutions
Solution 1:
A thermodynamic process is reversible if and only if both the system and surroundings can be returned to their initial states. This requires the process to be quasi-static (driving and opposing forces differ infinitesimally) and strictly non-dissipative (zero friction, viscosity, or electrical resistance).
Correct Option: (B)
Solution 2:
During reversible isothermal expansion, the gas pressure $P$ at each volume $V$ is given by $P = nRT/V$, which is strictly greater than the final pressure $P_2$ for all intermediate volumes $V < V_2$. Therefore, the area under the reversible curve $W_1 = \int_{V_1}^{V_2} P \, dV = nRT \ln(V_2/V_1)$ is strictly larger than the rectangular area under the constant external pressure line $W_2 = P_2(V_2 - V_1)$. Hence, $W_1 > W_2$.
Correct Option: (A)
Solution 3:
In adiabatic free expansion:
– Insulated container $\implies Q = 0$.
– Rigid container expanding into vacuum ($P_{\text{ext}} = 0$) $\implies W = 0$.
– First Law: $\Delta U = Q – W = 0$.
– Since the gas expands into double the volume with zero work, it cannot be returned to the initial state without external compressive work. The process generates entropy: $\Delta S_{\text{univ}} = n R \ln 2 > 0$.
Correct Option: (B)
Solution 4:
By the Clausius Inequality:
– For a reversible cycle: $\oint \frac{\delta Q}{T} = 0$.
– For an irreversible cycle: $\oint \frac{\delta Q}{T} < 0$.
– $\oint \frac{\delta Q}{T} > 0$ is thermodynamically impossible.
Correct Option: (C)
Solution 5:
1. In direct thermal contact, energy conservation gives $C(T_1 – T_f) = C(T_f – T_2) \implies T_{\text{irrev}} = \frac{T_1 + T_2}{2}$ (arithmetic mean).
2. In reversible heat engine operation, total entropy change is zero: $\Delta S_1 + \Delta S_2 = 0 \implies C \ln(T_f / T_1) + C \ln(T_f / T_2) = 0 \implies T_{\text{rev}} = \sqrt{T_1 T_2}$ (geometric mean).
Since the arithmetic mean is strictly greater than the geometric mean for distinct positive values ($\text{AM} > \text{GM}$), $T_{\text{irrev}} > T_{\text{rev}}$.
Correct Option: (A)
Solution 6:
(A) Correct: Heat transfer across a finite $\Delta T$ generates entropy and cannot reverse spontaneously.
(B) Correct: Unrestrained expansion into a vacuum delivers no work and generates entropy ($S_{\text{gen}} > 0$).
(C) Correct: Moving against mechanical friction degrades mechanical energy into heat; friction is dissipative regardless of speed.
(D) Correct: Mixing dissimilar gases produces mixing entropy ($\Delta S_{\text{mix}} > 0$) which cannot unmix spontaneously.
Correct Options: (A), (B), (C), (D)
Solution 7:
(A) Correct: Reversibility requires infinite slowness (quasi-static path) to maintain thermodynamic equilibrium.
(B) Incorrect: A quasi-static process with friction is irreversible because frictional dissipation cannot be recovered.
(C) Correct: By definition, a quasi-static process passes through a continuous series of equilibrium states.
(D) Correct: Friction, viscosity, or finite thermal gradients make an infinitely slow process irreversible.
Correct Options: (A), (C), (D)
Solution 8:
(A) Correct: Entropy $S$ is a state function; $\Delta S_{\text{sys}}$ depends solely on the initial and final states, not the path.
(B) Correct: For any reversible process, $\Delta S_{\text{univ}} = \Delta S_{\text{sys}} + \Delta S_{\text{surr}} = 0$.
(C) Correct: For any irreversible process, entropy is generated internally: $\Delta S_{\text{univ}} > 0$.
(D) Correct: Maximum work is always obtained along the reversible path ($W_{\text{rev}} > W_{\text{irrev}}$).
Correct Options: (A), (B), (C), (D)
Solution 9:
1. Reversible Isothermal Work:
$$W_{\text{rev}} = n R T_0 \ln\left(\frac{4V_0}{V_0}\right) = 2 \, n R T_0 \ln 2 = 2 \times 2500 \times 0.693 = 5000 \times 0.693 = \mathbf{3465\text{ J}}$$
2. Irreversible Expansion Work against $P_{\text{ext}} = P_f$:
Final pressure: $P_f = \frac{n R T_0}{4V_0} = \frac{P_0}{4}$.
$$W_{\text{irrev}} = P_f (4V_0 – V_0) = \frac{P_0}{4} (3V_0) = \frac{3}{4} P_0 V_0 = 0.75 \times 2500 = \mathbf{1875\text{ J}}$$
3. Lost Work ($W_{\text{lost}}$):
$$W_{\text{lost}} = W_{\text{rev}} – W_{\text{irrev}} = 3465\text{ J} – 1875\text{ J} = \mathbf{1590\text{ J}}$$
Answer: 1590
Solution 10:
For maximum work extraction via reversible Carnot engines, total entropy change of the system is zero:
$$\Delta S = C \ln\left(\frac{T_f}{T_1}\right) + C \ln\left(\frac{T_f}{T_2}\right) = 0 \implies T_f = \sqrt{T_1 T_2}$$
Substituting $T_1 = 900\text{ K}$ and $T_2 = 100\text{ K}$:
$$T_f = \sqrt{900 \times 100} = \sqrt{90000} = 300\text{ K}$$
The maximum work delivered equals the total decrease in internal energy of the two bodies:
$$W_{\text{max}} = -\Delta U = C(T_1 – T_f) + C(T_2 – T_f) = C(T_1 + T_2 – 2 T_f)$$
$$W_{\text{max}} = 100\text{ J/K} \times (900\text{ K} + 100\text{ K} – 2 \times 300\text{ K}) = 100 \times (1000 – 600) = 100 \times 400 = 40000\text{ J} = \mathbf{40\text{ kJ}}$$
Answer: 40