301. A frictionless piston contains a gas of instantaneous pressure \(P\). Consider three possible external pressures during an attempted expansion:
Process X: \(P_{\text{ext}}=P-\varepsilon\), where \(\varepsilon\) is positive and tends to an infinitesimal value.
Process Y: \(P_{\text{ext}}=0.9P\).
Process Z: \(P_{\text{ext}}=P\) exactly.
Which statement is correct?
ⓐ. X approaches reversible expansion; Y has a finite pressure imbalance; Z has none to drive expansion.
ⓑ. Y is the reversible process because it expands faster.
ⓒ. Z is the reversible expansion because equal pressures guarantee both equilibrium and finite expansion.
ⓓ. X and Y are equally reversible because both have \(P_{\text{ext}}\lt P\).
Correct Answer: X approaches reversible expansion; Y has a finite pressure imbalance; Z has none to drive expansion.
Explanation: For a mechanically reversible expansion, the pressure difference driving the piston must be infinitesimal. In Process X,
\[
P-P_{\text{ext}}=\varepsilon,
\]
and the reversible limit is approached as
\[
\varepsilon\rightarrow0^+
\]
This allows the piston to move while the gas remains arbitrarily close to mechanical equilibrium with the surroundings. In Process Y,
\[
P-P_{\text{ext}}=0.1P,
\]
which is a finite pressure difference. Such a finite driving force takes the system away from the reversible quasi-static limit. In Process Z, internal and external pressures are exactly equal at that instant, so there is no net mechanical pressure imbalance to initiate expansion. Reversible motion is therefore associated with an infinitesimal, not a finite or exactly zero, driving difference.
302. A body at \(300\,\text{K}\) receives heat from a reservoir at \(400\,\text{K}\). Two students make the following claims.
Student P: “If the heat transfer is made extremely slow, it becomes reversible even though the temperature difference remains finite.”
Student Q: “For reversible heat transfer, the temperature difference between system and surroundings must be infinitesimal at each stage.”
Which evaluation is correct?
ⓐ. Only Student P is correct.
ⓑ. Both students are incorrect.
ⓒ. Both students are correct.
ⓓ. Only Student Q is correct.
Correct Answer: Only Student Q is correct.
Explanation: A process can be slow without being thermodynamically reversible. The important issue for reversible heat transfer is the thermal driving difference. A finite temperature difference causes heat to flow in a definite spontaneous direction and introduces irreversibility. Merely reducing the rate of heat transfer while retaining that finite difference does not remove the underlying thermodynamic asymmetry. In the reversible idealisation, the system and surrounding reservoir differ in temperature only infinitesimally at every stage. Reversing the infinitesimal temperature ordering then reverses the heat-flow direction without requiring a finite nonequilibrium jump. Student P therefore confuses slow transfer with reversible transfer. Student Q identifies the relevant quasi-static thermal condition. Hence only Student Q is correct.
303. Match each process condition in List I with the most appropriate conclusion in List II.
| List I | List II |
|---|
| P. Frictionless piston, infinitesimal pressure imbalance, thermally insulated cylinder | 1. Reversible adiabatic idealisation |
| Q. Very slow piston motion with appreciable friction | 2. Quasi-static-looking but irreversible because of dissipation |
| R. Heat transfer through a finite temperature difference | 3. Irreversible thermal transfer |
| S. Frictionless quasi-static isothermal expansion with infinitesimal thermal driving difference | 4. Reversible isothermal idealisation |
ⓐ. P-1, Q-2, R-3, S-4
ⓑ. P-4, Q-1, R-2, S-3
ⓒ. P-2, Q-3, R-4, S-1
ⓓ. P-3, Q-4, R-1, S-2
Correct Answer: P-1, Q-2, R-3, S-4
Explanation: Process P combines thermal insulation with quasi-static frictionless piston motion. It therefore provides the standard reversible adiabatic idealisation, giving P-1. Process Q may occur very slowly, but friction remains a dissipative effect. Slow motion therefore does not make it reversible, so Q-2. Process R transfers heat through a finite temperature difference, which is a finite thermodynamic driving gradient and hence an irreversible thermal process, giving R-3. Process S combines quasi-static mechanical motion, absence of friction and an infinitesimal thermal driving difference while maintaining constant temperature. It therefore represents the reversible isothermal idealisation, so S-4. The matching demonstrates that process labels such as adiabatic or isothermal are not sufficient by themselves. Reversibility requires both near-equilibrium progression and absence of dissipation. Hence the correct matching is P-1, Q-2, R-3, S-4.
304. A process carries a system through the equilibrium-state sequence
\[
A\rightarrow B\rightarrow C\rightarrow D
\]
In the reverse operation, the system can be made to follow
\[
D\rightarrow C\rightarrow B\rightarrow A
\]
Which additional condition is required before the original process can be called thermodynamically reversible?
ⓐ. The reverse path must occur faster than the forward path.
ⓑ. The pressure must remain constant during both directions.
ⓒ. The reversed path must restore the surroundings too, with no other net change.
ⓓ. The heat transferred in the forward process must be zero.
Correct Answer: The reversed path must restore the surroundings too, with no other net change.
Explanation: Retracing the system states in reverse order is an important feature of a reversible path, but it is not the complete thermodynamic criterion. Reversibility concerns the combined system and surroundings. Suppose the system returns through
\[
D\rightarrow C\rightarrow B\rightarrow A
\]
but frictional heating remains in the surroundings. The system would then be restored while the universe outside it would not be. Similarly, a weight, reservoir or other work-storage device might retain a net change. Such a process would fail the reversibility test even though the system itself followed the reverse sequence of states. A reversible process must permit restoration of every relevant environmental change as well. No requirement of zero heat or constant pressure follows from the definition. Hence option C supplies the missing condition.
305. During one infinitesimal stage of a reversible process from \(A\) to \(B\), the system absorbs an amount of heat \(\delta Q\gt 0\) and does an amount of work \(\delta W\gt 0\). When the exact reverse path is traversed through the corresponding neighbouring states, what are the heat and work transfers for the system at that reversed stage?
ⓐ. \(+\delta Q,\ +\delta W\)
ⓑ. \(-\delta Q,\ +\delta W\)
ⓒ. \(-\delta Q,\ -\delta W\)
ⓓ. \(0,\ 0\)
Correct Answer: \(-\delta Q,\ -\delta W\)
Explanation: A reversible process can be retraced through the corresponding equilibrium states in the opposite direction. At a given infinitesimal stage of the forward path, \(\delta Q\gt 0\) means that heat enters the system. On the exact reverse stage, that same energy transfer must leave the system, so the heat changes sign to \(-\delta Q\). Likewise, \(\delta W\gt 0\) means that the system does work on the surroundings during the forward stage. Exact reversal requires the surroundings to do the corresponding work on the system, so the work done by the system becomes \(-\delta W\). The magnitudes match in the ideal reversible limit because the reverse retraces the same neighbouring states. Simply keeping one of the signs unchanged would fail to undo the corresponding change in the surroundings. Reversibility therefore concerns both the system path and restoration of the surroundings. Hence both infinitesimal transfers reverse sign, making option C correct.
306. Four processes involving a gas are described.
| Process | Description |
|---|
| I | Isothermal expansion through a frictionless piston, but heat is supplied across a finite temperature difference |
| II | Thermally insulated expansion through a frictionless piston with infinitesimal pressure imbalance at every stage |
| III | Extremely slow compression with appreciable piston friction |
| IV | Thermally insulated free expansion into vacuum |
Which process is reversible in the idealised limit?
ⓐ. Process I only
ⓑ. Process II only
ⓒ. Processes I and II only
ⓓ. Processes II and III only
Correct Answer: Process II only
Explanation: Process I is isothermal, but heat crosses a finite temperature difference. That finite thermal driving difference makes the heat transfer irreversible even though the piston is frictionless. Process II is thermally insulated and therefore adiabatic. Its piston is frictionless and the mechanical pressure imbalance is infinitesimal throughout, so it satisfies the quasi-static and non-dissipative requirements for a reversible adiabatic idealisation. Process III is slow, but friction introduces dissipation and prevents reversibility. Process IV is adiabatic because no heat crosses the boundary, but free expansion into vacuum proceeds through non-equilibrium conditions and is irreversible. Thus neither “isothermal,” “adiabatic,” nor “slow” alone determines reversibility. The complete mechanical, thermal and dissipative conditions must be examined together. Only Process II meets the required idealised conditions.
307. Assertion: If a thermodynamic system is restored to its initial state after some process, the original process must have been reversible.
Reason: A reversible process requires restoration of both the system and the surroundings to their initial states with no other change anywhere.
ⓐ. Both Assertion and Reason are correct, and Reason is the correct explanation of Assertion.
ⓑ. Both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.
ⓒ. Assertion is correct, but Reason is incorrect.
ⓓ. Assertion is incorrect, but Reason is correct.
Correct Answer: Assertion is incorrect, but Reason is correct.
Explanation: The Assertion is incorrect because restoration of the system alone is not sufficient for thermodynamic reversibility. An irreversible process can be followed by some additional process that returns the system to its starting state. In doing so, however, the surroundings may acquire net changes such as heat transfer, dissipated mechanical energy or displacement of a work source. The original process has not thereby become reversible. The Reason states the actual thermodynamic criterion. Both the system and the surroundings must be restorable, and the combined forward-plus-reverse operation must leave no other change anywhere. This distinction is especially important for processes such as free expansion or frictional motion. Thus the Assertion is false while the Reason is correct.
308. Consider the following statements about reversible thermodynamic processes.
Statement I: Reversibility requires complete restoration of both the system and the surroundings.
Statement II: In the reversible limit, the pressure and temperature differences driving mechanical and thermal changes are infinitesimal.
Statement III: Any quasi-static process is reversible even if friction is present.
Statement IV: A reversible adiabatic process requires \(Q=0\) together with quasi-static, non-dissipative behaviour.
Which combination is correct?
ⓐ. I and III only
ⓑ. II and IV only
ⓒ. I, II and IV only
ⓓ. I, II, III and IV
Correct Answer: I, II and IV only
Explanation: Statement I is correct because complete restoration is the defining thermodynamic criterion for reversibility. Statement II is also correct. A reversible mechanical change is driven by only an infinitesimal pressure imbalance, while reversible heat transfer requires only an infinitesimal temperature difference at each stage. Statement III is incorrect because quasi-static behaviour is necessary but not sufficient. Friction, viscosity or another dissipative effect can make an otherwise slow process irreversible. Statement IV is correct because zero heat transfer establishes the adiabatic condition but does not by itself establish reversibility. The process must additionally proceed quasi-statically and without dissipative effects in the idealised limit. Free expansion provides a direct counterexample to the claim that every adiabatic process is reversible. Therefore Statements I, II and IV only are correct.
309. An ideal gas is to undergo an isothermal expansion while remaining arbitrarily close to equilibrium. Which set of conditions represents the reversible idealisation most closely?
ⓐ. Frictionless piston, finite pressure drop across the piston, and heat supplied from a reservoir at a substantially higher temperature
ⓑ. Frictionless piston, infinitesimal pressure imbalance, and heat transfer through an infinitesimal temperature difference
ⓒ. Piston with friction, infinitesimal pressure imbalance, and exactly zero heat transfer
ⓓ. Sudden removal of the external load while the cylinder is maintained at constant temperature
Correct Answer: Frictionless piston, infinitesimal pressure imbalance, and heat transfer through an infinitesimal temperature difference
Explanation: An isothermal ideal-gas expansion generally requires heat input because the gas does positive work while its internal energy remains unchanged. Mechanical reversibility requires the gas pressure and external pressure to differ only infinitesimally during piston motion. A finite pressure difference would drive a non-equilibrium mechanical change. Thermal reversibility similarly requires the temperature difference responsible for heat transfer to be infinitesimal. A finite temperature difference produces irreversible heat transfer even if the piston itself is frictionless. Friction would also introduce dissipation and destroy reversibility. Sudden load removal creates a finite driving pressure difference and therefore does not represent the reversible limit. Thus reversibility requires both quasi-static driving and absence of dissipative effects. Option B supplies both mechanical and thermal conditions simultaneously.
310. During a reversible isothermal expansion of an ideal gas, the gas absorbs \(500\,\text{J}\) from an ideal thermal reservoir and does \(500\,\text{J}\) of work by raising a weight. Which operation would represent the exact thermodynamic reverse?
ⓐ. Lower the weight to do \(500\,\text{J}\) of work on the gas while it rejects \(500\,\text{J}\) as heat.
ⓑ. Compress the gas without work while it absorbs another \(500\,\text{J}\) from the reservoir.
ⓒ. Allow the weight to remain raised and cool the gas until its original volume is restored.
ⓓ. Compress the gas adiabatically by \(500\,\text{J}\) and leave the reservoir unchanged.
Correct Answer: Lower the weight to do \(500\,\text{J}\) of work on the gas while it rejects \(500\,\text{J}\) as heat.
Explanation: In the forward reversible isothermal expansion,
\[
\Delta U=0
\]
Therefore,
\[
Q=W=+500\,\text{J}
\]
The gas receives heat from the reservoir and raises the weight, changing both parts of the surroundings. An exact reverse must restore both of those changes. The weight must descend and return the same mechanical energy to the gas, so work done by the gas becomes
\[
W_{\text{rev}}=-500\,\text{J}
\]
The reverse process remains isothermal, hence
\[
\Delta U_{\text{rev}}=0
\]
Therefore,
\[
Q_{\text{rev}}=W_{\text{rev}}=-500\,\text{J}
\]
The rejected heat restores the thermal reservoir while the descending weight restores the mechanical surroundings.
\(\textbf{Final answer:}\) Option A.
311. A system undergoes two successive reversible processes:
\[
A\xrightarrow{P}B\xrightarrow{Q}C
\]
Which sequence is required for the exact reverse of the complete transformation?
ⓐ. First reverse \(Q\), then reverse \(P\)
ⓑ. First reverse \(P\), then reverse \(Q\)
ⓒ. Reverse either \(P\) or \(Q\) in any order because both are individually reversible
ⓓ. Return directly from \(C\) to \(A\) by any convenient path
Correct Answer: First reverse \(Q\), then reverse \(P\)
Explanation: The forward sequence changes the system from \(A\) to \(B\) through process \(P\), and then from \(B\) to \(C\) through process \(Q\). To retrace the same complete history, the last forward stage must be undone first. Therefore the reverse begins with
\[
C\rightarrow B
\]
by the exact reverse of \(Q\). Only after state \(B\) has been restored can the exact reverse of \(P\) take the system from
\[
B\rightarrow A
\]
This is analogous to reversing any ordered sequence of changes: the operations must be undone in reverse order. Merely choosing some different path from \(C\) to \(A\) may restore the system but need not restore the surroundings. Individual reversibility of \(P\) and \(Q\) does not make their order irrelevant. Hence the exact reverse sequence is \(Q^{-1}\) followed by \(P^{-1}\).
312. A system goes from state \(A\) to state \(B\) along reversible Path P, absorbing \(800\,\text{J}\) of heat and doing \(300\,\text{J}\) of work. It reaches the same state \(B\) from the same state \(A\) along an irreversible Path Q on which it does \(100\,\text{J}\) of work. How much heat is supplied along Path Q?
ⓐ. \(300\,\text{J}\)
ⓑ. \(500\,\text{J}\)
ⓒ. \(600\,\text{J}\)
ⓓ. \(800\,\text{J}\)
Correct Answer: \(600\,\text{J}\)
Explanation: \(\textbf{Fix the state-function change from Path P:}\)
\[
\Delta U_{AB}=Q_P-W_P
\]
Therefore,
\[
\Delta U_{AB}=800-300=500\,\text{J}
\]
Reversibility does not alter the fact that internal energy is a state function. Path Q has the same initial and final states, so
\[
\Delta U_Q=500\,\text{J}
\]
\(\textbf{Apply the first law to the irreversible path:}\)
\[
\Delta U_Q=Q_Q-W_Q
\]
Thus,
\[
500=Q_Q-100
\]
Hence,
\[
Q_Q=600\,\text{J}
\]
The two paths can have different heat and work values while producing the same endpoint change in internal energy. Reversibility is therefore a property of the process, not of the pair of endpoint states.
\(\textbf{Final answer:}\)
\[
\boxed{Q_Q=600\,\text{J}}
\]
313. A body is to be heated from \(300\,\text{K}\) to \(400\,\text{K}\). Method P places it directly in contact with a \(400\,\text{K}\) reservoir. Method Q successively uses reservoirs at \(301\,\text{K},302\,\text{K},\ldots,400\,\text{K}\). Which statement is most accurate?
ⓐ. P is reversible but Q is irreversible because Q involves many reservoirs.
ⓑ. Both are exactly reversible because the final temperature is the same.
ⓒ. Q is nearer reversibility than P, but even its \(1\,\text{K}\) heat-transfer steps remain irreversible.
ⓓ. Q is exactly reversible because \(1\,\text{K}\) is numerically small.
Correct Answer: Q is nearer reversibility than P, but even its \(1\,\text{K}\) heat-transfer steps remain irreversible.
Explanation: Method P begins with a large finite temperature difference of \(100\,\text{K}\), so the heat transfer is clearly irreversible. Method Q greatly reduces the thermal driving difference at each stage. However, each \(1\,\text{K}\) difference is still finite rather than infinitesimal. Therefore Method Q should not be called exactly reversible merely because the steps are small. The reversible limit is approached when the temperature difference at each stage tends toward an infinitesimal value. In that limiting construction the system remains arbitrarily close to thermal equilibrium with the surroundings. Increasing the number of intermediate reservoirs can therefore make the process progressively closer to that idealisation. Exact endpoint equality does not determine reversibility. Hence option C correctly distinguishes a finite-step approximation from the reversible limit.
314. A frictionless piston confines a gas at pressure \(P\). Which operation best illustrates the mechanical reversal criterion for a reversible expansion?
ⓐ. Remove a finite load so that the gas expands rapidly, then replace the same load suddenly.
ⓑ. Remove and then replace an infinitesimal load, retracing almost the same equilibrium states.
ⓒ. Remove the entire load, allow free expansion, then compress the gas along any path.
ⓓ. Keep the external pressure exactly equal to \(P\) and expect a finite expansion without changing any condition.
Correct Answer: Remove and then replace an infinitesimal load, retracing almost the same equilibrium states.
Explanation: A reversible mechanical process is driven by an infinitesimal departure from mechanical equilibrium. If a tiny load is removed, the external pressure becomes only infinitesimally smaller than the gas pressure. The gas then expands through a correspondingly small change while remaining arbitrarily close to equilibrium. Replacing that same infinitesimal load changes the pressure ordering by an infinitesimal amount in the opposite direction. The piston can then move backward through nearly the same equilibrium states. A finite load change creates a finite pressure imbalance and an irreversible departure from quasi-static behaviour. Free expansion is still more strongly irreversible because the gas expands without a controlled opposing pressure. Exact equality of pressures gives no net pressure force to drive finite motion. Thus option B captures the reversible limiting idea.
315. Consider the following statements.
Statement I: A reversible process may involve nonzero heat transfer.
Statement II: A reversible process may involve nonzero work.
Statement III: Every adiabatic process is reversible.
Statement IV: An isothermal process can be irreversible if heat transfer occurs through a finite temperature difference.
Which combination is correct?
ⓐ. I and II only
ⓑ. I, II, III and IV
ⓒ. II, III and IV only
ⓓ. I, II and IV only
Correct Answer: I, II and IV only
Explanation: Statement I is correct because reversible isothermal expansion is a standard example involving heat transfer. Statement II is also correct because the same reversible expansion can perform mechanical work. Reversibility therefore does not mean that all energy transfers vanish. Statement III is incorrect because adiabatic only means
\[
Q=0.
\]
Free expansion inside an insulated enclosure is adiabatic but irreversible. Statement IV is correct because constant system temperature does not remove the irreversibility produced by a finite temperature difference across which heat flows. A process must satisfy both near-equilibrium and non-dissipative conditions to be reversible. Hence Statements I, II and IV only are correct.
316. Two bodies at \(360\,\text{K}\) and \(300\,\text{K}\) are placed in direct thermal contact inside an otherwise isolated enclosure and eventually reach a common temperature. Which statement about the process is correct?
ⓐ. It is reversible because total energy of the two-body system is conserved.
ⓑ. It is irreversible: heat transfer across a finite temperature difference cannot spontaneously undo itself.
ⓒ. It is reversible because the final state is thermal equilibrium.
ⓓ. It is irreversible only if the two bodies have different masses.
Correct Answer: It is irreversible: heat transfer across a finite temperature difference cannot spontaneously undo itself.
Explanation: The total energy of the isolated two-body composite is conserved, so the first law is satisfied. However, the initial temperature difference is finite:
\[
360\,\text{K}-300\,\text{K}=60\,\text{K}
\]
Heat therefore flows spontaneously from the hotter body to the colder body. The bodies approach a common equilibrium temperature. Once equilibrium has been reached, the system does not spontaneously recreate the original \(360\,\text{K}\) and \(300\,\text{K}\) states while leaving everything else unchanged. Restoring that temperature separation would require additional external effects. Therefore the original heat-transfer process fails the reversibility criterion. Energy conservation and arrival at equilibrium do not make a process reversible. Hence option B is correct.
317. A gas in a thermally insulated, frictionless piston-cylinder initially has pressure \(5P_0\), while the external pressure is suddenly changed to \(2P_0\). The gas expands until a new equilibrium is reached. Which statement is correct?
ⓐ. The process is reversible because the piston is frictionless and the cylinder is insulated.
ⓑ. The process is reversible because both the initial and final states are equilibrium states.
ⓒ. It is irreversible because a finite pressure difference drives expansion despite zero friction and heat flow.
ⓓ. The process is isothermal because no heat is transferred.
Correct Answer: It is irreversible because a finite pressure difference drives expansion despite zero friction and heat flow.
Explanation: Thermal insulation gives
\[
Q=0,
\]
so the process is adiabatic. A frictionless piston removes one possible dissipative effect, but it does not guarantee reversibility. Immediately after the external pressure is changed, the mechanical driving difference is
\[
5P_0-2P_0=3P_0,
\]
which is finite. The gas therefore undergoes a non-quasi-static expansion rather than passing through states arbitrarily close to mechanical equilibrium. The fact that equilibrium states exist before and after the process does not determine the character of the path between them. Zero heat transfer also does not imply constant temperature. Thus the finite pressure jump makes the process irreversible, and option C is correct.
318. A thermally insulated gas is compressed very slowly by a piston, but friction between the piston and cylinder is appreciable. During the compression, \(400\,\text{J}\) of net mechanical work is done on the gas. Which statement is correct?
ⓐ. \(\Delta U=-400\,\text{J}\), and the process is reversible because it is slow.
ⓑ. \(\Delta U=+400\,\text{J}\), and the process can still be irreversible because friction is dissipative.
ⓒ. \(\Delta U=0\), because thermal insulation prevents any energy transfer.
ⓓ. \(\Delta U=+400\,\text{J}\), so the process must be reversible by the first law.
Correct Answer: \(\Delta U=+400\,\text{J}\), and the process can still be irreversible because friction is dissipative.
Explanation: \(\textbf{Energy balance:}\)
Thermal insulation gives
\[
Q=0.
\]
Work is done on the gas, so under the convention that work done by the gas is positive,
\[
W=-400\,\text{J}
\]
The first law gives
\[
\Delta U=Q-W
\]
Hence,
\[
\Delta U=0-(-400)=+400\,\text{J}
\]
\(\textbf{Process classification:}\)
The numerical energy balance says nothing by itself about reversibility. Very slow motion can make the process approximately quasi-static, but appreciable friction still causes dissipation. Reversing the piston motion would not automatically undo that frictional degradation while restoring the surroundings. Therefore the process can be quasi-static-looking yet irreversible.
\(\textbf{Final answer:}\) Option B.
319. Two different ideal gases occupy separate halves of a rigid container at the same temperature and pressure. The partition is removed and the gases spontaneously mix. Which statement is correct?
ⓐ. The mixing is reversible because there is no initial temperature difference.
ⓑ. Mixing is irreversible because the gases do not spontaneously separate into their original compartments.
ⓒ. The mixing is reversible because the container is rigid and therefore \(W=0\).
ⓓ. The mixing is necessarily adiabatic and therefore reversible.
Correct Answer: Mixing is irreversible because the gases do not spontaneously separate into their original compartments.
Explanation: No temperature difference is needed for diffusion to occur when different gases are initially separated. Once the partition is removed, the gases spontaneously spread through the available volume and mix. The reverse event, in which the mixed gases spontaneously separate into exactly their original compartments, is not observed as the natural reverse process. Restoring the original separated state would require some additional external action or compensating change. Therefore the mixing fails the complete-restoration criterion for reversibility. The facts that the container is rigid or that pressure-volume work may vanish do not determine reversibility. Similarly, even if the arrangement were thermally insulated, adiabatic does not mean reversible. The essential feature is spontaneous mixing and the absence of a spontaneous exact reverse. Hence option B is correct.
320. Match each process in List I with the principal feature responsible for its irreversibility in List II.
| List I | List II |
|---|
| P. Heat conduction between bodies at substantially different temperatures | 1. Finite temperature gradient |
| Q. Sudden gas expansion against a much lower external pressure | 2. Finite pressure imbalance |
| R. Slow piston motion with rubbing surfaces | 3. Frictional dissipation |
| S. Spontaneous mixing of initially separated gases | 4. Diffusion and spontaneous mixing |
ⓐ. P-1, Q-2, R-3, S-4
ⓑ. P-2, Q-1, R-4, S-3
ⓒ. P-3, Q-4, R-1, S-2
ⓓ. P-4, Q-3, R-2, S-1
Correct Answer: P-1, Q-2, R-3, S-4
Explanation: Process P is driven by a finite temperature difference, so its relevant source of irreversibility is a finite thermal gradient. Process Q is driven by a finite pressure difference between the gas and surroundings. That causes a rapid non-equilibrium expansion rather than a reversible quasi-static change. Process R may be slow, but rubbing surfaces introduce friction and dissipate mechanical energy. Thus its irreversibility arises from dissipation rather than from speed alone. Process S is spontaneous intermixing of initially separated gases and represents diffusion-driven irreversibility. These examples show that irreversible processes have several physical mechanisms rather than one universal appearance. The correct matching is therefore P-1, Q-2, R-3, S-4.
