401. The same ideal gas starts from \((P_0,V_0,T_0)\) and is expanded until its pressure becomes \(P_0/4\) by two reversible paths.
Path I: isothermal expansion.
Path II: adiabatic expansion with \(\gamma=2\).
The two final gas samples are then placed in thermal contact. Which statement is correct?
ⓐ. Both final temperatures equal \(T_0\), so no heat initially flows.
ⓑ. Both final pressures are equal, so their final temperatures must also be equal.
ⓒ. Path I ends at \(T_0/4\), Path II ends at \(T_0\), so heat initially flows from Path II to Path I.
ⓓ. Path I ends at \(T_0\), Path II at \(T_0/2\); heat initially flows from I to II.
Correct Answer: Path I ends at \(T_0\), Path II at \(T_0/2\); heat initially flows from I to II.
Explanation: \(\textbf{Isothermal path:}\)
By definition,
\[
T_{I,f}=T_0.
\]
\(\textbf{Adiabatic path:}\)
For \(\gamma=2\),
\[
P_0V_0^2=\frac{P_0}{4}V_f^2.
\]
Therefore,
\[
V_f=2V_0.
\]
Using the ideal-gas relation,
\[
\frac{T_{II,f}}{T_0}
=\frac{P_fV_f}{P_0V_0}.
\]
Thus,
\[
\frac{T_{II,f}}{T_0}
=\frac{(P_0/4)(2V_0)}{P_0V_0}
=\frac12.
\]
Hence,
\[
T_{II,f}=\frac{T_0}{2}.
\]
Although the two final pressures are the same, their volumes and temperatures are different. When the samples are brought into thermal contact,
\[
T_{I,f}\gt T_{II,f},
\]
so heat initially flows from the isothermal-path sample to the adiabatic-path sample.
\(\textbf{Final answer:}\) Heat initially flows from Path I to Path II.
402. A cyclic device follows a clockwise closed path on a \(P-V\) diagram and the enclosed area is \(300\,\text{J}\). The device returns to its original thermodynamic state after every cycle. It is claimed that its only external thermal interaction is absorption of \(300\,\text{J}\) from a single reservoir, all of which appears as work output. Which assessment is correct?
ⓐ. The device violates the first law because a cycle cannot produce work.
ⓑ. The device satisfies the first-law energy balance but violates the Kelvin-Planck statement of the second law.
ⓒ. The device violates the Clausius statement because heat flows from hot to cold.
ⓓ. The device satisfies both laws because the area enclosed by the cycle equals the absorbed heat.
Correct Answer: The device satisfies the first-law energy balance but violates the Kelvin-Planck statement of the second law.
Explanation: A clockwise \(P-V\) cycle has positive net work equal to its enclosed area:
\[
W_{\text{net}}=+300\,\text{J}.
\]
Because the device returns to its original state,
\[
\Delta U_{\text{cycle}}=0.
\]
The first law therefore requires
\[
Q_{\text{net}}=W_{\text{net}}.
\]
The claimed heat input is
\[
Q_{\text{net}}=+300\,\text{J},
\]
so the numerical energy balance is perfectly consistent. The problem appears only when the complete external result is examined. The sole claimed result is absorption of heat from one reservoir and complete conversion of that heat into work. That is exactly the cyclic outcome prohibited by the Kelvin-Planck statement. Thus first-law consistency does not guarantee second-law possibility.
403. An ideal gas sample \(G\) is initially in thermal equilibrium with a large reservoir \(R\) at temperature \(T_0\) and has volume \(V_0\). The gas then undergoes the following sequence:
Stage I: reversible isothermal expansion to \(2V_0\);
Stage II: insulated free expansion to \(4V_0\);
Stage III: reversible adiabatic compression back to \(V_0\), with \(\gamma=3/2\).
Immediately after Stage III, \(G\) is again placed in thermal contact with reservoir \(R\). Which statement is correct?
ⓐ. \(T_G=T_0/2\), so heat initially flows from \(R\) to \(G\).
ⓑ. \(T_G=T_0\), so no heat initially flows.
ⓒ. \(T_G=2T_0\), so heat initially flows from \(G\) to \(R\).
ⓓ. \(T_G=4T_0\), so heat initially flows from \(G\) to \(R\).
Correct Answer: \(T_G=2T_0\), so heat initially flows from \(G\) to \(R\).
Explanation: \(\textbf{Stage I — isothermal expansion:}\)
The temperature remains
\[
T_1=T_0.
\]
\(\textbf{Stage II — insulated free expansion:}\)
For free expansion into vacuum,
\[
Q_2=0,\qquad W_2=0.
\]
Therefore,
\[
\Delta U_2=0.
\]
For the ideal gas,
\[
T_2=T_0.
\]
At the end of Stage II,
\[
V_2=4V_0.
\]
\(\textbf{Stage III — reversible adiabatic compression:}\)
Using
\[
TV^{\gamma-1}=\text{constant},
\]
\[
T_0(4V_0)^{1/2}=T_3(V_0)^{1/2}.
\]
Hence,
\[
T_3=2T_0.
\]
The reservoir remains at \(T_0\). Therefore, when thermal contact is restored,
\[
T_G\gt T_R,
\]
so spontaneous heat transfer initially occurs from the gas to the reservoir.
\(\textbf{Final answer:}\)
\[
T_G=2T_0,
\]
and heat initially flows from \(G\) to \(R\).
404. A fixed amount of ideal gas is initially at \(A:(P_0/2,\,2V_0,\,T_0)\) and is compressed isothermally to \(B:(P_0,\,V_0,\,T_0)\). Process R is a reversible isothermal compression. Process I is an irreversible compression against a constant external pressure \(P_0\). Which comparison is correct?
ⓐ. \(|W_R|\gt |W_I|\) and \(|Q_R|\gt |Q_I|\)
ⓑ. \(Q_R=Q_I=0\) because both processes are isothermal
ⓒ. \(W_R=W_I\) because the endpoints are the same
ⓓ. \(|W_R|\lt |W_I|\) and \(|Q_R|\lt |Q_I|\)
Correct Answer: \(|W_R|\lt |W_I|\) and \(|Q_R|\lt |Q_I|\)
Explanation: For both processes the ideal gas starts and ends at the same temperature, so
\[
\Delta U=0.
\]
Also,
\[
nRT_0=P_0V_0.
\]
\(\textbf{Reversible compression:}\)
\[
W_R=nRT_0\ln\left(\frac{V_0}{2V_0}\right).
\]
Hence,
\[
W_R=-P_0V_0\ln2.
\]
\(\textbf{Constant-pressure irreversible compression:}\)
\[
W_I=P_{\text{ext}}(V_f-V_i).
\]
Therefore,
\[
W_I=P_0(V_0-2V_0)=-P_0V_0.
\]
Since
\[
1\gt \ln2,
\]
we have
\[
|W_I|\gt |W_R|.
\]
For each isothermal ideal-gas process,
\[
Q=W.
\]
Hence the irreversible compression also rejects a larger magnitude of heat:
\[
|Q_I|\gt |Q_R|.
\]
The finite pressure difference driving Process I is the source of its mechanical irreversibility.
\(\textbf{Final answer:}\)
\[
|W_R|\lt |W_I|,\qquad |Q_R|\lt |Q_I|.
\]
405. Two different ideal gases initially occupy separate halves of a rigid container at the same temperature and pressure. The partition is removed and the gases mix spontaneously. After complete mixing, the same partition is simply reinserted at its original position. Which statement is correct?
ⓐ. Reinserting the partition exactly reverses the mixing because the original volumes are restored.
ⓑ. The original state is restored because \(Q=W=0\) throughout.
ⓒ. The partition cannot unmix the gases, so each compartment retains a mixture.
ⓓ. The mixing becomes reversible if the partition is reinserted sufficiently slowly.
Correct Answer: The partition cannot unmix the gases, so each compartment retains a mixture.
Explanation: The rigid outer container permits no external pressure-volume work. If it is also thermally insulated, the mixing may occur with no heat exchanged through the outer wall. Neither fact establishes reversibility. Once the partition is removed, the two gases diffuse spontaneously throughout the available volume. Reinserting an ordinary partition only divides the already mixed gas into two portions. It does not sort one gas back into the left compartment and the other gas back into the right compartment. Therefore the original separated state is not restored. Achieving that separation would require an additional physical process with compensating changes elsewhere. Simply reversing the mechanical control action of “remove partition” to “insert partition” is not the thermodynamic reverse of diffusion. This is a standard example of irreversibility even when external \(Q\) and \(W\) may both be zero.
406. A fixed amount of ideal gas is contained in a rigid vessel. Initially it is at temperature \(T_0\). Stage I is carried out with the vessel thermally insulated while an external paddle wheel does \(500\,\text{J}\) of work on the gas. In Stage II the paddle is disconnected and the rigid vessel is placed in contact with a large reservoir at \(T_0\) until the gas returns to its initial state. Which statement is correct?
ⓐ. Stage II supplies \(+500\,\text{J}\) of heat, and the complete operation is reversible.
ⓑ. Stage II rejects \(500\,\text{J}\); restoring the gas does not restore the surroundings.
ⓒ. Stage II requires no heat because the vessel is rigid.
ⓓ. The complete operation has \(Q_{\text{net}}=W_{\text{net}}=0\).
Correct Answer: Stage II rejects \(500\,\text{J}\); restoring the gas does not restore the surroundings.
Explanation: \(\textbf{Paddle-work stage:}\)
The vessel is insulated, so
\[
Q_1=0.
\]
Work is done on the gas:
\[
W_1=-500\,\text{J}.
\]
Therefore,
\[
\Delta U_1=Q_1-W_1=+500\,\text{J}.
\]
The ideal-gas temperature rises above \(T_0\).
\(\textbf{Return to the original gas state:}\)
Stage II occurs at constant volume, so
\[
W_2=0.
\]
To return the gas to its initial temperature and internal energy,
\[
\Delta U_2=-500\,\text{J}.
\]
Hence,
\[
Q_2=-500\,\text{J}.
\]
The gas has returned to its original state, but the work source has lost \(500\,\text{J}\) and the reservoir has gained \(500\,\text{J}\). The surroundings have therefore not been restored. Restoring the system alone does not make the original dissipative history reversible.
\(\textbf{Final answer:}\)
\[
Q_2=-500\,\text{J}.
\]
407. A gas undergoes a reversible adiabatic expansion from \(A\) to \(B\), doing \(500\,\text{J}\) of work. It is then returned from \(B\) to the same state \(A\) by a slow compression in a piston-cylinder with friction. During this return path, the work done by the gas is \(-700\,\text{J}\). What are the heat transfer during \(B\rightarrow A\) and the net work over the complete cycle?
ⓐ. \(Q_{BA}=+200\,\text{J},\ W_{\text{cycle}}=-200\,\text{J}\)
ⓑ. \(Q_{BA}=-200\,\text{J},\ W_{\text{cycle}}=-200\,\text{J}\)
ⓒ. \(Q_{BA}=-700\,\text{J},\ W_{\text{cycle}}=+200\,\text{J}\)
ⓓ. \(Q_{BA}=0,\ W_{\text{cycle}}=0\)
Correct Answer: \(Q_{BA}=-200\,\text{J},\ W_{\text{cycle}}=-200\,\text{J}\)
Explanation: \(\textbf{Forward reversible adiabatic leg:}\)
\[
Q_{AB}=0
\]
The gas does
\[
W_{AB}=+500\,\text{J}
\]
Hence,
\[
\Delta U_{AB}=Q_{AB}-W_{AB}=-500\,\text{J}
\]
\(\textbf{Return to the original state:}\)
Because the system returns from \(B\) to \(A\),
\[
\Delta U_{BA}=+500\,\text{J}
\]
The work done by the gas on the return path is
\[
W_{BA}=-700\,\text{J}
\]
Using
\[
Q_{BA}=\Delta U_{BA}+W_{BA},
\]
we obtain
\[
Q_{BA}=500-700=-200\,\text{J}
\]
The net cycle work is
\[
W_{\text{cycle}}=500-700=-200\,\text{J}
\]
Since
\[
\Delta U_{\text{cycle}}=0,
\]
the cycle also has
\[
Q_{\text{cycle}}=-200\,\text{J}
\]
The extra mechanical input associated with the frictional return is ultimately rejected as heat, so the cycle is irreversible.
\(\textbf{Final answer:}\)
\[
\boxed{Q_{BA}=-200\,\text{J},\qquad W_{\text{cycle}}=-200\,\text{J}}
\]
408. A gas initially in mechanical equilibrium under a piston has pressure \(P_0\). The external pressure is suddenly reduced to \(0.6P_0\), causing expansion. After a new equilibrium is reached, the external pressure is suddenly restored to \(P_0\). A student claims that the second pressure change is the exact thermodynamic reverse of the first. Which statement is correct?
ⓐ. The student is correct because the external pressure ultimately returns to its original value.
ⓑ. The student is correct provided the piston is frictionless.
ⓒ. The student is incorrect: reversing the external pressure change cannot undo a finite-imbalance path.
ⓓ. The student is incorrect only if heat crosses the cylinder wall.
Correct Answer: The student is incorrect: reversing the external pressure change cannot undo a finite-imbalance path.
Explanation: The first change produces a finite pressure imbalance:
\[
P_{\text{gas}}-P_{\text{ext}}=0.4P_0
\]
immediately after the pressure reduction. The resulting expansion is therefore not in the reversible quasi-static limit. Restoring the external pressure to its original numerical value does not undo the detailed nonequilibrium history of the expansion. The return motion is itself driven by another finite pressure imbalance. The system need not pass through the same intermediate thermodynamic states in reverse order. Any mechanical energy dissipated during settling or other environmental changes are also not automatically restored. Reversibility requires an infinitesimal reversal of the driving conditions together with complete restoration of both system and surroundings. Simply reversing the direction of a finite control change is therefore not enough.
409. A metal body initially at \(300\,\text{K}\) is placed directly in contact with a large reservoir at \(350\,\text{K}\) until the body reaches \(350\,\text{K}\). It is then placed directly in contact with another large reservoir at \(300\,\text{K}\) until it returns to its original temperature. Which statement is correct?
ⓐ. The complete operation is reversible because the body returns to \(300\,\text{K}\).
ⓑ. The complete operation is reversible because the heat gained in the first stage can equal the heat lost in the second.
ⓒ. The second stage exactly reverses the first because the direction of heat transfer is reversed.
ⓓ. Both finite-temperature-difference transfers are irreversible, even though the body is restored.
Correct Answer: Both finite-temperature-difference transfers are irreversible, even though the body is restored.
Explanation: The first stage begins with a finite temperature difference of \(50\,\text{K}\). Heat therefore flows spontaneously from the \(350\,\text{K}\) reservoir to the colder body. The second stage again begins with a finite temperature difference, now driving heat from the body to the \(300\,\text{K}\) reservoir. Returning the body to \(300\,\text{K}\) restores only the system state. The first reservoir has lost energy and the second reservoir has gained energy. Those reservoir changes do not cancel merely because the body ends where it started. An exact reversible return would have to restore the reservoirs as well, with no other net change. Reversing the heat-flow direction in a second irreversible transfer is not the same as reversing the original thermodynamic process. Hence the complete operation is irreversible.
410. Two chambers of a rigid, thermally insulated vessel contain the same ideal gas at the same temperature but at different pressures. A valve between the chambers is opened, and the gas flows until a new equilibrium is established. Which statement is correct?
ⓐ. The process is reversible because the initial temperatures are equal.
ⓑ. The process is reversible because the outer vessel has \(Q=0\) and \(W=0\).
ⓒ. The finite pressure imbalance driving the flow makes the process irreversible.
ⓓ. The process violates the first law because the pressure changes without external work.
Correct Answer: The finite pressure imbalance driving the flow makes the process irreversible.
Explanation: The rigid outer vessel has no boundary-volume change, so external pressure-volume work is zero. Thermal insulation also gives zero heat exchange through the outer boundary. These conditions are fully compatible with conservation of the total energy of the gas. However, the two chambers begin with a finite pressure difference. Opening the valve therefore produces a directed gas flow through a finite mechanical driving gradient. During the equalisation, the system is not a sequence of states infinitesimally close to mechanical equilibrium. Real flow may also involve viscous and turbulent dissipation. After equilibrium is reached, simply reversing the valve operation does not spontaneously recreate the original pressure difference. Therefore the process is irreversible even though the outer-boundary values \(Q=0\) and \(W=0\) may hold.
411. In an experiment on an insulated frictionless piston-cylinder, pressure sensors show that throughout an expansion the gas pressure exceeds the external pressure by a constant finite amount \(0.2P_0\). Which conclusion is correct?
ⓐ. The process is reversible because there is no friction.
ⓑ. The process is reversible because \(Q=0\).
ⓒ. The process is adiabatic but irreversible owing to a finite pressure difference.
ⓓ. The process is isothermal because the pressure difference is constant.
Correct Answer: The process is adiabatic but irreversible owing to a finite pressure difference.
Explanation: Insulation establishes the condition
\[
Q=0,
\]
so the process is adiabatic. The frictionless piston removes one possible source of dissipation. However, the measured pressure difference is
\[
P_{\text{gas}}-P_{\text{ext}}=0.2P_0,
\]
which is finite rather than infinitesimal. The piston is therefore driven by a finite mechanical force imbalance. That means the gas is not passing through a succession of states arbitrarily close to mechanical equilibrium with its surroundings. Mechanical reversibility requires the pressure difference to approach zero while retaining the appropriate infinitesimal sign to drive the motion. Adiabaticity and absence of friction do not by themselves guarantee reversibility. Hence the process is adiabatic but irreversible.
412. Consider the following statements about a reversible thermodynamic process.
Statement I: Nonzero heat transfer is possible if it occurs through an infinitesimal temperature difference in the reversible limit.
Statement II: Nonzero work is possible if mechanical motion occurs through an infinitesimal pressure imbalance without dissipation.
Statement III: Any quasi-static process is reversible even when friction is present.
Statement IV: Any adiabatic process is reversible because \(Q=0\).
Which combination is correct?
ⓐ. I and II only
ⓑ. I, II and III only
ⓒ. II, III and IV only
ⓓ. I, II, III and IV
Correct Answer: I and II only
Explanation: Statement I is correct. Reversible processes need not have zero heat transfer; reversible isothermal expansion is a standard example. What is required is that the thermal driving temperature difference be infinitesimal in the reversible idealisation. Statement II is also correct because reversible mechanical processes can perform finite total work through a succession of infinitesimal pressure imbalances. Statement III is incorrect. Quasi-static behaviour keeps the system close to equilibrium, but friction can still dissipate mechanical energy and make the process irreversible. Statement IV is also incorrect because adiabaticity means only
\[
Q=0.
\]
Insulated free expansion is a direct example of an adiabatic but irreversible process. Hence only Statements I and II are correct.
413. A cyclic device absorbs \(800\,\text{J}\) of heat from a hot reservoir, rejects \(500\,\text{J}\) to a colder reservoir and produces \(300\,\text{J}\) of work. Measurements also show appreciable friction in its moving parts. Which conclusion is most appropriate?
ⓐ. The device is reversible because its first-law balance is exact.
ⓑ. The device is reversible because it exchanges heat with two reservoirs.
ⓒ. The device violates the Kelvin-Planck statement because it converts some heat into work.
ⓓ. The first law holds, but dissipative friction makes the actual process irreversible.
Correct Answer: The first law holds, but dissipative friction makes the actual process irreversible.
Explanation: The cyclic energy balance is
\[
800=500+300.
\]
Thus,
\[
Q_{\text{net}}=300\,\text{J}=W_{\text{net}},
\]
which is exactly what the first law requires for a cycle with
\[
\Delta U_{\text{cycle}}=0.
\]
These numerical flows therefore do not violate energy conservation. They also do not have the forbidden Kelvin-Planck form because the device rejects heat to a colder reservoir rather than converting all absorbed heat into work. However, first-law consistency does not prove reversibility. The observed friction is a dissipative mechanism. It prevents complete restoration of both device and surroundings through an infinitesimal reversal of the operating conditions. Therefore the actual device is irreversible despite having a perfectly valid energy balance.
414. A reversible cyclic device operates forward by absorbing \(700\,\text{J}\) from a hot reservoir, rejecting \(400\,\text{J}\) to a cold reservoir and producing \(300\,\text{J}\) of work. What must occur when the same device is operated through its exact reverse cycle?
ⓐ. It absorbs \(700\,\text{J}\) from the hot reservoir and consumes \(300\,\text{J}\) of work.
ⓑ. The device absorbs \(400\,\text{J}\) of heat from cold, consumes \(300\,\text{J}\) of work, and rejects \(700\,\text{J}\) to hot.
ⓒ. It transfers \(400\,\text{J}\) from cold to hot with no work input.
ⓓ. It produces \(300\,\text{J}\) of work while transferring \(400\,\text{J}\) from cold to hot.
Correct Answer: The device absorbs \(400\,\text{J}\) of heat from cold, consumes \(300\,\text{J}\) of work, and rejects \(700\,\text{J}\) to hot.
Explanation: A reversible cycle can be retraced exactly. Every energy interaction therefore reverses direction while retaining the same magnitude. In the forward cycle, the device receives
\[
700\,\text{J}
\]
from the hot reservoir. In the reverse cycle it must return
\[
700\,\text{J}
\]
to that reservoir. The forward device rejects
\[
400\,\text{J}
\]
to the cold reservoir, so the reverse device absorbs
\[
400\,\text{J}
\]
from it. The forward work output of
\[
300\,\text{J}
\]
becomes a reverse work input of the same magnitude. The reverse balance is therefore
\[
400+300=700\,\text{J}.
\]
Cold-to-hot transfer here is accompanied by work input, so it is not the Clausius-forbidden sole result.
415. An ideal gas is initially at \(P_1=3.0\times10^5\,\text{Pa}\), \(V_1=1.0\,\text{L}\) and fixed temperature \(T\). It expands isothermally to \(V_f=3.0\,\text{L}\). Compare three methods:
Method X: one sudden expansion against constant external pressure \(1.0\times10^5\,\text{Pa}\);
Method Y: first expand against \(1.5\times10^5\,\text{Pa}\) until equilibrium, then against \(1.0\times10^5\,\text{Pa}\) until \(3.0\,\text{L}\);
Method Z: reversible isothermal expansion.
Take \(\ln3=1.099\). Which work ordering and values are correct?
ⓐ. \(W_Z=329.7\,\text{J}\gt W_Y=250\,\text{J}\gt W_X=200\,\text{J}\)
ⓑ. \(W_X=600\,\text{J}\gt W_Y=450\,\text{J}\gt W_Z=329.7\,\text{J}\)
ⓒ. \(W_Y=329.7\,\text{J}\gt W_Z=250\,\text{J}\gt W_X=200\,\text{J}\)
ⓓ. \(W_X=W_Y=W_Z=300\,\text{J}\)
Correct Answer: \(W_Z=329.7\,\text{J}\gt W_Y=250\,\text{J}\gt W_X=200\,\text{J}\)
Explanation: The isothermal constant is
\[
nRT=P_1V_1.
\]
Thus,
\[
nRT=(3.0\times10^5)(1.0\times10^{-3})
=300\,\text{J}.
\]
\(\textbf{One-step irreversible expansion X:}\)
\[
W_X=P_{\text{ext}}(V_f-V_1).
\]
Hence,
\[
W_X=(1.0\times10^5)(2.0\times10^{-3})
=200\,\text{J}.
\]
\(\textbf{Two-step expansion Y:}\)
At
\[
P=1.5\times10^5\,\text{Pa},
\]
the isothermal equilibrium volume is
\[
V=\frac{300}{1.5\times10^5}=2.0\times10^{-3}\,\text{m}^3=2.0\,\text{L}.
\]
The first step gives
\[
W_{Y1}=(1.5\times10^5)(1.0\times10^{-3})
=150\,\text{J}.
\]
The second step gives
\[
W_{Y2}=(1.0\times10^5)(1.0\times10^{-3})
=100\,\text{J}.
\]
Therefore,
\[
W_Y=250\,\text{J}.
\]
\(\textbf{Reversible limit Z:}\)
\[
W_Z=nRT\ln3.
\]
Thus,
\[
W_Z=300(1.099)=329.7\,\text{J}.
\]
Using more, smaller pressure steps moves the process toward the reversible path and increases the work obtained from the expansion.
\(\textbf{Final answer:}\)
\[
W_Z\gt W_Y\gt W_X.
\]
416. A gas is driven very slowly from equilibrium state \(A\) to equilibrium state \(B\) along a well-defined \(P-V\) path while its piston experiences friction. An external controller then drives the gas very slowly from \(B\) back to \(A\) along the same sequence of gas \(P,V\) states in reverse. Which statement is correct?
ⓐ. The complete process is necessarily reversible because the gas retraces its \(P-V\) curve.
ⓑ. The process is reversible because slow motion eliminates frictional dissipation.
ⓒ. The process is reversible provided the net change in gas internal energy is zero.
ⓓ. Friction can make the process irreversible despite retracing the gas's \(P-V\) path.
Correct Answer: Friction can make the process irreversible despite retracing the gas's \(P-V\) path.
Explanation: Retracing the thermodynamic states of the system is necessary for an exact reversible return, but it is not by itself sufficient. The piston experiences friction in both directions. That friction converts organised mechanical energy into dissipative heating. An external controller can force the gas through the same \(P,V\) states in reverse, but it must supply or remove additional mechanical energy associated with the frictional resistance. Thus the surroundings do not automatically return to their original condition. The gas may have
\[
\Delta U_{\text{cycle}}=0,
\]
yet the external work source and thermal surroundings can retain net changes. Thermodynamic reversibility demands complete restoration of system and surroundings, not merely retracing a curve on a state diagram. Hence the described cycle can be geometrically retraced by the gas and still remain irreversible.
417. A rigid thermally insulated container has two compartments separated by a removable partition. The left compartment contains gas \(A\) at \(400\,\text{K}\), and the right compartment contains a different gas \(B\) at \(300\,\text{K}\). The partition is removed and the gases are allowed to reach complete equilibrium. Which statement best describes the process?
ⓐ. It is reversible because the outer container has \(Q_{\text{ext}}=W_{\text{ext}}=0\).
ⓑ. It is irreversible only because the gases have different temperatures; mixing itself is reversible.
ⓒ. Both finite-temperature heat flow and spontaneous mixing make the isolated process irreversible.
ⓓ. It violates the first law because two irreversible mechanisms act simultaneously.
Correct Answer: Both finite-temperature heat flow and spontaneous mixing make the isolated process irreversible.
Explanation: The rigid outer container performs no external pressure-volume work. Thermal insulation also gives
\[
Q_{\text{ext}}=0.
\]
Therefore the total energy of the enclosed composite can remain conserved. Internally, however, two distinct irreversible mechanisms are present. First, the gases begin at \(400\,\text{K}\) and \(300\,\text{K}\), so energy transfers through a finite temperature difference. Second, the different gases spontaneously diffuse and mix after the partition is removed. Neither process spontaneously reverses when equilibrium is reached. Restoring the original state would require recreating both the temperature separation and the compositional separation, with compensating changes elsewhere. Thus a perfectly first-law-consistent isolated process can contain multiple sources of irreversibility.
418. Match each source of irreversibility in List I with the modification in List II that approaches the corresponding reversible idealisation.
| List I | List II |
|---|
| P. Heat transfer through a finite temperature difference | 1. Reduce the temperature difference at each stage toward an infinitesimal value |
| Q. Expansion driven by a finite pressure imbalance | 2. Reduce the pressure imbalance toward an infinitesimal value |
| R. Piston friction | 3. Eliminate the dissipative rubbing interaction |
| S. Free expansion into vacuum | 4. Replace the uncontrolled expansion by a quasi-static expansion against an opposing pressure |
ⓐ. 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: For P, reversible heat transfer is approached by making the thermal driving difference infinitesimal at each stage, so P-1. For Q, mechanical reversibility requires the pressure difference driving the boundary motion to become infinitesimal, so Q-2. For R, merely slowing the process is not enough while friction remains present. The dissipative rubbing interaction itself must be removed in the reversible idealisation, giving R-3. Free expansion into vacuum has no controlled opposing pressure and passes through non-equilibrium intermediate conditions. Replacing it by a quasi-static expansion against an external pressure that differs only infinitesimally from the gas pressure provides the reversible limiting construction, so S-4. These modifications address different physical causes of irreversibility rather than applying one universal recipe. The correct matching is
\[
P-1,\quad Q-2,\quad R-3,\quad S-4.
\]
