201. A reversible process is best described as an ideal process that
ⓐ. must occur suddenly so that the system has no time to exchange heat
ⓑ. always has zero work and zero heat transfer
ⓒ. retraceable through the same equilibrium states
ⓓ. can occur only when pressure is zero throughout
Correct Answer: retraceable through the same equilibrium states
Explanation: A reversible process is an ideal limiting process. It proceeds through equilibrium states so that the system and surroundings can be restored to their original states by an infinitesimal change. Such a process is closely related to the quasistatic idea but also requires absence of dissipative effects such as friction. Real processes are usually irreversible to some extent. The key idea is retraceability through equilibrium states, not simply slowness by itself.
202. A gas suddenly expands into an evacuated chamber. This process is usually classified as irreversible mainly because
ⓐ. the gas pressure remains uniform at every instant
ⓑ. the gas passes through a sequence of well-defined equilibrium states
ⓒ. not retraceable by an infinitesimal change
ⓓ. the gas does maximum quasistatic work during expansion
Correct Answer: not retraceable by an infinitesimal change
Explanation: Sudden expansion into vacuum is not a quasistatic process. During the expansion, the gas is not in a sequence of well-defined equilibrium states. The process naturally proceeds in one direction and cannot be reversed by an infinitesimal change in external conditions. Such free expansion is a standard example of an irreversible process. The absence of a controllable opposing pressure also means it is not the same as slow piston expansion.
203. Friction makes a mechanical thermodynamic process irreversible because it
ⓐ. changes the unit of work from \(\text{J}\) to \(\text{K}\)
ⓑ. keeps the system in exact equilibrium at every instant
ⓒ. makes heat transfer impossible in all cases
ⓓ. dissipates mechanical energy as internal energy
Correct Answer: dissipates mechanical energy as internal energy
Explanation: Friction is a dissipative effect. It converts organized mechanical energy into internal energy of the bodies in contact and their surroundings. Because of this dissipation, simply reversing the direction of motion does not restore everything to the original condition. A reversible process requires no such dissipative loss. Friction therefore prevents ideal retraceability even if the motion is made slow.
204. Assertion: A reversible thermodynamic process is an idealization.
Reason: Real processes usually involve effects such as friction, finite temperature difference, turbulence, or uncontrolled expansion.
ⓐ. Both Assertion and Reason are true, and Reason explains Assertion
ⓑ. Both Assertion and Reason are true, but Reason does not explain Assertion
ⓒ. Assertion is true, but Reason is false
ⓓ. Assertion is false, but Reason is true
Correct Answer: Both Assertion and Reason are true, and Reason explains Assertion
Explanation: The Assertion is true because perfectly reversible processes are limiting ideal cases. The Reason is also true because real processes usually contain dissipative or non-equilibrium effects. Friction, turbulence, finite temperature difference in heat transfer, and free expansion all introduce irreversibility. These effects prevent exact retracing through the same equilibrium states. The Reason explains why reversible behaviour is treated as an ideal reference rather than as a common real process.
205. Study the table and identify the row that gives a correctly matched process type.
| Row | Process description | Classification |
|---|
| P | Slow frictionless compression through near-equilibrium states | Reversible idealization |
| Q | Free expansion into vacuum | Reversible idealization |
| R | Heat flow through a large finite temperature difference | Perfectly reversible |
| S | Turbulent mixing of two gases | Exactly retraceable by infinitesimal change |
ⓐ. Row P
ⓑ. Row Q
ⓒ. Row R
ⓓ. Row S
Correct Answer: Row P
Explanation: Row P gives the correct match because slow frictionless compression through near-equilibrium states is the ideal kind of process that can approach reversibility. Free expansion into vacuum is irreversible because it is uncontrolled and non-quasistatic. Heat flow through a large finite temperature difference is irreversible because it has a natural direction. Turbulent mixing is also irreversible because the mixed state does not separate back by an infinitesimal change. Reversibility requires much stricter conditions than simply having a process occur.
206. Use the arrangement described below: a gas is enclosed in a cylinder with a frictionless movable piston. In Case 1, the piston is moved outward very slowly by reducing the external pressure in tiny steps. In Case 2, the piston is suddenly released against a much lower external pressure. The better comparison is
ⓐ. neither case can involve work because the piston is frictionless
ⓑ. both cases are equally reversible because the gas expands in both
ⓒ. Case 1 is closer to reversible behaviour, while Case 2 is more irreversible
ⓓ. Case 2 is closer to reversible behaviour, while Case 1 is more irreversible
Correct Answer: Case 1 is closer to reversible behaviour, while Case 2 is more irreversible
Explanation: In Case 1, the external pressure is changed in tiny steps and the piston moves very slowly. This allows the gas to remain close to equilibrium throughout the expansion. If friction and other dissipative effects are absent, the process can approach reversibility. In Case 2, sudden release causes a rapid non-equilibrium expansion, so the process cannot be retraced through the same states by an infinitesimal change. A frictionless piston removes one source of irreversibility, but sudden uncontrolled motion can still make a process irreversible.
207. Specific heat capacity \(c\) of a substance is the heat required
ⓐ. to raise the temperature of unit mass by \(1\,\text{K}\)
ⓑ. to raise the temperature of one mole by \(1\,\text{K}\) only
ⓒ. to double the pressure at constant volume only
ⓓ. to convert all heat into work in a cycle
Correct Answer: to raise the temperature of unit mass by \(1\,\text{K}\)
Explanation: Specific heat capacity \(c\) is defined per unit mass. It tells how much heat is needed to raise the temperature of \(1\,\text{kg}\) of a substance by \(1\,\text{K}\). The relation is \(Q=mc\Delta T\) when \(c\) is constant over the temperature range. Molar heat capacity is different because it is defined per mole. The unit of \(c\) is \(\text{J kg}^{-1}\text{K}^{-1}\), which shows the “per kilogram per kelvin” meaning.
208. Molar heat capacity \(C\) is related to heat \(Q\), number of moles \(n\), and temperature change \(\Delta T\) by
ⓐ. \(C=nQ\Delta T\)
ⓑ. \(C=\frac{Q}{m\Delta T}\)
ⓒ. \(C=\frac{Q}{n\Delta T}\)
ⓓ. \(C=\frac{n\Delta T}{Q}\)
Correct Answer: \(C=\frac{Q}{n\Delta T}\)
Explanation: Molar heat capacity \(C\) is heat required per mole per kelvin temperature change. If \(n\) moles of a substance receive heat \(Q\) and undergo temperature change \(\Delta T\), then \(Q=nC\Delta T\). Rearranging gives \(C=\frac{Q}{n\Delta T}\). The expression \(C=\frac{Q}{m\Delta T}\) represents specific heat capacity \(c\), not molar heat capacity. The symbols \(m\) and \(n\) must not be interchanged because one refers to mass and the other to amount in moles.
209. A \(0.50\,\text{kg}\) metal block has specific heat capacity \(400\,\text{J kg}^{-1}\text{K}^{-1}\). How much heat is needed to raise its temperature by \(20\,\text{K}\)?
ⓐ. \(4000\,\text{J}\)
ⓑ. \(8000\,\text{J}\)
ⓒ. \(10000\,\text{J}\)
ⓓ. \(20000\,\text{J}\)
Correct Answer: \(4000\,\text{J}\)
Explanation: \( \textbf{Given data:} \) \(m=0.50\,\text{kg}\), \(c=400\,\text{J kg}^{-1}\text{K}^{-1}\), and \(\Delta T=20\,\text{K}\).
\( \textbf{Required quantity:} \) Heat \(Q\).
\( \textbf{Specific heat relation:} \)
\[
Q=mc\Delta T
\]
\( \textbf{Substitution:} \)
\[
Q=(0.50)(400)(20)\,\text{J}
\]
\( \textbf{First product:} \)
\[
0.50\times400=200
\]
\( \textbf{Final multiplication:} \)
\[
Q=200\times20=4000\,\text{J}
\]
\( \textbf{Unit check:} \) \(\text{kg}\) and \(\text{kg}^{-1}\) cancel, and \(\text{K}^{-1}\) cancels with \(\text{K}\), leaving \(\text{J}\).
\( \textbf{Final answer:} \) The required heat is \(4000\,\text{J}\).
210. A \(2.0\,\text{mol}\) sample of a substance requires \(1000\,\text{J}\) of heat for a temperature rise of \(10\,\text{K}\). Its molar heat capacity is
ⓐ. \(50\,\text{J mol}^{-1}\text{K}^{-1}\)
ⓑ. \(25\,\text{J mol}^{-1}\text{K}^{-1}\)
ⓒ. \(100\,\text{J mol}^{-1}\text{K}^{-1}\)
ⓓ. \(200\,\text{J mol}^{-1}\text{K}^{-1}\)
Correct Answer: \(50\,\text{J mol}^{-1}\text{K}^{-1}\)
Explanation: \( \textbf{Given data:} \) \(n=2.0\,\text{mol}\), \(Q=1000\,\text{J}\), and \(\Delta T=10\,\text{K}\).
\( \textbf{Required quantity:} \) Molar heat capacity \(C\).
\( \textbf{Molar heat relation:} \)
\[
Q=nC\Delta T
\]
\( \textbf{Solve for \(C\):} \)
\[
C=\frac{Q}{n\Delta T}
\]
\( \textbf{Substitution:} \)
\[
C=\frac{1000}{(2.0)(10)}
\]
\( \textbf{Denominator:} \)
\[
(2.0)(10)=20
\]
\( \textbf{Calculation:} \)
\[
C=\frac{1000}{20}=50\,\text{J mol}^{-1}\text{K}^{-1}
\]
\( \textbf{Final answer:} \) The molar heat capacity is \(50\,\text{J mol}^{-1}\text{K}^{-1}\).
211. For gases, heat capacity depends on the process because
ⓐ. required heat can differ when work is done
ⓑ. gases cannot have internal energy
ⓒ. heat capacity is fixed by temperature alone
ⓓ. temperature change has no meaning for gases
Correct Answer: required heat can differ when work is done
Explanation: For gases, heating may occur at constant volume, constant pressure, or along another path. At constant volume, the gas does no expansion work, so heat mainly changes internal energy. At constant pressure, the gas expands and does work while its temperature rises. Therefore, the heat required for the same temperature rise can depend on the process. This is why gases have different molar heat capacities such as \(C_V\) and \(C_P\).
212. Consider the following statements about heat capacities.
I. \(c=\frac{Q}{m\Delta T}\) is specific heat capacity.
II. \(C=\frac{Q}{n\Delta T}\) is molar heat capacity.
III. The unit of \(C\) is \(\text{J kg}^{-1}\text{K}^{-1}\).
ⓐ. I, II, and III
ⓑ. II and III only
ⓒ. I and III only
ⓓ. I and II only
Correct Answer: I and II only
Explanation: Statement I is true because specific heat capacity is defined per unit mass. Statement II is true because molar heat capacity is defined per mole. Statement III is false because the unit of molar heat capacity is \(\text{J mol}^{-1}\text{K}^{-1}\), not \(\text{J kg}^{-1}\text{K}^{-1}\). The unit \(\text{J kg}^{-1}\text{K}^{-1}\) belongs to specific heat capacity \(c\). The denominator tells whether the heat capacity is being measured per kilogram or per mole.
213. Use the graph description below.
For a fixed \(n\)-mole sample undergoing a specified process, a graph of heat supplied \(Q\) on the vertical axis against temperature rise \(\Delta T\) on the horizontal axis is a straight line through the origin.
If the process has molar heat capacity \(C\), the slope of the graph is
ⓐ. \(\frac{1}{nC}\)
ⓑ. \(n\Delta T\)
ⓒ. \(nC\)
ⓓ. \(\frac{C}{n}\)
Correct Answer: \(nC\)
Explanation: \( \textbf{Heat relation:} \)
\[
Q=nC\Delta T
\]
\( \textbf{Graph axes:} \) The vertical axis is \(Q\), and the horizontal axis is \(\Delta T\).
\( \textbf{Graph form:} \)
\[
Q=(nC)\Delta T
\]
\( \textbf{Slope comparison:} \) This has the straight-line form \(y=mx\).
\( \textbf{Slope:} \)
\[
m=nC
\]
\( \textbf{Meaning:} \) A larger amount \(n\) or larger molar heat capacity \(C\) makes the \(Q\)-\(\Delta T\) line steeper.
\( \textbf{Final answer:} \) The slope is \(nC\).
214. A gas sample is heated by \(30\,\text{K}\) once at constant volume and once at constant pressure. The heat required is not generally the same in the two cases because
ⓐ. constant-volume heating requires greater expansion work
ⓑ. constant pressure means \(\Delta T=0\)
ⓒ. constant-pressure heating includes expansion work
ⓓ. heat is measured in kelvin during heating
Correct Answer: constant-pressure heating includes expansion work
Explanation: At constant volume, the boundary does not move, so the pressure-volume work is zero. The heat supplied then changes internal energy for the gas. At constant pressure, heating usually makes the gas expand, so it does work on the surroundings. Because some supplied heat is used for expansion work, the heat needed for a given temperature rise is different from the constant-volume case. This process dependence is why \(C_P\) and \(C_V\) are separated for gases.
215. The molar heat capacity at constant volume, \(C_V\), is defined by
ⓐ. \(C_V=\left(\frac{Q}{m\Delta T}\right)_P\)
ⓑ. \(C_V=P(V_f-V_i)\)
ⓒ. \(C_V=\frac{W}{Q}\)
ⓓ. \(C_V=\left(\frac{Q}{n\Delta T}\right)_V\)
Correct Answer: \(C_V=\left(\frac{Q}{n\Delta T}\right)_V\)
Explanation: \(C_V\) is the molar heat capacity measured under the condition of constant volume. The subscript \(V\) indicates that the process is carried out at fixed volume. Since it is molar heat capacity, the amount is measured in moles \(n\), not mass \(m\). The relation can be written as \(Q_V=nC_V\Delta T\). The definition is not a work formula; it describes heat needed per mole per kelvin at constant volume.
216. In a constant-volume process for a gas, the work \(W\) is zero because
ⓐ. \(dV=0\), so \(dW=P\,dV=0\)
ⓑ. \(Q=0\) in every constant-volume process
ⓒ. \(P=0\) throughout the process
ⓓ. \(T=0\,\text{K}\) throughout the process
Correct Answer: \(dV=0\), so \(dW=P\,dV=0\)
Explanation: Constant volume means that the system boundary does not move outward or inward. Therefore, the small volume change is \(dV=0\). Since quasistatic pressure-volume work is \(dW=P\,dV\), the work is zero even if pressure is not zero. Heat may still enter or leave at constant volume. In fact, when heat is supplied at constant volume, it changes the internal energy rather than producing expansion work.
217. For an ideal gas heated at constant volume, the first law gives
ⓐ. \(Q_V=P\Delta V\) with \(\Delta V\ne0\)
ⓑ. \(Q_V=W\) with \(\Delta U=0\)
ⓒ. \(Q_V=0\) for every temperature change
ⓓ. \(Q_V=\Delta U=nC_V\Delta T\)
Correct Answer: \(Q_V=\Delta U=nC_V\Delta T\)
Explanation: At constant volume, the gas does no pressure-volume work because \(\Delta V=0\). Thus, \(W=0\). The first law \(Q=\Delta U+W\) becomes \(Q_V=\Delta U\). For an ideal gas, the internal-energy change for a temperature change is written as \(\Delta U=nC_V\Delta T\). Therefore, heat supplied at constant volume directly equals the increase in internal energy for the ideal gas.
218. A \(1.5\,\text{mol}\) ideal gas sample is heated at constant volume. If \(C_V=20\,\text{J mol}^{-1}\text{K}^{-1}\) and the temperature increases by \(40\,\text{K}\), the heat supplied is
ⓐ. \(1200\,\text{J}\)
ⓑ. \(1600\,\text{J}\)
ⓒ. \(600\,\text{J}\)
ⓓ. \(900\,\text{J}\)
Correct Answer: \(1200\,\text{J}\)
Explanation: \( \textbf{Given data:} \) \(n=1.5\,\text{mol}\), \(C_V=20\,\text{J mol}^{-1}\text{K}^{-1}\), and \(\Delta T=40\,\text{K}\).
\( \textbf{Process condition:} \) The gas is heated at constant volume.
\( \textbf{Work at constant volume:} \)
\[
W=0
\]
\( \textbf{Heat relation:} \)
\[
Q_V=nC_V\Delta T
\]
\( \textbf{Substitution:} \)
\[
Q_V=(1.5)(20)(40)\,\text{J}
\]
\( \textbf{Intermediate product:} \)
\[
(1.5)(20)=30
\]
\( \textbf{Calculation:} \)
\[
Q_V=30\times40=1200\,\text{J}
\]
\( \textbf{Energy meaning:} \) Since \(W=0\), this heat increases internal energy.
\( \textbf{Final answer:} \) \(Q_V=1200\,\text{J}\).
219. A constant-volume heating process supplies \(750\,\text{J}\) to \(3.0\,\text{mol}\) of an ideal gas and raises its temperature by \(25\,\text{K}\). The value of \(C_V\) is
ⓐ. \(10\,\text{J mol}^{-1}\text{K}^{-1}\)
ⓑ. \(20\,\text{J mol}^{-1}\text{K}^{-1}\)
ⓒ. \(15\,\text{J mol}^{-1}\text{K}^{-1}\)
ⓓ. \(5\,\text{J mol}^{-1}\text{K}^{-1}\)
Correct Answer: \(10\,\text{J mol}^{-1}\text{K}^{-1}\)
Explanation: \( \textbf{Given heat:} \) \(Q_V=750\,\text{J}\).
\( \textbf{Amount of gas:} \) \(n=3.0\,\text{mol}\).
\( \textbf{Temperature rise:} \) \(\Delta T=25\,\text{K}\).
\( \textbf{Constant-volume relation:} \)
\[
Q_V=nC_V\Delta T
\]
\( \textbf{Solve for \(C_V\):} \)
\[
C_V=\frac{Q_V}{n\Delta T}
\]
\( \textbf{Substitution:} \)
\[
C_V=\frac{750}{(3.0)(25)}
\]
\( \textbf{Denominator:} \)
\[
(3.0)(25)=75
\]
\( \textbf{Calculation:} \)
\[
C_V=10\,\text{J mol}^{-1}\text{K}^{-1}
\]
\( \textbf{Final answer:} \) \(C_V=10\,\text{J mol}^{-1}\text{K}^{-1}\).
220. A gas at constant volume receives heat \(Q_V\). A note says, “Since the gas has pressure, some of \(Q_V\) must become expansion work.” The best correction is:
ⓐ. Expansion work occurs whenever temperature changes, even if volume is fixed.
ⓑ. Constant volume means pressure is zero.
ⓒ. Heat supplied at constant volume must become work.
ⓓ. Boundary displacement is needed for volume work.
Correct Answer: Boundary displacement is needed for volume work.
Explanation: Pressure-volume work requires volume change. At constant volume, the boundary does not move, so \(\Delta V=0\). Even if the gas has pressure, the work \(W=\int P\,dV\) is zero because \(dV=0\). Heat supplied at constant volume can increase the internal energy of the gas. The important distinction is that pressure must act through a displacement to transfer energy as boundary work.
