Chemical Thermodynamics MCQs with Answers – Part 1 (NEET Chemistry)

Explanation: A thermodynamic system is the specified portion of the universe selected for investigation. Its boundary separates the chosen system from everything outside it. In this case, the liquid reaction mixture, rather than the entire beaker, constitutes the system. The portions of the inner glass wall touching the liquid form part of its boundary. The exposed liquid–air interface forms the remaining boundary. The glass wall is a physical boundary, whereas the liquid–air interface can be treated as an imaginary thermodynamic surface separating the chosen liquid from the air. A boundary need not be a rigid, physically constructed wall. The choice of system determines which surfaces must be included in its boundary. Thus, the correct boundary includes both the glass–liquid contact and the liquid–air interface.

Explanation: A closed system does not exchange matter with its surroundings, although it may exchange energy. The impermeable cylinder and piston prevent gas molecules from crossing the boundary. Therefore, the gas constitutes a closed system. Thermal insulation prevents heat transfer through the boundary, making the system adiabatic. However, the piston is movable, so the surroundings can compress the gas by doing mechanical work on it. Mechanical work transfers energy even when heat transfer is prevented. Consequently, the gas cannot be classified as isolated. A system is isolated only when both matter transfer and every form of energy transfer across its boundary are absent. The gas volume can also change during compression, so the process is not necessarily isochoric. Thus, the system is closed and adiabatic, but not isolated.

Explanation: The classification of a thermodynamic system depends on what crosses its specified boundary. In the original open beaker, vapour leaves the selected system and enters the atmosphere. Matter therefore crosses the boundary, so the system is open. After the gas-tight membrane is fitted, the vapour remains within the selected system. Vaporisation may continue internally, but an internal phase change does not itself constitute matter transfer across the system boundary. The system is consequently closed rather than open. Heating can still supply energy through the beaker wall. The flexible membrane can also move against external pressure, allowing pressure-volume work. Therefore, preventing matter transfer does not make the system isolated. The membrane's motion also shows that impermeability does not imply constant volume. The correct classification changes from open to closed, with energy exchange still possible.

Explanation: \(\textbf{First system boundary:}\) When the entire assembly is selected, the relevant boundary is the outer container. The outer container is specified to be perfectly insulated, rigid and impermeable. Neither matter nor energy can cross this boundary. The entire assembly therefore constitutes an isolated system. Heat exchanged internally between the reacting solution and the water bath does not violate isolation of the combined assembly. \(\textbf{Redefined system boundary:}\) When only the reacting solution is selected, the boundary shifts to the interior of the sealed flask. The sealed flask prevents matter exchange with the water bath. However, heat can pass through its thin wall. The reacting solution is therefore a closed system, not an isolated system. The water bath, flask wall and remaining assembly now belong to its surroundings. The physical apparatus has not changed; only the chosen system and its boundary have changed. Thus, isolation of a complete assembly does not imply isolation of every subsystem inside it.

Explanation: \(\textbf{Heat-transfer convention:}\) Heat entering the selected system is positive, while heat leaving it is negative. The metal loses \(0.80\,\mathrm{kJ}\) to the water. Hence, for the metal as the system, \[ q_M=-0.80\,\mathrm{kJ} \] The water receives exactly the heat lost by the metal, since no other energy transfer occurs. Thus, for the water as the system, \[ q_W=+0.80\,\mathrm{kJ} \] \(\textbf{Combined assembly:}\) When the metal and water are included within the complete insulated assembly, their heat exchange is internal. The heat passing from the metal to the water does not cross the outer boundary. Therefore, \[ q_{\mathrm{assembly}}=0 \] The energy transferred between the two components cancels in the combined accounting: \[ q_M+q_W=-0.80+0.80=0\,\mathrm{kJ} \] \(\textbf{Physical check:}\) The metal and water individually exchange heat, but the complete assembly exchanges none with its external surroundings. This illustrates why heat-transfer signs and magnitudes must always refer to a clearly specified system. \(\textbf{Final answer:}\) \[ \boxed{(q_M,q_W,q_{\mathrm{assembly}})=(-0.80,+0.80,0)\,\mathrm{kJ}} \]

Explanation: \(\textbf{Amount-dependent properties:}\) Volume and internal energy are extensive properties for the specified independent portions. For an otherwise identical homogeneous substance at unchanged thermodynamic conditions, these quantities scale with the amount of substance. The larger portion contains three-fourths of the original amount. Its volume is therefore \[ V_{\mathrm{large}}=\frac{3}{4}V \] Its internal energy, using the same energy reference, is \[ U_{\mathrm{large}}=\frac{3}{4}U \] \(\textbf{Density comparison:}\) Density is the ratio of mass to volume. Both mass and volume are reduced by the same fraction: \[ \rho_{\mathrm{large}}=\frac{3m/4}{3V/4} \] \[ \rho_{\mathrm{large}}=\frac{m}{V}=\rho \] Thus, density remains unchanged although the total mass, volume and internal energy decrease. This conclusion requires the two portions to retain the original composition and thermodynamic state. Simply reducing sample size does not reduce an intensive property in the same proportion. \(\textbf{Final answer:}\) \[ \boxed{\left(V_{\mathrm{large}},U_{\mathrm{large}},\rho_{\mathrm{large}}\right)=\left(\frac{3V}{4},\frac{3U}{4},\rho\right)} \]

Explanation: \(\textbf{Molar-volume comparison:}\) Molar volume is the total volume divided by the amount of gas. \[ V_m=\frac{V}{n} \] Using the table, \[ V_{m,P}=\frac{10}{1}=10\,\mathrm{L\,mol^{-1}} \] \[ V_{m,Q}=\frac{20}{2}=10\,\mathrm{L\,mol^{-1}} \] \[ V_{m,R}=\frac{30}{2}=15\,\mathrm{L\,mol^{-1}} \] Hence, P and Q have equal molar volumes, while R has a greater molar volume. \(\textbf{Density comparison:}\) For the same gas, let the molar mass be \(M\). \[ \rho=\frac{nM}{V}=\frac{M}{V_m} \] Because the molar mass is identical, density is inversely proportional to molar volume. \[ \rho_P=\rho_Q=\frac{M}{10} \] \[ \rho_R=\frac{M}{15} \] Therefore, \[ \rho_P=\rho_Q\gt \rho_R \] Although Q contains twice as much gas as P, it occupies twice the volume at the same temperature and pressure, so its density remains unchanged. Sample R is at a higher temperature under the same pressure and has a larger molar volume. An intensive property is independent of proportional sample scaling, but it need not remain constant when temperature changes. Thus, both comparisons in option C are correct.

Explanation: \(\textbf{Composition of the mixtures:}\) An ideal-gas mixture has a mean molar mass determined by its mole fractions. For mixture P, the mole fractions are \[ x_{A,P}=\frac{1}{3},\qquad x_{B,P}=\frac{2}{3} \] For Q, both component amounts have been doubled: \[ x_{A,Q}=\frac{2}{6}=\frac{1}{3},\qquad x_{B,Q}=\frac{4}{6}=\frac{2}{3} \] Therefore, P and Q have the same composition and mean molar mass. For R, \[ x_{A,R}=\frac{4}{6}=\frac{2}{3},\qquad x_{B,R}=\frac{2}{6}=\frac{1}{3} \] The composition of R differs from that of P. \(\textbf{Density relation:}\) For an ideal-gas mixture, \[ \rho=\frac{P\overline M}{RT} \] Here, \(\overline M\) denotes mean molar mass. Since temperature and pressure are common, density depends on the mean molar mass. Mixtures P and Q therefore have equal densities. However, interchanging the unequal mole fractions of gases having different molar masses changes the mean molar mass in R. Consequently, R has a different density from P. Increasing the total amount at fixed composition does not itself increase density. Changing composition can change density even when temperature and pressure remain unchanged. The correct comparison is \(\rho_Q=\rho_P\) and \(\rho_R\ne\rho_P\).

Explanation: The Assertion is true because internal energy is a thermodynamic state function. A state function has a definite value for every specified equilibrium state. Consequently, its change depends only on the initial and final states. For internal energy, \[ \Delta U=U_f-U_i \] Two processes starting and ending at the same states must therefore have identical changes in internal energy. The Reason is also true because heat is a mode of energy transfer, not a property stored in the system. The amount of heat exchanged may depend on how the process is carried out. Work is also path-dependent, so it can differ between the two processes. The first law ensures that the sum of heat and work equals the same internal-energy change for both paths. Thus, different values of heat can be compensated by different values of work without changing the endpoint-dependent internal-energy difference. The Reason correctly identifies the state-function and path-function distinction responsible for the Assertion. Hence, both are true, and the Reason correctly explains the Assertion.

Explanation: \(\textbf{Energy accounting along path I:}\) The first law of thermodynamics is \[ \Delta U=q+w \] Heat absorbed by the system is positive, whereas work performed by the system is negative. For path I, \[ q_I=+450\,\mathrm{J} \] \[ w_I=-150\,\mathrm{J} \] Substituting, \[ \Delta U_I=450-150 \] \[ \Delta U_I=+300\,\mathrm{J} \] \(\textbf{Path-independent quantity:}\) Both paths connect the same equilibrium states A and B. Therefore, because internal energy is a state function, \[ \Delta U_{II}=\Delta U_I=+300\,\mathrm{J} \] \(\textbf{Heat required along path II:}\) The work done by the system along path II is \[ w_{II}=-260\,\mathrm{J} \] Applying the first law to path II, \[ 300=q_{II}-260 \] \[ q_{II}=300+260 \] \[ q_{II}=560\,\mathrm{J} \] \(\textbf{Physical check:}\) Path II involves more work done by the system than path I. For the same increase in internal energy, the system must therefore receive more heat along path II. \(\textbf{Final answer:}\) \[ \boxed{q_{II}=+560\,\mathrm{J}} \]

Explanation: \(\textbf{State-function principle:}\) Enthalpy is a state function, so its change between two specified states is independent of the route taken. \[ \Delta H_{i\rightarrow f}=H_f-H_i \] \(\textbf{Relating states A, B and C:}\) From the first relation, \[ H_B-H_A=+35\,\mathrm{kJ} \] The second relation is written in the direction \(C\rightarrow B\). \[ H_B-H_C=-25\,\mathrm{kJ} \] Rearranging, \[ H_C=H_B+25\,\mathrm{kJ} \] Substituting the first relation, \[ H_C=H_A+60\,\mathrm{kJ} \] \(\textbf{Locating state D:}\) For the transition \(D\rightarrow A\), \[ H_A-H_D=-15\,\mathrm{kJ} \] Therefore, \[ H_D=H_A+15\,\mathrm{kJ} \] \(\textbf{Required transition:}\) \[ \Delta H_{D\rightarrow C}=H_C-H_D \] \[ \Delta H_{D\rightarrow C}=(H_A+60)-(H_A+15) \] \[ \Delta H_{D\rightarrow C}=+45\,\mathrm{kJ} \] The reference value \(H_A\) cancels because only a difference in enthalpy is required. \(\textbf{Direction check:}\) State C has a higher enthalpy than state D, so the positive sign is consistent. \(\textbf{Final answer:}\) \[ \boxed{\Delta H_{D\rightarrow C}=+45\,\mathrm{kJ}} \]

Explanation: \(\textbf{Stage 1:}\) For \(A\rightarrow B\), apply the first law: \[ \Delta U_{AB}=q_{AB}+w_{AB} \] \[ \Delta U_{AB}=80+20 \] \[ \Delta U_{AB}=+100\,\mathrm{J} \] \(\textbf{Stage 2:}\) For \(B\rightarrow C\), \[ \Delta U_{BC}=q_{BC}+w_{BC} \] \[ \Delta U_{BC}=-50-10 \] \[ \Delta U_{BC}=-60\,\mathrm{J} \] \(\textbf{Cyclic state-function condition:}\) A complete cycle returns the system to its original state. Since internal energy is a state function, \[ \Delta U_{AB}+\Delta U_{BC}+\Delta U_{CA}=0 \] Substituting the known changes, \[ 100-60+\Delta U_{CA}=0 \] \[ \Delta U_{CA}=-40\,\mathrm{J} \] \(\textbf{Final-stage heat calculation:}\) For \(C\rightarrow A\), \[ \Delta U_{CA}=q_{CA}+w_{CA} \] \[ -40=q_{CA}-30 \] \[ q_{CA}=-10\,\mathrm{J} \] \(\textbf{Cycle verification:}\) The net heat supplied during the complete cycle is \[ q_{\mathrm{cycle}}=80-50-10=+20\,\mathrm{J} \] The net work is \[ w_{\mathrm{cycle}}=20-10-30=-20\,\mathrm{J} \] Therefore, \[ q_{\mathrm{cycle}}+w_{\mathrm{cycle}}=0=\Delta U_{\mathrm{cycle}} \] The energy balance confirms the result and illustrates that cyclic heat and work need not separately vanish. \(\textbf{Final answer:}\) \[ \boxed{q_{CA}=-10\,\mathrm{J}} \]

Explanation: \(\textbf{Adiabatic compression:}\) During Stage 1, no heat crosses the system boundary. \[ q_1=0 \] Compression involves work done on the gas, so \(w_1\gt 0\). Consequently, its internal energy increases. For an ideal gas, internal energy depends only on temperature. The specified increase from \(T_1\) to \(T_2\) is consistent with this increase in internal energy. \(\textbf{Isothermal expansion:}\) Stage 2 occurs at constant temperature \(T_2\). For an ideal gas, \[ \Delta U_2=0 \] The gas expands against its surroundings, so pressure-volume work done by the gas is negative. \[ w_2\lt 0 \] The first law gives \[ 0=q_2+w_2 \] \[ q_2=-w_2\gt 0 \] Thus, heat must enter the gas during the isothermal expansion to compensate for the energy transferred as work. \(\textbf{Final-state comparison:}\) Although the final volume equals the original volume, the final temperature is \(T_2\), not \(T_1\). The total internal-energy change is therefore positive. Returning to the original volume does not mean returning to the original thermodynamic state. Hence, the final internal energy is higher, and Stage 2 requires positive heat transfer.

Explanation: \(\textbf{Constant-pressure heating:}\) For a fixed amount of ideal gas, \[ PV=nRT \] During Stage 1, pressure and amount remain constant. Therefore, volume is directly proportional to absolute temperature. \[ \frac{V_2}{V_0}=\frac{T_2}{T_0} \] Since \(V_2=2V_0\), \[ T_2=2T_0 \] The expansion produces negative work under the chemistry convention. \(\textbf{Constant-volume cooling:}\) During Stage 2, the volume remains \(2V_0\). The temperature decreases from \(2T_0\) to \(T_0\). At constant volume, pressure is directly proportional to temperature. \[ \frac{P_f}{P_0}=\frac{T_0}{2T_0} \] \[ P_f=\frac{P_0}{2} \] \(\textbf{Internal-energy comparison:}\) The final and initial temperatures are identical. For an ideal gas, internal energy depends only on temperature. Hence, \[ \Delta U_{\mathrm{total}}=0 \] \(\textbf{Net energy transfer:}\) Stage 1 involves expansion and therefore negative pressure-volume work. Stage 2 involves no pressure-volume work because its volume remains constant. Thus, \[ w_{\mathrm{total}}\lt 0 \] Using the first law for the complete two-stage process, \[ 0=q_{\mathrm{total}}+w_{\mathrm{total}} \] \[ q_{\mathrm{total}}=-w_{\mathrm{total}}\gt 0 \] The system returns to its initial temperature but not to its initial pressure or volume. Therefore, zero net internal-energy change is compatible with positive net heat absorption and negative net work.

Explanation: \(\textbf{Endpoint comparison:}\) The two paths begin and end at the same thermodynamic states. Internal energy is a state function, so its change is identical along both paths. \[ \Delta U_R=\Delta U_S \] \(\textbf{Pressure-volume work:}\) Under the chemistry sign convention, pressure-volume work is \[ w=-\int_{V_i}^{V_f}P_{\mathrm{ext}}\,dV \] For expansion, \(dV\) is positive in the direction of increasing volume. The area under each external-pressure curve gives the magnitude of its pressure-volume work. Because the external-pressure curve for R lies above that for S throughout the intermediate volume interval, \[ \int_{V_i}^{V_f}P_{\mathrm{ext},R}\,dV \gt \int_{V_i}^{V_f}P_{\mathrm{ext},S}\,dV \] The negative sign in the work expression reverses the inequality: \[ w_R\lt w_S \] Thus, R involves more negative work, meaning a greater amount of energy is transferred from the gas through expansion work. \(\textbf{Heat-transfer inference:}\) Applying the first law, \[ q=\Delta U-w \] Since both paths have the same \(\Delta U\), but \(w_R\lt w_S\), \[ q_R\gt q_S \] The additional heat supplied along R compensates for its greater work output while maintaining the same internal-energy change. The result does not require knowing the actual numerical value or sign of the common internal-energy change. It follows from path-independent internal energy and path-dependent pressure-volume work. Therefore, \(\Delta U_R=\Delta U_S\) and \(q_R\gt q_S\).

Explanation: \(\textbf{Process P — Free expansion:}\) Free expansion occurs when a gas expands against zero external pressure. The pressure-volume work is \[ w_P=-P_{\mathrm{ext}}\Delta V \] Since the external pressure is zero, \[ w_P=0 \] The surrounding container is perfectly insulated, so heat cannot cross its outer boundary. \[ q_P=0 \] Applying the first law, \[ \Delta U_P=q_P+w_P=0 \] For an ideal gas, internal energy depends only on temperature. Therefore, its temperature remains unchanged during this free expansion. \(\textbf{Process Q — Reversible isothermal expansion:}\) In the reversible process, the gas expands against an external pressure that remains infinitesimally below its own pressure. The work done by the expanding gas is negative. For an ideal gas, \[ w_Q=-nRT\ln\left(\frac{V_f}{V_i}\right) \] Here, \(n=1\) and \(V_f=2V_i\), so \[ w_Q=-RT\ln 2\lt 0 \] The temperature remains constant, making the internal-energy change zero. \[ \Delta U_Q=0 \] By the first law, \[ q_Q=-w_Q=RT\ln 2\gt 0 \] \(\textbf{Thermodynamic interpretation:}\) Both processes have the same initial and final equilibrium states and therefore the same internal-energy change. However, their heat and work transfers differ because these quantities depend on the process path. Free expansion transfers no energy as pressure-volume work, while reversible isothermal expansion requires heat absorption to compensate for the work performed. Thus, the correct comparison is \(q_P=w_P=0\), with \(q_Q\gt 0\) and \(w_Q\lt 0\).

Explanation: A thermodynamic system is defined by its selected boundary rather than necessarily by the walls of the apparatus. Here, only the gas occupying the left half of the vessel is selected as the system. The imaginary plane therefore forms part of the boundary separating this system from the gas in the right half. Although the macroscopic gas distribution is uniform, individual molecules continuously move across the imaginary plane. Matter crosses the selected boundary in both directions. Zero net matter transfer means that the opposing molecular transfers balance; it does not mean that no molecules cross the boundary. The left-half system is therefore treated as an open system when actual molecular exchange across its fixed imaginary boundary is considered. For the complete sealed vessel, however, no matter crosses the outer wall. The complete vessel contents consequently constitute a closed system, provided energy exchange through the outer wall remains possible. The distinction depends on which boundary is selected and whether matter can cross that boundary. Thus, a zero net flux does not automatically establish the absence of matter exchange.

Explanation: The classification of a system depends on matter and energy exchange across its selected boundary. For system P, the boundary surrounds only the remaining solid phase. During decomposition, material originally present in the solid is converted into gaseous products. These products leave the solid phase and enter the gas-filled region outside P's boundary. Thus, matter crosses the boundary of P, making it an open system. For system Q, both the remaining solid and all gaseous products belong to the selected system. The conversion of solid into gas occurs entirely within Q. Because the vessel is completely sealed, no matter crosses its outer boundary. System Q is therefore closed. However, the thermally conducting vessel wall allows heat exchange with the external surroundings. Consequently, Q is not necessarily isolated. The same decomposition reaction can therefore be described using different system classifications without changing the physical apparatus. The decisive distinction is whether the gaseous products cross the boundary of the selected system or remain within it.

Explanation: \(\textbf{Energy transfers for gas A:}\) The first law, using the chemistry sign convention, is \[ \Delta U=q+w \] Heat leaves A and enters B. Therefore, for gas A, \[ q_A=-40\,\mathrm{J} \] Gas B performs 90 J of work on A, so work is done on system A. \[ w_A=+90\,\mathrm{J} \] Its internal-energy change is \[ \Delta U_A=-40+90 \] \[ \Delta U_A=+50\,\mathrm{J} \] \(\textbf{Energy transfers for gas B:}\) Gas B receives the heat released by A. \[ q_B=+40\,\mathrm{J} \] Gas B performs work on A, so its work has the opposite sign. \[ w_B=-90\,\mathrm{J} \] Therefore, \[ \Delta U_B=40-90 \] \[ \Delta U_B=-50\,\mathrm{J} \] \(\textbf{Combined-system verification:}\) When both gases are considered as one system, the heat and work exchanged between them are internal transfers. Neither transfer crosses the insulated rigid outer boundary. Hence, \[ \Delta U_{A+B}=\Delta U_A+\Delta U_B \] \[ \Delta U_{A+B}=50-50=0\,\mathrm{J} \] The internal energies of the individual gases change, even though that of the complete isolated assembly remains unchanged. \(\textbf{Final answer:}\) \[ \boxed{(\Delta U_A,\Delta U_B)=(+50,-50)\,\mathrm{J}} \]

Explanation: Mass and volume are extensive properties because they depend on the amount of material and scale with the size of an otherwise identical system. Before mixing, the combined mass of the two liquids is \[ m_{\mathrm{total}}=100+100=200\,\mathrm{g} \] Since no material escapes during mixing, the final mass remains 200 g. The sum of the initial volumes is \[ V_{\mathrm{initial,total}}=120+100=220\,\mathrm{mL} \] However, the measured final solution volume is only 210 mL. The volume contraction is \[ \Delta V=210-220=-10\,\mathrm{mL} \] This does not contradict the extensive nature of volume. Extensivity means that volume scales with the amount of an otherwise identical system and is additive for appropriately separated portions in the same states. Mixing different liquids can alter intermolecular interactions and packing, producing a final state whose volume differs from the sum of the separate initial volumes. Density remains an intensive property even though its numerical value may change on mixing. The final density is \[ \rho_{\mathrm{final}}=\frac{200}{210}\approx0.952\,\mathrm{g\,mL^{-1}} \] A change in an intensive property's value does not make that property extensive. The correct conclusion is that mass remains additive and volume remains extensive, while mixing may produce volume contraction.