Chemical Thermodynamics MCQs with Answers – Part 2 (NEET Chemistry)

Explanation: \(\textbf{Energy change of the water bath:}\) The bath cools by \(2.0\,\mathrm{K}\). Its temperature change is \[ \Delta T_{\mathrm{bath}}=-2.0\,\mathrm{K} \] The heat transferred from the bath is \[ q_{\mathrm{bath}}=C_{\mathrm{bath}}\Delta T_{\mathrm{bath}} \] Substituting, \[ q_{\mathrm{bath}}=(200)(-2.0) \] \[ q_{\mathrm{bath}}=-400\,\mathrm{J} \] \(\textbf{Heat received by the gas:}\) The bath and gas exchange heat only with each other. Therefore, the heat lost by the bath equals the heat gained by the gas. \[ q_{\mathrm{gas}}=-q_{\mathrm{bath}} \] \[ q_{\mathrm{gas}}=+400\,\mathrm{J} \] \(\textbf{Work during gas compression:}\) The gas volume decreases by \(1.0\,\mathrm{L}\). Therefore, \[ \Delta V=-1.0\,\mathrm{L} \] At constant external pressure, \[ w=-P_{\mathrm{ext}}\Delta V \] \[ w=-(1.0)(-1.0) \] \[ w=+1.0\,\mathrm{L\,bar} \] \[ w=+100\,\mathrm{J} \] \(\textbf{Internal-energy balance:}\) The first law gives \[ \Delta U=q+w \] Substituting, \[ \Delta U=400+100 \] \[ \Delta U=+500\,\mathrm{J} \] \(\textbf{System-surroundings interpretation:}\) The water bath loses thermal energy while the gas receives heat. The gas also receives mechanical energy through compression work. Both transfers increase the gas internal energy. The bath cooling does not mean that the gas internal energy must decrease. The heat signs reverse when the system is changed from bath to gas, whereas the work contribution must be accounted for separately. \(\textbf{Final answer:}\) \[ \boxed{\Delta U_{\mathrm{gas}}=+500\,\mathrm{J}} \]

Explanation: \(\textbf{Identifying constant-pressure energy transfers:}\) The first experiment releases 50.0 kJ of heat. Thus, \[ q_P=-50.0\,\mathrm{kJ} \] It also receives electrical work. Therefore, \[ w_{\mathrm{electrical}}=+4.0\,\mathrm{kJ} \] \(\textbf{Recovering the reaction enthalpy:}\) At constant pressure, \[ \Delta_rH=q_P+w_{\mathrm{electrical}} \] This relation applies because pressure-volume and electrical work are the only allowed work contributions. Substituting, \[ \Delta_rH=-50.0+4.0 \] \[ \Delta_rH=-46.0\,\mathrm{kJ} \] \(\textbf{Gaseous mole correction:}\) For an ideal-gas reaction, \[ \Delta_rH=\Delta_rU+\Delta n_gRT \] The gaseous mole change is \[ \Delta n_g=-2 \] Therefore, \[ \Delta n_gRT=(-2)(8.314)(300) \] \[ \Delta n_gRT=-4988.4\,\mathrm{J} \] Converting, \[ \Delta n_gRT=-4.9884\,\mathrm{kJ} \] \(\textbf{Internal-energy change:}\) Rearranging, \[ \Delta_rU=\Delta_rH-\Delta n_gRT \] Substitution gives \[ \Delta_rU=-46.0-(-4.9884) \] \[ \Delta_rU=-41.0116\,\mathrm{kJ} \] \(\textbf{Heat in the second experiment:}\) The second experiment is conducted at constant volume without electrical or other non-pressure-volume work. Therefore, \[ w_V=0 \] Consequently, \[ q_V=\Delta_rU \] \[ q_V=-41.0116\,\mathrm{kJ} \] \(\textbf{Consistency check:}\) The constant-pressure heat in the first experiment cannot be identified directly with reaction enthalpy because electrical work is supplied. After correcting for electrical work, the enthalpy–internal-energy relation must be applied using the gaseous mole change. The negative final heat indicates that the second experiment also releases heat. Its heat-release magnitude is smaller than that of the first experiment. \(\textbf{Final answer:}\) \[ \boxed{q_V\approx-41.01\,\mathrm{kJ}} \]

Explanation: \(\textbf{Identifying the work-producing stages:}\) Only the two isothermal stages involve changes in volume. The constant-volume stages have zero pressure-volume work. Therefore, \[ w_{\mathrm{cycle}}=w_I+w_{III} \] \(\textbf{Work during the high-temperature expansion:}\) For reversible isothermal expansion, \[ w_I=-nRT_H\ln\left(\frac{2V_0}{V_0}\right) \] Thus, \[ w_I=-nRT_H\ln2 \] This work is negative because the gas expands and performs work on its surroundings. \(\textbf{Work during the low-temperature compression:}\) The gas is compressed from \(2V_0\) to \(V_0\). Therefore, \[ w_{III}=-nRT_C\ln\left(\frac{V_0}{2V_0}\right) \] Using \[ \ln\left(\frac12\right)=-\ln2 \] gives \[ w_{III}=+nRT_C\ln2 \] The compression work is positive. \(\textbf{Combining the contributions:}\) \[ w_{\mathrm{cycle}}=-nRT_H\ln2+nRT_C\ln2 \] Factoring, \[ w_{\mathrm{cycle}}=-nR(T_H-T_C)\ln2 \] Since \(T_H\gt T_C\), \[ T_H-T_C\gt 0 \] Therefore, \[ w_{\mathrm{cycle}}\lt 0 \] \(\textbf{Interpretation:}\) The expansion takes place at the higher temperature, producing a greater work magnitude than the compression at the lower temperature requires. The gas therefore performs net work on its surroundings. The net work depends on the temperature difference rather than the temperature sum. Returning to the initial volume does not make the net work zero because pressure follows different paths during expansion and compression. For the complete cycle, the internal-energy change is zero, so the net heat absorbed equals the magnitude of the net work delivered. The negative work sign is therefore consistent with net positive heat absorption.

Explanation: The supplied reaction enthalpy applies specifically at 300 K. For an ideal-gas reaction evaluated at a common temperature, \[ \Delta_rH(T)=\Delta_rU(T)+\Delta n_gRT \] The reaction has \[ \Delta n_g=2-1=+1 \] At 300 K, the corresponding internal-energy change could be calculated. However, the requested constant-volume heat refers to a different temperature, 600 K. At constant volume, with only pressure-volume work, \[ q_V(600\,\mathrm{K})=\Delta_rU(600\,\mathrm{K}) \] To calculate this value, the reaction enthalpy at 600 K is required. The reaction enthalpy generally varies with temperature because the heat capacities of reactants and products need not be identical. The supplied enthalpy at 300 K therefore does not automatically remain valid at 600 K. The ideal-gas correction \(\Delta n_gRT\) can relate enthalpy and internal-energy changes at a specified common temperature. It does not by itself determine how reaction enthalpy changes between two different temperatures. Without suitable additional heat-capacity information or the reaction enthalpy at 600 K, a unique numerical value cannot be obtained. Applying the 300 K enthalpy unchanged at 600 K would introduce an unsupported assumption. Therefore, the correct conclusion is that the requested heat is not uniquely determined.

Explanation: \(\textbf{Heat-transfer sign:}\) The reacting system releases heat to its surroundings. Therefore, \[ q=-2.0\,\mathrm{kJ} \] \(\textbf{Volume-change sign:}\) The volume decreases by 5.0 L. Thus, \[ \Delta V=-5.0\,\mathrm{L} \] This indicates compression rather than expansion. \(\textbf{Pressure-volume work:}\) At constant external pressure, \[ w=-P_{\mathrm{ext}}\Delta V \] Substitution gives \[ w=-(1.0)(-5.0) \] \[ w=+5.0\,\mathrm{L\,bar} \] Using the conversion, \[ w=+500\,\mathrm{J} \] \[ w=+0.5\,\mathrm{kJ} \] \(\textbf{Internal-energy change:}\) Applying the first law, \[ \Delta U=q+w \] \[ \Delta U=-2.0+0.5 \] \[ \Delta U=-1.5\,\mathrm{kJ} \] \(\textbf{Enthalpy change:}\) At constant pressure, with only pressure-volume work, \[ q_P=\Delta H \] Therefore, \[ \Delta H=-2.0\,\mathrm{kJ} \] \(\textbf{Independent verification:}\) Using the enthalpy definition, \[ \Delta H=\Delta U+P\Delta V \] Since \[ P\Delta V=-0.5\,\mathrm{kJ} \] we obtain \[ \Delta H=-1.5-0.5 \] \[ \Delta H=-2.0\,\mathrm{kJ} \] \(\textbf{Physical interpretation:}\) The surroundings perform positive compression work on the reacting system. This work partly compensates for the energy lost as heat. Consequently, the decrease in internal energy is smaller in magnitude than the heat released. A contracting exothermic system need not have an internal-energy change numerically equal to its enthalpy change. \(\textbf{Final answer:}\) \[ \boxed{(\Delta U,\Delta H)=(-1.5,-2.0)\,\mathrm{kJ}} \]

Explanation: \(\textbf{System P:}\) The process is isothermal expansion of a fixed amount of ideal gas. Therefore, the internal-energy change is zero. Because the gas expands against positive external pressure, work done on the gas is negative. The first law consequently requires heat absorption equal to the magnitude of work performed by the gas. Thus, P matches relation 4. \(\textbf{System Q:}\) The process is adiabatic, so no heat crosses the system boundary. Compression produces positive pressure-volume work on the gas. The first law gives \(\Delta U=w\gt 0\). Thus, Q matches relation 2. \(\textbf{System R:}\) The vessel is insulated, so \(q=0\). The electrical stirrer receives energy from an external source and transfers mechanical work to the gas. This shaft work is positive under the chemistry convention. The constant-volume condition eliminates pressure-volume work but not shaft work. Thus, R matches relation 3. \(\textbf{System S:}\) The process occurs at constant pressure, and only pressure-volume work is possible. Under these conditions, the heat supplied equals the enthalpy change. Thus, S matches relation 1. The matching requires the process constraints to be interpreted together with the permitted work mechanisms. Constant temperature, zero heat transfer and constant volume are not interchangeable thermodynamic conditions. The correct matching is P–4, Q–2, R–3 and S–1.

Explanation: \(\textbf{Stage A to B — Constant-volume heating:}\) At constant volume, pressure-volume work is zero. Therefore, \[ q_{AB}=\Delta U_{AB} \] The internal-energy change is \[ \Delta U_{AB}=C_V\Delta T_1 \] Substitution gives \[ \Delta U_{AB}=30(50) \] \[ \Delta U_{AB}=1500\,\mathrm{J} \] Hence, \[ q_{AB}=+1500\,\mathrm{J} \] \(\textbf{Stage B to C — Constant-pressure heating:}\) For an ideal gas, \[ \Delta H=\Delta U+nR\Delta T \] The internal-energy increase during this stage is \[ \Delta U_{BC}=C_V\Delta T_2 \] \[ \Delta U_{BC}=30(50) \] \[ \Delta U_{BC}=1500\,\mathrm{J} \] The pressure-volume contribution is \[ nR\Delta T_2=10(50) \] \[ nR\Delta T_2=500\,\mathrm{J} \] Therefore, \[ \Delta H_{BC}=1500+500 \] \[ \Delta H_{BC}=2000\,\mathrm{J} \] At constant pressure, with only pressure-volume work, \[ q_{BC}=\Delta H_{BC}=2000\,\mathrm{J} \] \(\textbf{Total heat absorbed:}\) \[ q_{\mathrm{total}}=q_{AB}+q_{BC} \] \[ q_{\mathrm{total}}=1500+2000 \] \[ q_{\mathrm{total}}=3500\,\mathrm{J} \] \(\textbf{Overall enthalpy change:}\) The total temperature rise is \[ \Delta T_{\mathrm{total}}=50+50=100\,\mathrm{K} \] The overall internal-energy change is \[ \Delta U_{\mathrm{total}}=30(100) \] \[ \Delta U_{\mathrm{total}}=3000\,\mathrm{J} \] For the fixed amount of ideal gas, \[ \Delta H_{\mathrm{total}}=\Delta U_{\mathrm{total}}+nR\Delta T_{\mathrm{total}} \] \[ \Delta H_{\mathrm{total}}=3000+10(100) \] \[ \Delta H_{\mathrm{total}}=4000\,\mathrm{J} \] \(\textbf{Interpretation:}\) The total heat absorbed differs from the overall enthalpy change because the complete process is not conducted entirely at constant pressure. The first stage supplies heat at constant volume, while the second supplies heat at constant pressure. Enthalpy is a state function, so its overall change is determined by the complete temperature change of the fixed amount of ideal gas. Heat is path-dependent and must be accumulated according to the conditions of each stage. \(\textbf{Final answer:}\) \[ \boxed{(q_{\mathrm{total}},\Delta H)=(3500,4000)\,\mathrm{J}} \]

Explanation: \(\textbf{Step 1 — Identifying the limiting reactant:}\) The balanced reaction is \[ \mathrm{2A(g)+B(g)\rightarrow2C(g)} \] According to the stoichiometric coefficients, \(2\) mol of A react with \(1\) mol of B. Initially, \[ n_{A,i}=2.0\,\mathrm{mol} \] \[ n_{B,i}=3.0\,\mathrm{mol} \] Complete consumption of \(2.0\) mol of A requires only \(1.0\) mol of B. Therefore, A is the limiting reactant. The reaction occurs once according to the balanced equation as written. \(\textbf{Step 2 — Final gaseous composition:}\) After the limiting reactant is exhausted, \[ n_{A,f}=2.0-2.0=0 \] The remaining amount of B is \[ n_{B,f}=3.0-1.0=2.0\,\mathrm{mol} \] The amount of C produced is \[ n_{C,f}=2.0\,\mathrm{mol} \] The inert gas does not participate in the reaction. Therefore, \[ n_{I,f}=1.0\,\mathrm{mol} \] \(\textbf{Step 3 — Initial and final gaseous amounts:}\) The initial total amount of gas is \[ n_i=2.0+3.0+1.0 \] \[ n_i=6.0\,\mathrm{mol} \] The final total amount is \[ n_f=0+2.0+2.0+1.0 \] \[ n_f=5.0\,\mathrm{mol} \] \(\textbf{Step 4 — Pressure ratio at constant volume:}\) The ideal-gas equation is \[ PV=nRT \] In Experiment V, temperature and volume remain constant. Consequently, pressure is directly proportional to the total gaseous amount. \[ \frac{P_f}{P_i}=\frac{n_f}{n_i} \] Substituting, \[ \frac{P_f}{P_i}=\frac{5.0}{6.0} \] \[ \frac{P_f}{P_i}=\frac56 \] The inert gas must be included when calculating the total initial and final gaseous amounts. \(\textbf{Step 5 — Change in gaseous stoichiometric amount:}\) For the reaction as written, \[ \Delta n_g=n_{g,\mathrm{products}}-n_{g,\mathrm{reactants}} \] Therefore, \[ \Delta n_g=2-(2+1) \] \[ \Delta n_g=-1 \] The inert gas contributes equally to the initial and final amounts, so it does not affect the net gaseous mole change. \(\textbf{Step 6 — Relating enthalpy and internal energy:}\) For an ideal-gas reaction under the specified conditions, \[ \Delta H=\Delta U+\Delta n_gRT \] Rearranging, \[ \Delta U=\Delta H-\Delta n_gRT \] The reaction proceeds to an extent corresponding to the balanced equation as written. Hence, \[ \Delta H=-80.0\,\mathrm{kJ} \] The gaseous-mole correction is \[ \Delta n_gRT=(-1)(8.314)(300) \] \[ \Delta n_gRT=-2494.2\,\mathrm{J} \] Converting into kilojoules, \[ \Delta n_gRT=-2.4942\,\mathrm{kJ} \] Therefore, \[ \Delta U=-80.0-(-2.4942) \] \[ \Delta U=-77.5058\,\mathrm{kJ} \] \(\textbf{Step 7 — Heat exchanged at constant volume:}\) In Experiment V, the volume remains constant. Because only pressure-volume work occurs, \[ w_V=0 \] Applying the first law, \[ \Delta U=q_V+w_V \] Therefore, \[ q_V=\Delta U \] \[ q_V=-77.5058\,\mathrm{kJ} \] Rounding appropriately, \[ q_V\approx-77.51\,\mathrm{kJ} \] The negative sign indicates heat released by the reacting system. \(\textbf{Consistency check:}\) At constant pressure, the heat exchanged would be \[ q_P=\Delta H=-80.0\,\mathrm{kJ} \] However, the constant-volume heat is different because the reaction decreases the gaseous mole number. The inert gas affects the pressure ratio through the total gaseous amount but does not affect the stoichiometric \(\Delta n_g\) correction. \(\textbf{Final answer:}\) \[ \boxed{\left(\frac{P_f}{P_i},q_V\right)=\left(\frac56,-77.51\,\mathrm{kJ}\right)} \]

Explanation: Heat capacity describes the heat required to produce a unit temperature change under specified conditions. For a sample with constant applicable heat capacity, \[ q=C\Delta T \] The total heat capacity depends on the amount of substance present. Therefore, heat capacity is an extensive property. Sample Q contains three times the amount present in P. Hence, \[ C_Q=3C_P \] Substituting, \[ C_Q=3(150) \] \[ C_Q=450\,\mathrm{J\,K^{-1}} \] Molar heat capacity is defined as heat capacity divided by the amount of substance. \[ C_m=\frac{C}{n} \] For sample Q, both C and n are three times their corresponding values for P. Their ratio therefore remains unchanged. Thus, the molar heat capacities of P and Q are equal. Molar heat capacity is intensive for a specified substance, physical state and thermodynamic condition. The total heat capacity increases with sample size, but the energy required per mole per kelvin does not. For the same temperature rise, Q requires three times the heat required by P. This distinction prevents confusion between total heat capacity and molar heat capacity.

Explanation: \(\textbf{Selecting the appropriate heat-capacity relation:}\) For a temperature interval over which heat capacity is constant, \[ q=C\Delta T \] However, the heat capacity changes between the two specified temperature intervals. Therefore, the entire heating process cannot be calculated using one arbitrarily selected heat-capacity value. \(\textbf{Heating from 300 K to 330 K:}\) The temperature change is \[ \Delta T_1=330-300=30\,\mathrm{K} \] The applicable heat capacity is \(40\,\mathrm{J\,K^{-1}}\). Hence, \[ q_1=C_1\Delta T_1 \] \[ q_1=40(30) \] \[ q_1=1200\,\mathrm{J} \] \(\textbf{Heat remaining for the second interval:}\) The total supplied heat is 2400 J. Therefore, \[ q_2=q_{\mathrm{total}}-q_1 \] \[ q_2=2400-1200 \] \[ q_2=1200\,\mathrm{J} \] \(\textbf{Temperature rise in the second interval:}\) The applicable heat capacity is now \(60\,\mathrm{J\,K^{-1}}\). \[ q_2=C_2\Delta T_2 \] \[ 1200=60\Delta T_2 \] \[ \Delta T_2=20\,\mathrm{K} \] The final temperature is \[ T_f=330+20 \] \[ T_f=350\,\mathrm{K} \] \(\textbf{Physical check:}\) The final temperature lies within the second specified interval, so the selected heat capacity is applicable. A higher heat capacity requires more heat per kelvin of temperature increase. Treating the entire heating range as having the lower heat capacity would overestimate the final temperature. The result also shows why a temperature-dependent heat capacity must be handled according to the intervals for which its value is valid. \(\textbf{Final answer:}\) \[ \boxed{T_f=350\,\mathrm{K}} \]

Explanation: \(\textbf{Energy conservation for the insulated assembly:}\) The complete assembly exchanges no heat with the external surroundings. Therefore, heat lost by the initially hot block equals the combined heat gained by the cold block and calorimeter. Let the final equilibrium temperature be \(T_f\). \(\textbf{Heat lost by block A:}\) The temperature of A decreases from \(80^\circ\mathrm{C}\) to \(T_f\). Its heat loss magnitude is \[ q_{A,\mathrm{lost}}=C_A(80-T_f) \] \[ q_{A,\mathrm{lost}}=120(80-T_f) \] \(\textbf{Heat gained by block B:}\) Block B warms from \(20^\circ\mathrm{C}\) to \(T_f\). \[ q_B=180(T_f-20) \] \(\textbf{Heat gained by the calorimeter:}\) The calorimeter also begins at \(20^\circ\mathrm{C}\). \[ q_{\mathrm{cal}}=60(T_f-20) \] The combined heat gain is \[ q_{\mathrm{gained}}=(180+60)(T_f-20) \] \[ q_{\mathrm{gained}}=240(T_f-20) \] \(\textbf{Setting up the heat balance:}\) \[ 120(80-T_f)=240(T_f-20) \] Expanding, \[ 9600-120T_f=240T_f-4800 \] \[ 14400=360T_f \] \[ T_f=40^\circ\mathrm{C} \] \(\textbf{Energy verification:}\) The heat lost by A is \[ q_{A,\mathrm{lost}}=120(80-40)=4800\,\mathrm{J} \] The heat gained by B and the calorimeter is \[ q_{\mathrm{gained}}=240(40-20)=4800\,\mathrm{J} \] The two magnitudes agree. The calorimeter must be included because it also absorbs heat during equilibration. Ignoring its heat capacity would produce an incorrect final temperature. The calculated temperature also lies between the initial temperatures, as expected for heat exchange without an external energy source. \(\textbf{Final answer:}\) \[ \boxed{T_f=40^\circ\mathrm{C}} \]

Explanation: \(\textbf{Calibrating the complete calorimetric assembly:}\) The supplied electrical energy raises the temperature of both the water and the calorimeter vessel. The effective total heat capacity is \[ C_{\mathrm{total}}=\frac{q_{\mathrm{calibration}}}{\Delta T} \] Substitution gives \[ C_{\mathrm{total}}=\frac{840}{2.0} \] \[ C_{\mathrm{total}}=420\,\mathrm{J\,K^{-1}} \] \(\textbf{Separating the calorimeter contribution:}\) The water contributes \(300\,\mathrm{J\,K^{-1}}\). Therefore, \[ C_{\mathrm{vessel}}=C_{\mathrm{total}}-C_{\mathrm{water}} \] \[ C_{\mathrm{vessel}}=420-300 \] \[ C_{\mathrm{vessel}}=120\,\mathrm{J\,K^{-1}} \] \(\textbf{Heat absorbed during combustion:}\) The temperature rises by 3.0 K. Therefore, the water and vessel together absorb \[ q_{\mathrm{absorbed}}=C_{\mathrm{total}}\Delta T \] \[ q_{\mathrm{absorbed}}=420(3.0) \] \[ q_{\mathrm{absorbed}}=1260\,\mathrm{J} \] \(\textbf{Heat exchanged by the reacting system:}\) The reacting system releases the heat absorbed by the calorimetric assembly. \[ q_{\mathrm{reaction}}=-1260\,\mathrm{J} \] The reaction occurs at constant volume, and only pressure-volume work is possible. Therefore, \[ \Delta U_{\mathrm{reaction}}=q_V \] \[ \Delta U_{\mathrm{reaction}}=-1260\,\mathrm{J} \] \(\textbf{Converting to molar combustion energy:}\) The amount of substance burned is \[ n=0.015\,\mathrm{mol} \] Hence, \[ \Delta_cU_m=\frac{-1260}{0.015} \] \[ \Delta_cU_m=-84000\,\mathrm{J\,mol^{-1}} \] \[ \Delta_cU_m=-84\,\mathrm{kJ\,mol^{-1}} \] \(\textbf{Physical interpretation:}\) The positive temperature rise belongs to the calorimeter and water, whereas the reacting system loses energy. The calorimeter vessel absorbs part of the released energy and must not be omitted. The electrical calibration determines the effective heat capacity of the complete assembly before the combustion measurement. Because the combustion occurs at constant volume, the measured reaction heat corresponds to internal-energy change rather than automatically to reaction enthalpy. \(\textbf{Final answer:}\) \[ \boxed{\Delta_cU_m=-84\,\mathrm{kJ\,mol^{-1}}} \]

Explanation: \(\textbf{Heat-capacity relations for the two experiments:}\) For a fixed amount of ideal gas, \[ q_P=nC_{p,m}\Delta T \] \[ q_V=nC_{V,m}\Delta T \] Subtracting, \[ q_P-q_V=n(C_{p,m}-C_{V,m})\Delta T \] For an ideal gas, \[ C_{p,m}-C_{V,m}=R \] Therefore, \[ q_P-q_V=nR\Delta T \] \(\textbf{Recovering the amount of gas:}\) The difference between measured heat transfers is \[ q_P-q_V=850-750 \] \[ q_P-q_V=100\,\mathrm{J} \] Hence, \[ n=\frac{100}{(8.0)(25)} \] \[ n=\frac{100}{200} \] \[ n=0.50\,\mathrm{mol} \] \(\textbf{Determining constant-volume molar heat capacity:}\) At constant volume, \[ q_V=nC_{V,m}\Delta T \] Rearranging, \[ C_{V,m}=\frac{q_V}{n\Delta T} \] Substitution gives \[ C_{V,m}=\frac{750}{(0.50)(25)} \] \[ C_{V,m}=\frac{750}{12.5} \] \[ C_{V,m}=60\,\mathrm{J\,mol^{-1}\,K^{-1}} \] \(\textbf{Independent pressure-heat check:}\) The constant-pressure molar heat capacity is \[ C_{p,m}=C_{V,m}+R \] \[ C_{p,m}=60+8=68\,\mathrm{J\,mol^{-1}\,K^{-1}} \] The predicted constant-pressure heat is \[ q_P=(0.50)(68)(25)=850\,\mathrm{J} \] This agrees with the measured value. The calculation distinguishes the amount-dependent total heat capacity from the molar heat capacity. The additional heat at constant pressure accounts for expansion work during heating. \(\textbf{Final answer:}\) \[ \boxed{(n,C_{V,m})=(0.50\,\mathrm{mol},60\,\mathrm{J\,mol^{-1}\,K^{-1}})} \]

Explanation: For constant-pressure heating of an ideal gas, \[ q_P=nC_{p,m}\Delta T \] Both samples contain the same gas and undergo heating under the same specified conditions. Therefore, their molar heat capacities are identical. Both also absorb equal quantities of heat. Thus, \[ n_P C_{p,m}\Delta T_P=n_Q C_{p,m}\Delta T_Q \] Cancelling the common molar heat capacity, \[ n_P\Delta T_P=n_Q\Delta T_Q \] Substituting the amounts, \[ 2\Delta T_P=5\Delta T_Q \] Therefore, \[ \frac{\Delta T_P}{\Delta T_Q}=\frac52 \] The smaller sample undergoes the larger temperature rise because it has the smaller total heat capacity. However, both processes occur at constant pressure with only pressure-volume work. Therefore, \[ \Delta H_P=q_P \] \[ \Delta H_Q=q_Q \] Since both samples absorb exactly 1000 J, \[ \Delta H_P=\Delta H_Q=+1000\,\mathrm{J} \] This does not imply equal molar enthalpy changes. The molar enthalpy change is greater for the smaller sample because the same total enthalpy change is distributed over fewer moles. The problem therefore requires distinguishing the extensive enthalpy change from the intensive molar heat capacity. The correct comparison is a greater temperature rise for P but equal total enthalpy changes.

Explanation: The Assertion is true for a fixed amount of ideal gas under the stated conditions. At constant volume, no pressure-volume work is performed. Consequently, the heat supplied equals the increase in internal energy. At constant pressure, heating causes the ideal gas to expand. The expanding gas performs work against the external pressure. The supplied heat must therefore account for the same internal-energy increase and the additional expansion work. For a fixed amount of ideal gas, \[ q_P=nC_{p,m}\Delta T \] \[ q_V=nC_{V,m}\Delta T \] The molar heat capacities satisfy \[ C_{p,m}-C_{V,m}=R \] Since R is positive, \[ C_{p,m}\gt C_{V,m} \] For the same positive temperature rise, \(q_P\gt q_V\). The difference is \[ q_P-q_V=nR\Delta T \] This additional energy corresponds to the pressure-volume work contribution during constant-pressure heating. The Reason therefore correctly explains the Assertion. The conclusion specifically requires ideal-gas behaviour and the stated work conditions. It should not be extended without qualification to arbitrary substances or processes involving additional forms of work.

Explanation: \(\textbf{Heat absorbed by the calorimetric assembly:}\) The heat capacity is \[ C_{\mathrm{cal}}=250\,\mathrm{J\,K^{-1}} \] The temperature rise is \[ \Delta T=+4.0\,\mathrm{K} \] Therefore, \[ q_{\mathrm{cal}}=C_{\mathrm{cal}}\Delta T \] \[ q_{\mathrm{cal}}=250(4.0) \] \[ q_{\mathrm{cal}}=+1000\,\mathrm{J} \] \(\textbf{Reaction heat at constant volume:}\) The calorimeter gains heat released by the reacting system. Therefore, \[ q_{V,\mathrm{reaction}}=-1000\,\mathrm{J} \] At constant volume with only pressure-volume work, \[ \Delta U_{\mathrm{actual}}=-1000\,\mathrm{J} \] \(\textbf{Reaction extent:}\) The balanced equation consumes 2 mol of NO per mole of reaction extent. The actual amount consumed is 0.20 mol. Therefore, \[ \xi=\frac{0.20}{2} \] \[ \xi=0.10\,\mathrm{mol} \] The internal-energy change per mole of reaction as written is \[ \Delta_rU=\frac{-1000}{0.10} \] \[ \Delta_rU=-10000\,\mathrm{J} \] \[ \Delta_rU=-10.00\,\mathrm{kJ} \] \(\textbf{Gaseous mole correction:}\) For the balanced reaction, \[ \Delta n_g=2-(2+1) \] \[ \Delta n_g=-1 \] The ideal-gas correction at 300 K is \[ \Delta n_gRT=(-1)(8.314)(300) \] \[ \Delta n_gRT=-2494.2\,\mathrm{J} \] \[ \Delta n_gRT=-2.4942\,\mathrm{kJ} \] \(\textbf{Reaction enthalpy:}\) \[ \Delta_rH=\Delta_rU+\Delta n_gRT \] \[ \Delta_rH=-10.00-2.4942 \] \[ \Delta_rH=-12.4942\,\mathrm{kJ} \] \(\textbf{Sign and stoichiometry check:}\) The temperature rise is measured for the calorimeter, which absorbs heat. The reacting system consequently has a negative heat transfer. The reaction extent must be calculated using the stoichiometric coefficient of NO. Because the reaction decreases the number of gaseous moles, the enthalpy change is more negative than the internal-energy change. \(\textbf{Final answer:}\) \[ \boxed{\Delta_rH\approx-12.49\,\mathrm{kJ}} \]

Explanation: \(\textbf{Initial thermal equilibration:}\) The complete outer enclosure is insulated. Therefore, during thermal equilibration, energy lost by the warmer sample equals energy gained by the cooler sample. Let the intermediate equilibrium temperature be \(T_e\). \[ C_A(60-T_e)=C_B(T_e-20) \] Substituting, \[ 200(60-T_e)=300(T_e-20) \] Expanding, \[ 12000-200T_e=300T_e-6000 \] \[ 18000=500T_e \] \[ T_e=36^\circ\mathrm{C} \] \(\textbf{Combined heat capacity:}\) After thermal equilibrium is established, the samples are heated together. Their combined heat capacity is \[ C_{\mathrm{total}}=C_A+C_B \] \[ C_{\mathrm{total}}=200+300 \] \[ C_{\mathrm{total}}=500\,\mathrm{J\,K^{-1}} \] \(\textbf{Electrical-work stage:}\) The outer enclosure remains insulated, so \[ q=0 \] The heater performs 1500 J of work on the system. Therefore, \[ w=+1500\,\mathrm{J} \] The first law gives \[ \Delta U=q+w \] \[ \Delta U=+1500\,\mathrm{J} \] The resulting temperature increase is \[ \Delta T=\frac{\Delta U}{C_{\mathrm{total}}} \] \[ \Delta T=\frac{1500}{500} \] \[ \Delta T=3\,\mathrm{K} \] \(\textbf{Final temperature:}\) \[ T_f=36+3 \] \[ T_f=39^\circ\mathrm{C} \] \(\textbf{Interpretation:}\) The first stage redistributes energy internally and does not change the total internal energy of the isolated assembly. The second stage adds energy through electrical work. A temperature increase of 3 K has the same numerical magnitude as an increase of \(3^\circ\mathrm{C}\). The heating is not caused by heat crossing the boundary; it results from electrical work. \(\textbf{Final answer:}\) \[ \boxed{T_f=39^\circ\mathrm{C}} \]

Explanation: \(\textbf{Heat transferred during cooling:}\) For constant heat capacity, \[ q=C\Delta T \] During the first stage, \[ \Delta T_1=30-80 \] \[ \Delta T_1=-50\,\mathrm{K} \] Therefore, \[ q_1=250(-50) \] \[ q_1=-12500\,\mathrm{J} \] The negative sign indicates that the sample releases heat. \(\textbf{Heat transferred during reheating:}\) During the second stage, \[ \Delta T_2=50-30 \] \[ \Delta T_2=+20\,\mathrm{K} \] Therefore, \[ q_2=250(20) \] \[ q_2=+5000\,\mathrm{J} \] The positive sign indicates heat absorbed by the sample. \(\textbf{Combining the transfers:}\) \[ q_{\mathrm{total}}=q_1+q_2 \] \[ q_{\mathrm{total}}=-12500+5000 \] \[ q_{\mathrm{total}}=-7500\,\mathrm{J} \] \[ q_{\mathrm{total}}=-7.50\,\mathrm{kJ} \] \(\textbf{Endpoint verification:}\) The overall temperature change is \[ \Delta T_{\mathrm{overall}}=50-80=-30\,\mathrm{K} \] Since the applicable heat capacity is the same throughout, \[ q_{\mathrm{total}}=250(-30) \] \[ q_{\mathrm{total}}=-7500\,\mathrm{J} \] This independently confirms the result. A Celsius temperature difference has the same numerical magnitude as the corresponding kelvin difference. Adding 273 to a temperature difference would be incorrect. The sample absorbs heat during one stage but releases more heat during the other, giving a net heat loss. \(\textbf{Final answer:}\) \[ \boxed{q_{\mathrm{total}}=-7.50\,\mathrm{kJ}} \]

Explanation: \(\textbf{Trial I — Empty calorimeter:}\) The electrical heater supplies 800 J to the calorimeter vessel. The temperature rise is 4.0 K. Therefore, the vessel heat capacity is \[ C_{\mathrm{vessel}}=\frac{800}{4.0} \] \[ C_{\mathrm{vessel}}=200\,\mathrm{J\,K^{-1}} \] \(\textbf{Trial II — Vessel containing liquid:}\) The heater supplies 1800 J. The combined vessel and liquid undergo a 3.0 K temperature rise. Thus, \[ C_{\mathrm{combined}}=\frac{1800}{3.0} \] \[ C_{\mathrm{combined}}=600\,\mathrm{J\,K^{-1}} \] \(\textbf{Separating the liquid contribution:}\) For components undergoing the same temperature change, their heat capacities are additive. Therefore, \[ C_{\mathrm{combined}}=C_{\mathrm{vessel}}+C_{\mathrm{liquid}} \] Rearranging, \[ C_{\mathrm{liquid}}=C_{\mathrm{combined}}-C_{\mathrm{vessel}} \] Substitution gives \[ C_{\mathrm{liquid}}=600-200 \] \[ C_{\mathrm{liquid}}=400\,\mathrm{J\,K^{-1}} \] \(\textbf{Independent energy check:}\) In Trial II, the vessel absorbs \[ q_{\mathrm{vessel}}=200(3.0)=600\,\mathrm{J} \] The remaining supplied energy is \[ q_{\mathrm{liquid}}=1800-600 \] \[ q_{\mathrm{liquid}}=1200\,\mathrm{J} \] The corresponding liquid heat capacity is \[ C_{\mathrm{liquid}}=\frac{1200}{3.0} \] \[ C_{\mathrm{liquid}}=400\,\mathrm{J\,K^{-1}} \] \(\textbf{Interpretation:}\) The second trial measures the heat capacity of the vessel and liquid together, not the liquid alone. The vessel contribution must be removed using the independent calibration. Different quantities of supplied energy can produce different temperature rises because the effective heat capacities differ. The result is therefore obtained by separating the additive heat-capacity contributions rather than directly comparing the trial temperatures. \(\textbf{Final answer:}\) \[ \boxed{C_{\mathrm{liquid}}=400\,\mathrm{J\,K^{-1}}} \]

Explanation: The matching requires distinguishing total heat capacity from molar heat capacity. Total heat capacity depends on the amount of gas and is extensive. Molar heat capacity is the heat capacity per mole under the specified conditions. \(\textbf{Entry P:}\) Total constant-volume heat capacity is obtained by multiplying the molar constant-volume heat capacity by the number of moles. \[ C_V=nC_{V,m} \] Therefore, P matches 2. \(\textbf{Entry Q:}\) For an ideal gas, the molar heat capacities satisfy \[ C_{p,m}-C_{V,m}=R \] Rearranging, \[ C_{p,m}=C_{V,m}+R \] Therefore, Q matches 3. \(\textbf{Entry R:}\) The difference between the molar heat capacities is expressed as \[ C_{p,m}-C_{V,m} \] Thus, R matches 1. Its value equals R for an ideal gas. \(\textbf{Entry S:}\) The total heat capacities are \[ C_P=nC_{p,m} \] \[ C_V=nC_{V,m} \] Therefore, \[ C_P-C_V=n(C_{p,m}-C_{V,m}) \] Using the ideal-gas relation, \[ C_P-C_V=nR \] Thus, S matches 4. The factor n appears in the difference between total heat capacities but not in the difference between molar heat capacities. This distinction is essential when interpreting experimental heat-capacity values for samples containing different amounts of gas. The correct matching is P–2, Q–3, R–1 and S–4.