Thermodynamics practice · Medium

Option B ($\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6$) is correct because the enthalpy of polymerization calculation leads to 6 formaldehyde units.

Option C (If Assertion is True but the Reason is False) is correct because while entropy increases with temperature, the provided reason is not universally true.

Option B (Cyclic process) is correct because the system returns to its initial state (P1, V1, T1) after the series of changes.

Option D is correct because entropy is not always constant for a closed system, and a closed system is not always reversible.

Option C (Assertion is True but the Reason is False) is correct because adiabatic free expansion is irreversible, but PV = constant is only true for reversible adiabatic processes.

Option A ($-300 \mathrm{R} \ln 10$) is correct because it is the work done calculated using the formula for isothermal reversible expansion.

Option A (8584 K) is correct because it is the calculated flame temperature based on the heat released and the heat capacity of the product, water vapor.

Option C ($20 \mathrm{~J} / \mathrm{K}$) is correct because the entropy change for the universe is the sum of the entropy changes of the source and the sink.

Option A (Calorie) is correct because a calorie represents a larger amount of energy compared to joule, erg, and electron-volt.

Option D (-550 kcal) is correct because it is calculated using the formula: ΔH = Σ(B.E.)R - Σ(B.E.)P, considering the bond energies of reactants and products.

Option D is correct because both the assertion and the reason are incorrect. Spontaneity depends on Δ G, not just exothermic nature or entropy increase.

Option A (Monoatomic) is correct because the calculated value of $\gamma$ (1.66) is characteristic of a monoatomic gas.

Option B (195 kJ mol-1) is correct because the heat of combustion is calculated by multiplying the heat liberated by the number of moles of benzene.

Option C ($431 \mathrm{~kJ} / \mathrm{mol}$) is correct because it is calculated using the given enthalpy of formation and bond energies of reactants, applying Hess's law to determine the bond dissociation energy of $\mathrm{HCl}$.

Option A ($(x+y)$) is correct because the heat of formation of ${SO}_3$ is obtained by adding the given reactions, resulting in the sum of their enthalpy changes.

Option A (x > y) is correct because the enthalpy change for the reaction $\mathrm{CH}_4+\frac{1}{2} \mathrm{O}_2 \rightarrow \mathrm{CH}_3 \mathrm{OH}$ is negative, implying that the enthalpy of combustion of $\mathrm{CH}_4$ (x) is greater than that of $\mathrm{CH}_3 \mathrm{OH}$ (y).

Option B (Constant temperature) is correct because an isothermal process is defined as a thermodynamic process that occurs at a constant temperature.

Option D (Reversible processes may be performed infinite time) is correct because a reversible process, by definition, occurs infinitely slowly, implying it takes an infinite amount of time, not that it can be performed in an infinite amount of time.

Option A (-23 kJ/mol) is correct because the enthalpy change is calculated by subtracting the sum of bond energies of products from the sum of bond energies of reactants.

Option D (-22.06 kcal) is correct because for a reaction where the change in the number of moles of gas is zero, the change in enthalpy is equal to the change in internal energy.

Option C ($2x-z=y$) is correct because it is derived by manipulating the given thermochemical equations and applying Hess's Law.

Option C (1000, 1500 and 3000 K) is correct because at these temperatures, the Gibbs free energy change ($\Delta G$) is negative, indicating a spontaneous reaction.

Option C (Movable) is correct because if a closed system has adiabatic boundaries (no heat exchange), energy can only be exchanged through work, requiring at least one movable boundary.

The correct answer is (y-2x)/2. The enthalpy of formation of CO is calculated using Hess's Law by manipulating the given equations.

Option B (375 kJ) is correct because it accurately calculates the C=O bond energy by using the enthalpy of atomization for both CCl₄ and COCl₂.

Option C is correct. The assertion that entropy of the system always increases for a spontaneous reaction is true, but the reason that entropy of the reaction always decreases for a spontaneous reaction is false.

Option C (Enthalpy remains unaffected) is correct because during isothermal expansion of an ideal gas, the change in temperature is zero, leading to no change in enthalpy.

Option A (Low T, High P) is correct because low temperature and high pressure favor the backward reaction, which leads to the formation of 'A' according to Le Chatelier's principle.

Option B ($37^{\circ} {C}, 99^{\circ} {C}$) is correct because solving the Carnot engine efficiency equations with the given conditions yields these temperatures for the sink and source.

Option C (0.5) is correct because the change in the number of moles (∆n) is calculated by subtracting the moles of gaseous reactants from the moles of gaseous products in the balanced combustion equation, resulting in +0.5.