Thermodynamics practice · Mixed difficulty

Option D (0.2 kcal) is the closest correct answer. The work done is calculated using $W = - \Delta n_{g}RT$, where $\Delta n_{g}$ is the change in the number of moles of gas.

Option B (${\mathrm{O}_3}$is less stable than${\mathrm{O}_2}$) is correct because the positive ∆H indicates that energy is absorbed to form O3, making it less stable than O2.

Option B (In the second case energy was spent to expand the gases) is correct because at constant pressure, the system does work against the external pressure as the volume increases, requiring energy.

Option B (-880430 J $\mathrm{mo}{\mathrm{l}^{ - 1}}$) is correct because it calculates the enthalpy change for the combustion of one mole of methane using the given energy change and accounting for the difference between $\Delta E$ and $\Delta H$.

Option A, $-2.303 n R T \log \left({V}_2 / {V}_1\right)$, correctly represents the work done in a reversible isothermal process.

Option B ($\Delta G = + \mathrm{ve}$ non spontaneous) is correct because at -15°C (below the freezing point), the melting of ice is a non-spontaneous process, hence ∆G is positive.

Option D (${\mathrm{N}_2} + 3{\mathrm{H}_2}\mathrm{\;} \to 2\mathrm{N}{\mathrm{H}_3}$) is correct because in this reaction, the number of moles of gaseous reactants is not equal to the number of moles of gaseous products, leading to a change in volume and thus $\Delta H \ne \Delta U$.

The correct answer is D (E and H are unaltered) because for an ideal gas in an isothermal process, internal energy (E) and enthalpy (H) depend only on temperature, which remains constant.

Option C ($10\,\mathrm{J}\,\,$) is correct because the change in internal energy is calculated as the heat added to the system minus the work done by the system.

Option A (Ice) is correct because ice has a highly ordered crystalline structure, making it the least random state of water compared to liquid water and steam.

Option B (warm on expansion) is correct because the gas has a higher pressure than expected for an ideal gas at the given temperature, indicating attractive forces dominate, leading to warming upon expansion (Joule-Thomson effect).

Option A (30.38 J) is correct because it is calculated using the formula for work done at constant pressure, converting volume to liters, and then to Joules.

Option A (R) is correct because at constant pressure, the work done is equal to R when the temperature of one mole of gas increases by 1 degree.

Option D (Sublimation) is correct because sublimation is the process of a solid changing directly into a gas.

Option B (200 kcal) is correct because the heat of combustion is calculated based on the molar mass of methane (16 g/mol) and the heat liberated during the combustion of 2.0 g of methane.

Option A ($\Delta {H}_2-\frac{1}{2} \Delta {H}_1$) is correct because it is obtained by manipulating the given reactions to arrive at the desired reaction.

Option C (Infinite) is correct because at constant temperature and pressure, a phase change occurs, and the heat capacity approaches infinity as temperature change is minimal for a significant heat input.

Option D (120 kJ) is correct because the Gibbs free energy change is calculated using ∆G = ∆H - T∆S.

Option A (2) is correct because the entropy change is calculated by dividing the heat of fusion by the temperature, and then adjusting for the mass of the ice.

Option D (-100 J) is correct because in an adiabatic process, the work done by the system is equal to the negative change in internal energy.

Option C ($2 \mathrm{A} \rightleftharpoons \mathrm{B}+\mathrm{C}$) is correct because by calculating the equilibrium constant K from the given Δ G° and comparing it with the reaction quotient Q, we can determine that the reaction shifts towards the products, which corresponds to the stoichiometry of the reaction $2 \mathrm{A} \rightleftharpoons \mathrm{B}+\mathrm{C}$.

Option D is correct because in a vacuum expansion, no external pressure opposes the expansion, so no work is done (W=0), no heat is exchanged (q=0), and the internal energy remains constant (∆U=0).

Option A (11.4 kJ) is correct because the heat released is determined by the limiting reactant (NaOH) and the heat of neutralization.

Option A ($\Delta \mathrm{G}=0$) is correct because at equilibrium, the Gibbs free energy change is zero, indicating no net change in the system.

Option C (Third law of thermodynamics) is correct because the third law specifically addresses the entropy of pure crystalline solids at absolute zero.

Option D (3224 cal) is correct because it involves calculating the work done during isothermal expansion, then finding the useful work using the Carnot efficiency.

Option C (Some amount of hot water enclosed in a closed container which is not insulated) is correct because a closed system allows energy exchange but not mass exchange.

Option A (Hess) is correct because Hess's Law states that the enthalpy change of a reaction is independent of the path taken.

Option A is correct because work done in an isobaric process is given by W = -PΔV, which is the same for both reversible and irreversible processes under constant pressure.

Option C ((II) and (III)) is correct because q (heat) and w (work) are path-dependent functions, while q+w (which equals Δ U) and H-TS (which equals Gibbs Free Energy) are state functions.