Thermodynamics practice · Easy

Option D is correct because entropy is an extensive property, meaning its value depends on the amount of substance, while boiling point, viscosity, and surface tension are intensive properties.

Option D (Enthalpy, internal energy, volume) is correct because all three properties are extensive, meaning they depend on the amount of substance.

Option B (Increases by 6 kJ) is correct because, according to the first law of thermodynamics, the change in internal energy is the sum of heat absorbed and work done by the system.

Option D (Standard conditions are denoted by adding the superscript ⦵ to the symbol ΔH e.g., -ΔH⦵) is correct because the standard state of solid iron at a specified temperature is its pure form at 1 bar, and the example given in the question is correct.

Option B ($\Delta \mathrm{G}<0$) is correct because a spontaneous process is characterized by a decrease in Gibbs free energy.

Option A (Zero) is correct because the work done during the expansion of an ideal gas into a vacuum is zero, as the external pressure is zero.

The correct answer is C (Spontaneous at all temperatures) because the decomposition of ozone has a negative enthalpy and positive entropy change, making Gibbs free energy negative at all temperatures.

Option C (Energy) is the state function because its value depends only on the initial and final states of the system, not on the path taken.

Option B (Both (A) & (R) are true but the (R) is not the correct explanation of the (A)) is correct because both statements are true, but the reason doesn't explain why work is a path function.

Option C (${Ne}$ and ${Cl}$) is correct because the electron gain enthalpy of Ne is positive (noble gas) and that of Cl is highly negative, resulting in the maximum difference.

Option B ($\frac{1}{2}{\Delta _f}\Delta H^\circ \left( {{\mathrm{B}_2}{\mathrm{O}_3}} \right)$) is correct because the standard heat of combustion of solid boron is half the standard heat of formation of B2O3.

Option D (Internal energy remains constant) is correct because for an ideal gas, internal energy depends only on temperature, and in an isothermal process, temperature is constant.

Option D ($\Delta_{\mathrm{eg}} \mathrm{H}(\mathrm{Te}) < \Delta_{\mathrm{eg}} \mathrm{H}(\mathrm{Po})$) is correct because electron gain enthalpy generally becomes less negative down the group due to increasing atomic size.

Option C (Isolated system) is correct because a well-stoppered thermos flask prevents both matter and energy exchange with the surroundings.

Option D (∆H= ∆U-RT) is correct because ∆H and ∆U are related by the equation ∆H = ∆U + ∆nRT, where ∆n is the change in the number of moles of gas.

Option B (Law of conservation of energy) is correct because Hess's law is a direct consequence of the law of conservation of energy, stating that the total enthalpy change is independent of the pathway.

Option B (G=H-TS) is correct because it represents the fundamental relationship between Gibbs free energy, enthalpy, temperature, and entropy.

Option A (The high bond energy of the nitrogen molecule) is correct because the high bond energy of $N_2$ requires a significant amount of energy to break the bond, making the formation of nitrogen oxides endothermic.

Option C (${O}-{H}$ bond will be easily broken is case of hydroxide of ' ${X}$ ') is correct because if X(OH) is more acidic than Y(OH), the O-H bond in X(OH) is more easily broken, indicating a stronger tendency to donate a proton.