Thermodynamics practice · Mixed difficulty

Option B (Temperature) is correct because the internal energy of an ideal gas is directly proportional to its temperature.

Option D (-40 kcal) is correct because the enthalpy change is calculated by subtracting the bond energies of products from the bond energies of reactants.

Option B ($w = - 2.303\,\,nRT\,\,\log \frac{{{V_f}}}{{{V_i}}}$) is correct because it is the correct formula for work done in an isothermal reversible process.

Option B (Its absolute value cannot be determined) is correct because, according to thermodynamics, only changes in internal energy can be measured, not the absolute value.

Option A ($2.41 \mathrm{~JK}^{-1} \mathrm{~mol}^{-1}$) is correct because the entropy change is calculated by dividing the heat exchanged during a reversible process at constant temperature by the temperature.

The correct answer is A because work done in an isobaric process is given by $W = -P\Delta V$, and since the initial and final states are the same, $\Delta V$ is the same for both reversible and irreversible processes.

Option A ($x$) is correct because the enthalpy of formation of $CO_2(g)$ is equal to the enthalpy of combustion of $C(graphite)$.

Option B is correct. The Gibbs energy change ($\Delta G$) and standard Gibbs energy change ($\Delta G^\circ$) are equal when the reaction quotient (Q) is equal to 1, as $\Delta G = \Delta G^\circ + RT \ln Q$.

Option C ($5:3$) is correct because both Helium and Neon are monoatomic gases, and the ratio of $C_p$ to $C_v$ for a monoatomic gas is always $5/2 : 3/2$, which simplifies to $5:3$.

Option D (I, III and IV) is correct because it correctly identifies the conditions where ΔSvap is greater than, less than, or equal to 100 J/K mol.

Option B is the correct answer because the weaker the acid, the less negative the enthalpy of neutralization (${\Delta _{neut}}{H^0}$) and the higher the pKa value.

Option B (Always negative) is correct because the heat of combustion is always an exothermic process, meaning heat is released, resulting in a negative enthalpy change.

Option C (zero) is correct because the third law of thermodynamics states that the entropy of a perfectly crystalline substance at absolute zero is zero.

The correct answer is A (-87.5 kJ mol⁻¹) because it is obtained by using the Gibbs free energy equation, considering the standard free energy change and the partial pressure of CO₂.

Option A ($58 \mathrm{gm}$) is correct because it is the mass of butane required to produce the heat needed to raise the water temperature.

Option D (Work) is correct because work is a path function, meaning its value depends on the path taken during a process, while Gibbs energy, pressure, and temperature are state functions.

Option B is correct because the reaction of a strong acid (HCl) with a strong base (NaOH) to form salt and water is specifically termed heat of neutralization.

Option A (5.260 cal/(mol K)) is correct because it is calculated by dividing the enthalpy of fusion by the temperature in Kelvin.

The correct answer is A (y – 2x) because it is derived by manipulating the given equations to find the enthalpy of formation of SO2.

Option B (Negative) is correct because for a spontaneous chemical change, the Gibbs free energy change (∆G) must be negative.

The correct answer is C (If Statement-I is true but Statement-II is false) because hydrocarbons require more moles of oxygen for complete combustion than the moles of hydrocarbon, but this is not universally true for all substances.

Option B ($L{T^{ - 1}}$) is correct because entropy change is defined as the heat absorbed divided by the temperature at which the process occurs.

Option A (760 J) is correct because the change in internal energy is calculated using the formula $\Delta U = nC_v\Delta T$ and the given isobaric conditions.

The correct answer is C (Isobaric) because an isobaric process is defined as a process that occurs at constant pressure.

Option A ($201.2 \,{cal}$) is correct because it is the change in internal energy calculated using the formula $\Delta E = nC_V \Delta T$.

Option B (1) is correct because when the reaction quotient Q is equal to 1, the Gibbs energy change equals the standard Gibbs energy change.

Option A (62 kJ) is correct because it involves using Kirchoff's relation to calculate the change in enthalpy by integrating the heat capacity equation over the given temperature range.

Option D (0.00 J/K) is correct because the entropy change of the system is zero for a reversible process where the heat gained and lost are equal, and the temperature difference is infinitesimally small.

The correct answer is B (Both Statement I and Statement II are correct and Statement II is not the correct explanation of Statement I) because increasing temperature increases entropy, and higher temperature leads to more disorder, but the second statement isn't directly explaining the first.

The correct answer is D ($0.1\mathrm{Jmo}{\mathrm{l}^{ - 1}}{\mathrm{K}^{ - 1}}$) because for an isothermal process, the change in internal energy (∆U) is zero.