Thermodynamics practice · Mixed difficulty

Option C (HI is a stable compound) is false because a positive enthalpy of formation indicates that HI is an endothermic and therefore unstable compound.

Option A ($1.215 \times {10^3}\mathrm{J}$) is correct because it is calculated using the formula $W = -P_{ext} \Delta V$, converting liters to cubic meters.

Option D (q=0) is correct because in an isothermal free expansion, no work is done by the gas against external pressure, and since it's isothermal, there's no change in internal energy, hence no heat transfer.

The correct answer is Graphite is the stabler allotrope. Since the combustion of graphite releases slightly more heat than the combustion of diamond, graphite is at a lower energy state and therefore more stable.

Option C ($1\mathrm{\;cal} > 1\;J > 1\;erg$) is correct because it accurately represents the relative magnitudes of calorie, joule, and erg.

Option D (Trouton’s rule) is correct because it states that the ratio of molar heat of vaporization to the boiling point is approximately constant for many liquids.

Option C is correct because, according to the third law of thermodynamics, the entropy of a perfectly crystalline substance at absolute zero is zero.

Option B ($58.55 \mathrm{~kg}$) is correct because the work done by the gas during isothermal reversible expansion is used to lift the mass against gravity.

Option D (Does not violates 1st law of thermodynamics) is correct because the first law of thermodynamics states that energy is conserved, and the work done by the engine can be equal to the sum of the heat absorbed.

Option A ((i),(ii) and (iii)) is correct because work, heat, and energy all have the same units of energy, typically Joules (J).

Option C is correct. The assertion is correct because the enthalpy of neutralization of a weak acid and strong base is less than 57.1 kJ due to the energy required to ionize the weak acid. The reason is incorrect because strong bases are completely ionized.

Option A ($- 200\mathrm{ }cal$) is correct because the change in internal energy (∆E) is calculated as heat absorbed (q) plus work done on the system (w), where w is negative since the system does work on the surroundings.

Option B (+ 800kJ) is correct because the enthalpy change for the reaction is calculated by manipulating the given equations using Hess's Law.

Option D is correct because the slope depends on the adiabatic index (gamma), and He has the highest gamma, leading to the steepest slope, while O3 has the lowest.

Option B (Diamond) is correct because diamond has a highly ordered, rigid crystal structure, resulting in the least entropy compared to the other options.

Option D (Fe (1 mol,400 K)⟶Fe(1 mol,300 K)) is correct because the entropy decreases as the temperature decreases, leading to a more ordered state.

Option B ($- 100.5$) is correct because it represents the heat released per mole of FeS formed, calculated from the given data.

Option C (0) is correct because for an isothermal process with an ideal gas, the enthalpy change (∆H) is zero since enthalpy depends only on temperature.

Option D (Assertion is false but reason is true) is correct. Heat absorbed at constant temperature and volume is Δ U, not -Δ G, but a negative Δ G indicates spontaneity.

Work is done by the system equal to -179 J because, according to the first law of thermodynamics, $\Delta U = q + w$, so $w = \Delta U - q = -125 J - 54 J = -179 J$.

Option D (Saturated vapour) is correct because saturated vapor can have a negative specific heat under certain conditions, meaning adding heat can cause condensation and a decrease in temperature.

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

The correct answer is B (Cyclic process) because the system returns to its initial state after a series of changes.

Option A (-196 KJ) is the correct answer because it is calculated using the enthalpy of formation values for the products and reactants.

Option C (< 57.2 kJ) is correct because acetic acid is a weak acid, and some energy is used for its ionization, resulting in a lower heat of neutralization than that of a strong acid.

Option C (Low) is correct because molecular solids have weak intermolecular forces, requiring less energy to overcome during fusion.

Option B (Negative) is correct because a spontaneous chemical change is characterized by a decrease in Gibbs free energy, hence a negative value.

Option B ($\frac{3}{2} \mathrm{R}$) is correct because for a monatomic gas, the only degrees of freedom are translational, leading to this specific heat at constant volume.

Option A (Br₂(l)) is correct because the standard molar enthalpy of formation is zero for an element in its standard state.

Option A (An increase in enthalpy) is correct because melting is an endothermic process, requiring heat input and increasing the enthalpy of the substance.