In an adiabatic process, \(Q = 0\), so \(\Delta U = -W\), which means internal energy changes, so Statement I is incorrect. Statement II is correct since change in internal energy corresponds directly to the work done on or by the gas.

Temperature is the physical quantity that determines thermal equilibrium. Two systems are in thermal equilibrium if and only if they are at the same temperature.

From the ideal gas law \(PV = nRT\), the volume is proportional to \(\frac{T}{P}\) for equal masses of the same gas. Thus, \(\frac{V_1}{V_2} = \frac{T_1}{T_2} \times \frac{P_2}{P_1} = \frac{400}{600} \times \frac{2P_0}{P_0} = \frac{4}{3}\).

Number of moles \(n = \frac{m}{M} = \frac{14\text{ g}}{28\text{ g/mol}} = 0.5\text{ mol}\). Substituting \(n = \frac{1}{2}\) in \(PV = nRT\) gives \(PV = \frac{1}{2}RT\).

By first law, $Q = \Delta U + W$. In an isobaric process, the work done is $W = P\Delta V = nR\Delta T$. Therefore, both expressions are correct.

Heat and work depend on the path taken between states, whereas internal energy is a state function. Therefore, (a) and (c) are path functions.

Efficiency is given by $\eta = 1 - \frac{T_{\text{sink}}}{T_{\text{source}}}$. Substituting the given values: $\eta = 1 -\frac{200}{600} =\frac{2}{3} \approx 66.67%$.

The molecular mass of nitrogen gas \((\text{N}_2)\) is \(28\text{ g/mol}\). The number of moles is \(n = \frac{14}{28} = 0.5\). Thus, using \(PV = nRT\), we get \(PV = \frac{1}{2}RT\).

For an isothermal process, the change in internal energy is \( \Delta U = 0 \). According to the first law of thermodynamics, \( Q = \Delta U + W \), which gives \( Q = 0 + 150\text{ J} = 150\text{ J} \).

Isothermal process has constant temperature (a-iv). Isobaric has constant pressure (b-ii). Isochoric has constant volume (c-i). Adiabatic has no heat flow (d-iii). Matching these gives option D.