According to the first law of thermodynamics, \(Q = \Delta U + W\). For an isobaric process, the work done is \(W = P\Delta V = nR\Delta T\). Therefore, both equations (1) and (2) are correct representation.

Heat and work depend on the path taken by the system during a thermodynamic process, making them path functions. Internal energy is a state function as it depends only on the initial and final states of the system.

The maximum efficiency is given by the Carnot efficiency formula: \(\eta = 1 - \frac{T_2}{T_1}\). Here, \(T_1 = 600\text{ K}\) and \(T_2 = 200\text{ K}\), so \(\eta = 1 -\frac{200}{600} = 1 - \frac{1}{3} = \frac{2}{3} \approx 66.67%\).

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.

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 \( Q = 0 + 150\text{ J} = 150\text{ J} \).