(a) Which of the thermodynamic quantities \(p, H, q, w,\) and \(G\) are state functions? (b) Consider a system going from state 1 to state 2 in a reversible and an irreversible way. Compare \(q_{\text {rev }}\) and \(q_{\text {irtev }}\) (c) Consider a system going from state 1 to state 2 in a reversible and an irreversible way. Compare \(w_{\text {rev }}\) and \(w_{\text {trev }}\). (d) For a reversible isothermal process, write an expression for \(\Delta H\) and an expression for \(\Delta G\) in terms of \(q, w\) and \(T, p\) and \(\Delta V\).

State functions are properties that depend only on the current state of the system and not on the path taken to reach that state. From the given thermodynamic quantities, we can identify state functions as follows: 1. \(p\) - Pressure is a state function, as it depends only on the current state of the system. 2. \(H\) - Enthalpy is a state function, as it depends only on the internal energy, pressure, and volume of the system. 3. \(q\) - Heat is not a state function, as it depends on the path taken by the system during a change of state. 4. \(w\) - Work is not a state function, as it also depends on the path taken by the system during a change of state. 5. \(G\) - Gibbs free energy is a state function, as it depends only on enthalpy, temperature, and entropy.

In a reversible process, the system is in thermal and mechanical equilibrium with its surroundings at each moment. In an irreversible process, the system is not in complete equilibrium during the change of state. Since the heat transfer \(q\) depends on the path taken by the system during a change of state, \(q_{rev}\) and \(q_{irrev}\) can have different values.

Similar to heat, work also depends on the path taken. In a reversible process, the work done (\(w_{rev}\)) is maximum, while in an irreversible process, the work done (\(w_{irrev}\)) is less. This is because there is more energy lost as heat in an irreversible process, which reduces the amount of work done.

For a reversible isothermal process, the temperature \(T\) remains constant. The change in enthalpy (\(\Delta H\)) can be expressed as: \[\Delta H = q_p\] where \(q_p\) is the heat transfer at constant pressure. The change in Gibbs free energy (\(\Delta G\)) can be expressed as: \[\Delta G = \Delta H - T \Delta S\] Since the process is isothermal, we can relate the pressure, volume, and temperature changes: \[w = -nRT \ln \frac{V_2}{V_1}\] where \(n\) is the number of moles, \(R\) is the gas constant, and \(V_1\) and \(V_2\) are the initial and final volumes, respectively. As \(\Delta G\) is equal to the maximum non-expansion work done by the system, we have: \[\Delta G = -w_{rev}\] Thus, in a reversible isothermal process, we can write the change in enthalpy as: \[\Delta H = q_p\] and the change in Gibbs free energy as: \[\Delta G = -w_{rev} = nRT \ln \frac{V_2}{V_1}\]