Which of the following relations are correct?
(A) $$\Delta U = q + p\Delta V$$
(B) $$\Delta G = \Delta H - T\Delta S$$
(C) $$\Delta S = \frac{q_{rev}}{T}$$
(D) $$\Delta H = \Delta U - \Delta nRT$$
Choose the most appropriate answer from the options given below:

We need to identify which thermodynamic relations are correct.

(A) $$\Delta U = q + p\Delta V$$

The first law of thermodynamics states: $$\Delta U = q + w$$.

For work done on the system at constant pressure: $$w = -p\Delta V$$ (IUPAC convention).

So $$\Delta U = q - p\Delta V$$, not $$q + p\Delta V$$.

Statement (A) is incorrect.

(B) $$\Delta G = \Delta H - T\Delta S$$

The Gibbs free energy is defined as $$G = H - TS$$.

At constant temperature: $$\Delta G = \Delta H - T\Delta S$$.

Statement (B) is correct.

(C) $$\Delta S = \frac{q_{rev}}{T}$$

By definition, the entropy change for a reversible process at constant temperature is:

$$\Delta S = \frac{q_{rev}}{T}$$

Statement (C) is correct.

(D) $$\Delta H = \Delta U - \Delta nRT$$

The correct relation between enthalpy and internal energy for ideal gases is:

$$\Delta H = \Delta U + \Delta n_g RT$$

Statement (D) has a minus sign instead of a plus sign, so it is incorrect.

Conclusion: Only statements B and C are correct.

The correct answer is Option B: B and C only.