1. Process (A): Freezing water at $273\,\text{K}$ and $1\,\text{atm}$

At $273\,\text{K}$ and $1\,\text{atm}$, water and ice are in equilibrium.

• During freezing, the system becomes more ordered, so $\Delta S_{\text{sys}} < 0$ Hence, $(r)$ is correct.

• At phase equilibrium, $\Delta G = 0$ for the phase change at the transition temperature and pressure. Hence, $(t)$ is correct.

So, $A \to r,t$

2. Process (B): Expansion of 1 mol of an ideal gas into a vacuum under isolated conditions

This is free expansion into vacuum.

• Since external pressure is zero, $w = -\int P_{\text{ext}}\,dV = 0$ So $(q)$ is correct.

• The container is isolated, so no heat exchange: $q=0$ So $(p)$ is correct.

• For an ideal gas, internal energy depends only on temperature. In free expansion of an ideal gas in an isolated system, $\Delta U = q+w = 0+0=0$ So $(s)$ is correct.

Thus, $B \to p,q,s$

3. Process (C): Mixing of equal volumes of two ideal gases at constant temperature and pressure in an isolated container

For mixing in an isolated container:

• Isolated system implies $q=0$ So $(p)$ is correct.

• Since gases mix on their own in the container, no useful work is done; for ideal gases mixing in a rigid isolated container, effectively $w=0$ So $(q)$ is correct.

• For ideal gases, internal energy depends only on temperature. Since temperature is constant, $\Delta U = 0$ So $(s)$ is correct.

Thus, $C \to p,q,s$

4. Process (D): Reversible heating of $\mathrm{H_2(g)}$ at $1\,\text{atm}$ from $300\,\text{K}$ to $600\,\text{K}$, followed by reversible cooling to $300\,\text{K}$ at $1\,\text{atm}$

This is a cyclic process: initial and final states are same.

Therefore, all state functions return to original values:

• $\Delta U = 0$ so $(s)$ is correct.

• Since the process is cyclic, $\Delta G = 0$ so $(t)$ is correct.

Now check $q$ and $w$:

At constant pressure,

• heating: gas expands, so work is done,
• cooling: gas contracts, work is done on gas.

For 1 mol ideal gas at constant pressure, $w = -P\Delta V = -nR\Delta T$ for each step.

Heating $300 \to 600$ K: $w_1 = -nR(600-300) = -300R$

Cooling $600 \to 300$ K: $w_2 = -nR(300-600)=+300R$

Hence, $w_{\text{net}} = w_1+w_2=0$ So $(q)$ is correct.

From first law over complete cycle, $\Delta U = q+w = 0$ Since $w=0$, we get $q=0$ So $(p)$ is also correct.

Thus, $D \to p,q,s,t$

5. Now match with options

We obtained:

• $A \to r,t$
• $B \to p,q,s$
• $C \to p,q,s$
• $D \to p,q,s,t$

This exactly matches Option C.

6. Comparison with stored answer

Stored correct answer: C

Our derived answer: C

So they agree.