Matching Thermodynamic Processes with System Property Relations

To solve this problem, we analyze each thermodynamic process given in Column I and evaluate its corresponding system property changes and energy terms given in Column II.

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(A) Freezing of water at $273\text{ K}$ and $1\text{ atm}$

1. Phase Equilibrium ($\Delta G = 0$): At $273\text{ K}$ ($0^\circ\text{C}$) and $1\text{ atm}$, liquid water and ice exist in phase equilibrium. For a reversible phase transition at equilibrium: $\Delta G = 0$ This corresponds to option (T).

2. Entropy Change ($\Delta S_{\text{sys}} < 0$): Freezing represents a transition from a disordered liquid phase to a highly ordered solid crystalline structure. Therefore, the entropy of the system decreases: $\Delta S_{\text{sys}} < 0$ This corresponds to option (R).

3. Other Quantities:

   • Freezing is an exothermic process, releasing latent heat of fusion ($\Delta H_{\text{freeze}} < 0$), so $q \neq 0$.
   • The volume increases slightly upon freezing ($\Delta V \neq 0$), so expansion/compression work $w = -P\Delta V \neq 0$.
   • Consequently, $\Delta U = q + w \neq 0$.

Conclusion for (A): Matches with (R), (T).

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(B) Expansion of $1\text{ mol}$ of an ideal gas into a vacuum under isolated conditions

1. Heat Exchange ($q = 0$): Since the container is isolated from the surroundings, no heat transfer occurs: $q = 0$ This corresponds to option (P).

2. Work Done ($w = 0$): Free expansion into a vacuum implies that the external pressure $P_{\text{ext}} = 0$. Thus, work done by or on the gas is: $w = -\int P_{\text{ext}} dV = 0$ This corresponds to option (Q).

3. Internal Energy Change ($\Delta U = 0$): By the First Law of Thermodynamics: $\Delta U = q + w = 0 + 0 = 0$ This corresponds to option (S).

4. Entropy and Gibbs Energy:

   • Free expansion is an irreversible process for which $\Delta S_{\text{sys}} = n R \ln\left(\frac{V_2}{V_1}\right) > 0$ (since $V_2 > V_1$).
   • Since $\Delta U = 0$ for an ideal gas, $\Delta T = 0$, leading to $\Delta G = -T\Delta S_{\text{sys}} < 0$.

Conclusion for (B): Matches with (P), (Q), (S).

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

1. Heat Exchange ($q = 0$): The container is thermally isolated, so no heat exchange occurs with the surroundings: $q = 0$ This corresponds to option (P).

2. Work Done ($w = 0$): During the isothermal and isobaric mixing of ideal gases in a closed system without external pressure changes or movement of boundaries: $w = 0$ This corresponds to option (Q).

3. Internal Energy Change ($\Delta U = 0$): By the First Law of Thermodynamics: $\Delta U = q + w = 0$ This corresponds to option (S).

4. Entropy and Gibbs Energy:

   • Mixing is a spontaneous, irreversible process resulting in increased spatial disorder: $\Delta S_{\text{mix}} = -R \sum n_i \ln x_i > 0$
   • For ideal gases, enthalpy of mixing is zero ($\Delta H_{\text{mix}} = 0$), so $\Delta G_{\text{mix}} = -T\Delta S_{\text{mix}} < 0$.

Conclusion for (C): Matches with (P), (Q), (S).

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(D) Reversible heating of $\text{H}_2(g)$ at $1\text{ atm}$ from $300\text{ K}$ to $600\text{ K}$, followed by reversible cooling to $300\text{ K}$ at $1\text{ atm}$

1. Cyclic Nature of State Functions ($\Delta U = 0$, $\Delta G = 0$): The system starts at $(300\text{ K}, 1\text{ atm})$ and returns back to the exact initial thermodynamic state $(300\text{ K}, 1\text{ atm})$. For any cyclic process, state functions depend solely on the initial and final states: $\Delta U = U_{\text{final}} - U_{\text{initial}} = 0$ $\Delta G = G_{\text{final}} - G_{\text{initial}} = 0$ These correspond to options (S) and (T).

2. Total Heat and Work Transferred ($q = 0$, $w = 0$):

   • Step 1 (Heating from $300\text{ K}$ to $600\text{ K}$ at $1\text{ atm}$): $q_1 = n C_p (600 - 300) = 300 \, n C_p$ $w_1 = -n R (600 - 300) = -300 \, n R$
   • Step 2 (Cooling from $600\text{ K}$ to $300\text{ K}$ at $1\text{ atm}$): $q_2 = n C_p (300 - 600) = -300 \, n C_p$ $w_2 = -n R (300 - 600) = +300 \, n R$
   • Net Heat and Work: $q_{\text{net}} = q_1 + q_2 = 0$ $w_{\text{net}} = w_1 + w_2 = 0$ These correspond to options (P) and (Q).

Conclusion for (D): Matches with (P), (Q), (S), (T).

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Final Correct Matches:

• (A) $\rightarrow$ (R), (T)
• (B) $\rightarrow$ (P), (Q), (S)
• (C) $\rightarrow$ (P), (Q), (S)
• (D) $\rightarrow$ (P), (Q), (S), (T)