Match Gas Conditions with Properties and Thermodynamic Laws

To match each condition in Column I with the corresponding properties/laws in Column II, we analyze the behavior of gases using the van der Waals equation of state:

$\left(P + \frac{an^2}{V^2}\right)(V - nb) = nRT$

where $a$ is the measure of intermolecular attractive forces and $b$ accounts for the finite size of the gas molecules. The compressibility factor is defined as $Z = \frac{PV}{nRT}$.

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(A) Hydrogen gas ($P = 200\text{ atm}, T = 273\text{ K}$)

1. Hydrogen ($\text{H}_2$) has an extremely low polarizability, making intermolecular attractive forces negligible ($a \approx 0$) at $273\text{ K}$.
2. At high pressure ($P = 200\text{ atm}$), the molar volume $V/n$ is small, so the molecular volume term $nb$ cannot be neglected compared to $V$.
3. Substituting $a \approx 0$ into the van der Waals equation:
   $P(V - nb) = nRT$
4. Rearranging for the compressibility factor $Z$:
   $PV - nPb = nRT \implies Z = \frac{PV}{nRT} = 1 + \frac{Pb}{RT} > 1 \implies Z \neq 1$

Thus, (A) matches with (p) and (s).

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(B) Hydrogen gas ($P \sim 0, T = 273\text{ K}$)

1. As the pressure approaches zero ($P \sim 0$), the molar volume becomes extremely large ($V \to \infty$).
2. Under these conditions, both intermolecular forces ($\frac{an^2}{V^2} \approx 0$) and the co-volume term ($nb \ll V$) become completely negligible.
3. The equation reduces to the ideal gas law:
   $PV = nRT$
4. Here, $Z = 1$.

Thus, (B) matches with (r).

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(C) $\text{CO}_2$ ($P = 1\text{ atm}, T = 273\text{ K}$)

1. For carbon dioxide ($\text{CO}_2$), $273\text{ K}$ is below its critical temperature ($T_c = 304.1\text{ K}$) and well below its Boyle temperature ($T_b \approx 714\text{ K}$).
2. At $P = 1\text{ atm}$, attractive intermolecular forces are significant, meaning the attractive term $\frac{an^2}{V^2}$ dominates while the molecular volume $nb$ is negligible compared to $V$.
3. Under dominant attractive forces, the pressure is lower than ideal pressure, making $Z < 1$, which implies $Z \neq 1$.

Thus, (C) matches with (p) and (q).

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(D) Real gas with very large molar volume

1. When the molar volume $V_m = \frac{V}{n}$ is extremely large ($V_m \to \infty$), the gas density is extremely low.
2. The pressure correction term $\frac{a}{V_m^2}$ drops off as $O(1/V_m^2)$, and the volume correction $\frac{b}{V_m}$ drops off as $O(1/V_m)$.
3. If $\frac{a}{V_m^2}$ is neglected first, the equation takes the form $P(V - nb) = nRT$.
4. At very large molar volume, both corrections become completely negligible ($\frac{an^2}{V^2} \to 0$ and $nb \ll V$), so the real gas obeys the ideal gas equation:
   $PV = nRT$

Thus, (D) matches with (r).

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Final Matching Summary:

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