Question

Match gases under specified conditions listed in Column I with their properties/laws in Column II.

Column I		Column II
A	Hydrogen gas $(P=200$ atm, $T=273K)$	P	Compression factor $\ne 1$
B	Hydrogen gas $(P=0,$, $T=273K)$	Q	Attractive forces are dominant
C	$C{O}_2(P=1atm,T=273K)$	R	$PV=nRT$
D	Real gas with very large molar volume	S	$P(V-nb)=nRT$

• A. $(A)\to (P),(R);(B)\to (R);(C)\to (P),(Q);(D)\to (Q)(S)$
• B. $(A)\to (P),(Q);(B)\to (R);(C)\to (P),(Q);(D)\to (P)(S)$
• C. $(A)\to (P),(S);(B)\to (R);(C)\to (P),(Q);(D)\to (P)(S)$
• D. $(A)\to (P),(S);(B)\to (R);(C)\to (P),(Q);(D)\to (Q)(S)$

Answer 1

Answer: (C)

(A) Hydrogen gas at high pressure and low temperature does not behave as an ideal gas, so $Z\ne 1$. The van der Waal's equation is
$\left(P+\frac{a{n}^2}{V_m^2}\right)\left(V_m-b\right)=nRT$
At high pressure, the term $a{n}^2/V_m{}^2$ can be ignored in comparison to value of $P$.
Therefore, the equation reduces to $P(V-nb)=nRT$
(B) For $H_2$ gas at pressure $=0$ atm, the gas behaves as ideal gas,
so value of $Z=1$ at $P=0$ and it increases continuously on increasing $P$.
The ideal gas equation is obeyed, so $PV=nRT$.

(C) $C{O}_2$ molecules have larger attractive forces under normal conditions.
The gas is at temperature, close to its liquefaction temperature, so deviation from ideal gas behaviour is high.
(D) At very large molar volume $Z$ is not equal to 1 , instead $Z=$ $P{V}_m/RT$.