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(D) For a van der Waals gas,the critical constants are related to the parameters $a$ and $b$ as follows:$V_C = 3b \implies b = \frac{V_C}{3} = \frac{0

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The gas cannot be liquefied at $200 \ K$

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(D) For a van der Waals gas,the critical constants are related to the parameters $a$ and $b$ as follows:$V_C = 3b \implies b = \frac{V_C}{3} = \frac{0.123}{3} = 0.041 \ L \ mol^{-1}$. Thus,option $B$ is incorrect.The volume of a single molecule is related to the excluded volume $b$ by $b = 4 	imes N_A 	imes V_{molecule}$. Therefore,$V_{molecule} = \frac{b}{4 	imes N_A} = \frac{0.123}{3 	imes 4 	imes 6.023 	imes 10^{23}} \ L$. Converting to $cc$ $(1 \ L = 1000 \ cc)$,we get $V_{molecule} = \frac{0.123 	imes 1000}{12 	imes 6.023 	imes 10^{23}} \ cc$. Thus,option $A$ is incorrect.The Boyle temperature $T_B$ is given by $T_B = \frac{a}{Rb}$. Since $T_C = \frac{8a}{27Rb}$,we have $T_B = \frac{27}{8} T_C = 3.375 	imes 180 \ K = 607.5 \ K$. Since $T_B > T_C$,option $C$ is incorrect.By definition,a gas cannot be liquefied above its critical temperature $T_C$. Since $T_C = 180 \ K$,the gas cannot be liquefied at $200 \ K$. Thus,option $D$ is correct.

$Gases$	$Critical \ temperature \ in \ Kelvin$
$P$	$33.2$
$Q$	$5.3$
$R$	$154.3$
$S$	$126$

Critical constants for a real gas are given as:
$T_C = 180 \ K$; $V_C = 0.123 \ L \ mol^{-1}$
$P_C = 45 \ atm$; $R = 0.082 \ L \ atm \ mol^{-1} \ K^{-1}$
Which of the following statements is correct for this real gas?

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Critical constants for a real gas are given as:$T_C = 180 \ K$; $V_C = 0.123 \ L \ mol^{-1}$$P_C = 45 \ atm$; $R = 0.082 \ L \ atm \ mol^{-1} \ K^{-1}$Which of the following statements is correct for this real gas?