For real gases,the relation between $p$,$V$,and $T$ is given by the van der Waals equation:
$(p + \frac{an^2}{V^2})(V - nb) = nRT$
Where $a$ and $b$ are van der Waals constants,$nb$ is approximately equal to the total volume of the molecules of a gas,and $a$ is the measure of the magnitude of intermolecular attraction.
$(i)$ Arrange the following gases in the increasing order of $b$. Give reason: $O_2, CO_2, H_2, He$
$(ii)$ Arrange the following gases in the decreasing order of magnitude of $a$. Give reason: $CH_4, O_2, H_2$
$(p + \frac{an^2}{V^2})(V - nb) = nRT$
Where $a$ and $b$ are van der Waals constants,$nb$ is approximately equal to the total volume of the molecules of a gas,and $a$ is the measure of the magnitude of intermolecular attraction.
$(i)$ Arrange the following gases in the increasing order of $b$. Give reason: $O_2, CO_2, H_2, He$
$(ii)$ Arrange the following gases in the decreasing order of magnitude of $a$. Give reason: $CH_4, O_2, H_2$
(A) $(i)$ The van der Waals constant $b$ represents the excluded volume per mole,which is related to the size of the gas molecules. Larger molecules occupy more volume. Thus,the order of increasing $b$ is $H_2 < He < O_2 < CO_2$.
$(ii)$ The van der Waals constant $a$ is a measure of the magnitude of intermolecular attraction. Larger molecules with larger electron clouds exhibit stronger London dispersion forces. Thus,the order of decreasing magnitude of $a$ is $CH_4 > O_2 > H_2$.
$(ii)$ The van der Waals constant $a$ is a measure of the magnitude of intermolecular attraction. Larger molecules with larger electron clouds exhibit stronger London dispersion forces. Thus,the order of decreasing magnitude of $a$ is $CH_4 > O_2 > H_2$.
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