States of Matter - Result Question 36
####36. For real gases, van der Waals equation is written as
$ (p+\frac{a n^{2}}{V^{2}})(V-n b)=n R T $
where ’ $a$ ’ and ’ $b$ ’ are van der Waals constants.
Two sets of gases are :
(I) $O_2, CO_2, H_2$ and $He$ (II) $CH_4, O_2$ and $H_2$ The gases given in set-I in increasing order of ’ $b$ ’ and gases given in set-II in decreasing order of ’ $a$ ‘, are arranged below. Select the correct order from the following :
[2012 M]
(a) (I) $He<H_2<CO_2<O_2$ (II) $CH_4>H_2>O_2$
(b) (I) $O_2<He<H_2<CO_2$ (II) $H_2>O_2>CH_4$
(c) (I) $H_2<He<O_2<CO_2$ (II) $CH_4>O_2>H_2$
(d) (I) $H_2<O_2<He<CO_2$ (II) $O_2>CH_4>H_2$
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Solution:
- (c) Set-I : $O_2, CO_2, H_2, He$ for ‘b’ values Set-II : $CH_4, O_2, H_2, \quad$ for ‘a’ values Value of van der Waals constant ‘b’ increases with increase in molecular volume. Clearly, the increasing order of molecular volume (size of molecule) is:
$He<H_2<O_2<CO_2$
Value of ‘a’ increases with increase in intermolecular attraction. $CH_4$ is a polar molecule, thus, it will possess highest value of ’ $a$ ‘. $O_2$ and $H_2$, both are non-polar molecules. The van der Waals force $\propto$ molecular mass. Hence, the correct order of value of ‘b’ is : $CH_4>O_2>H_2$. From the given options, (c) is most suitable.
