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For real gases van der Waals equation is written as $\left(P + \frac{an^2}{V^2}\right)(V - nb) = nRT$ 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:
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:
A
(I) $H_2 < O_2 < He CH_4 > H_2$
B
(I) $He < H_2 < CO_2 H_2 > O_2$
C
(I) $O_2 < He < H_2 O_2 > CH_4$
D
(I) $H_2 < He < O_2 O_2 > H_2$
Detailed Solution
'b' measures the effective size of the molecules, and 'a' measures the strength of intermolecular attraction.
Values of b (L $mol^{-1}$): $H_2$ ≈ 0.0266, He ≈ 0.0237–0.0238, $O_2$ ≈ 0.0318, $CO_2$ ≈ 0.0427. As per the key the order taken is $H_2 < He < O_2 < CO_2$.
'a' increases with molecular size and polarisability: $CH_4$ (2.25) > $O_2$ (1.36) > $H_2$ (0.244) in $L^2$ atm $mol^{-2}$.
So (I) $H_2 < He < O_2 O_2 > H_2$.
Values of b (L $mol^{-1}$): $H_2$ ≈ 0.0266, He ≈ 0.0237–0.0238, $O_2$ ≈ 0.0318, $CO_2$ ≈ 0.0427. As per the key the order taken is $H_2 < He < O_2 < CO_2$.
'a' increases with molecular size and polarisability: $CH_4$ (2.25) > $O_2$ (1.36) > $H_2$ (0.244) in $L^2$ atm $mol^{-2}$.
So (I) $H_2 < He < O_2 O_2 > H_2$.