Question

For real gases the relation between $p,V$ and $T$ is given by the Van der Waals equation:

$(P+\frac{a{n}^2}{V^2})(V–nb)=nRT$

where ${}^{\prime }{a}^{\prime }$ and ${}^{\prime }{b}^{\prime }$ are van der Waals constants, ${}^{\prime }n{b}^{\prime }$ is approximately equal to the total volume of the molecules of a gas. ${}^{\prime }{a}^{\prime }$ is the measure of the magnitude of intermolecular attraction.

A Arrange the following gases in the increasing order of ${}^{\prime }{b}^{\prime }$. Give a reason. $O_2,C{O}_2,H_2,He$.

B Arrange the following gases in the decreasing order of magnitude of ${}^{\prime }{a}^{\prime }$. Give a reason. $C{H}_4,O_2,H_2$

Answer 1

${}^{\prime }{b}^{\prime }$ denotes molar volume occupied by the gas. The greater the size of the molecules, the greater is the volume occupied by the molecules.

Therefore, the increasing order of $b$ is: $H_2<He<O_2<C{O}_2$.

As hydrogen gas is of smallest size and carbon dioxide of largest size. On the other hand, helium is having size greater than hydrogen gas but smaller than oxygen gas.


Understanding van der Waals constant ${}^{\prime }{a}^{\prime }$:

${}^{\prime }{a}^{\prime }$ presents intermolecular attractions of the molecules.

All of the given molecules $C{H}_4,O_2$ and $H_2$ are nonpolar. Since, the surface area of $C{H}_4$ is highest and it contains highest number of electrons. So, it has the highest Van Der Waals force of attraction. So, it will have the highest value of ${}^{\prime }{a}^{\prime }$.

Therefore, the decreasing order of magnitude ${}^{\prime }{a}^{\prime }$ is:
$C{H}_4>O_2>H_2$

Hydrogen gas have smallest size (or surface area ) and it contains lowest number of electrons. So, it will have weakest van der waals forces of attraction and thus lowest value of ${}^{\prime }{a}^{\prime }$.