$${{a{n^2}} \over {{V^2}}} = atm \Rightarrow a = atm \times {{d{m^6}} \over {mo{l^2}}}$$

For London dispersion forces.

$$E \propto {1 \over {{r^6}}}$$

Hence, x = $$-$$6

According to Dalton’s law of partial pressure, p_i = x_i × P_T

p_i = partial pressure of the ith component

x_i = mole fraction of the ith component

p_T = total pressure of mixture

$$ \Rightarrow $$ 2 atm = $$\left( {{{{n_{{H_2}}}} \over {{n_{{H_2}}} + {n_{He}} + {n_{{O_2}}}}}} \right)$$ $$ \times $$ p_T

$$ \Rightarrow $$ p_T = $$\left( {{{1 + 1 + 1} \over 1}} \right)$$ $$ \times $$ 2 = 6 atm

For ideal gas,

PM = dRT [As V = $${M \over d}$$ ]

$$ \Rightarrow $$ d = $${{PM} \over {RT}}$$

$$ \therefore $$ d $$ \propto $$ $${1 \over T}$$ $$ \Rightarrow $$ d v/s $${1 \over T}$$ graph is Straight line.

d v/s T $$ \Rightarrow $$ graph is Hyperbolic.

As d $$ \propto $$ p

$$ \therefore $$ d v/s p graph is Straight line.