Gaseous State

$\mathrm{\left(P+\frac{a}{V^2}\right)(V-b)=R T}$

$\mathrm{\rho=\frac{PM}{RT}}$

For maximum density, pressure should be maximum and temperature should be minimum.

Boyle's law is defined at constant temperature for an ideal gas. $\mathrm{P} \propto \frac{1}{\mathrm{~V}}$

$\mathrm{P}=(\mathrm{nRT})\left(\frac{1}{\mathrm{~V}}\right)$ [straight line equation]

slope of P versus $\frac{1}{\mathrm{~V}}$ curve is $\mathrm{n R T}$

$\Rightarrow$ Slope $\uparrow ~\Rightarrow \mathrm{T} \uparrow~ \therefore \mathrm{T}_{3} > \mathrm{T}_{2} > \mathrm{T}_{1}$

As from the given data NH₃ has highest critical temperature which suggests maximum attractive forces hence NH₃ will get liquified first.

$\bullet$ Dalton's Law of partial pressure states that the total pressure by the mixture of non-reactive gases is equal to the sum of the partial pressures of individual gases.

$\bullet$ ${p_{Total}} = {p_1} + {p_2} + {p_3}$

$\bullet$ Also, ${p_i} = {\chi _i}p$; where p_i and $\chi$_i are the partial pressure and mole fraction of i^th gas respectively and p is the total pressure.

$\bullet$ ${p_{Total}} = {p_1} + {p_2} + {p_3}$

$ = {n_1}{{RT} \over V} + {n_2}{{RT} \over V} + {n_3}{{RT} \over V}$

$ = ({n_1} + {n_2} + {n_3}){{RT} \over V}$

We know for ideal gas

PV = nRT

P = n${{RT} \over V}$

P = ${{64} \over {32}} \times {{0.0831 \times 300} \over {10}}$

P = 4.9 bar

Pressure of O₂ gas inside the flask = 4.9 bar