$$\eqalign{ & {\text{Given}}\,\,P = \frac{{nRT}}{{v - nb}} - a{\left( {\frac{n}{v}} \right)^2} \cr & {\text{Which can also be written as}} \cr & \left[ {P + \frac{{{n^2}a}}{{{V^2}}}} \right]\left( {V - nb} \right) = nRT \cr} $$
At low pressure and high temperature the effect of $$\frac{a}{{{V^2}}}$$  and $$b$$ is negligible hence $$PV = nRT.$$

$${P_1}{V_1} = {P_2}{V_2}\,\,{\text{hence,}}\,\,{V_2} = \frac{{{P_1}{V_1}}}{{{P_2}}}$$

As the temperature is constant, Boyle’s law is applicable.
$$\eqalign{ & {p_1}{V_1} = {p_2}{V_2} \cr & {V_1} = 380\,mL,\,{p_1} = 730\,mm,\,{V_2} = ? \cr & {p_2} = 760\,mm \cr & 730 \times 380 = 760 \times {V_2} \cr & {V_2} = \frac{{730 \times 380}}{{760}} = 365\,mL \cr} $$

According to Graham's law, $$r \propto \frac{1}{{\sqrt M }}$$
As all conditions are identical for $$X$$  and $$Y,$$
$$\eqalign{ & \frac{{{r_X}}}{{{r_Y}}} = \sqrt {\frac{{{M_Y}}}{{{M_X}}}} \Rightarrow \frac{d}{{24 - d}} = \sqrt {\frac{{40}}{{10}}} = 2 \cr & d = 48 - 2d \Rightarrow 3d = 48 \Rightarrow d = 16\,cm \cr} $$

All the given phenomenon occurs due to surface tension

$$\eqalign{ & n,T\,{\text{same hence}}\,P \propto \frac{1}{V}, \cr & {V_1} = 1000\,c{m^3} \cr & {V_2} = \pi {\left( {10} \right)^2} \times 10 = 1000\,\pi \,c{m^3} \cr & {V_3} = \frac{4}{3}\pi {\left( {10} \right)^3} = \frac{4}{3}\pi \,1000\,c{m^3} \cr} $$
∴ Pressure of the gas is minimum in (III) container, Pressure of the gas is maximum in (I),

Given, edge length $$= 361\,pm$$
Four metal atoms in one unit cell
i.e. effective number in unit cell $$(z)$$  $$= 4$$ (given)
It is a $$FCC$$  structure
$$\eqalign{ & \therefore \,\,\,{\text{Face diagonal}} = 4r \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\sqrt 2 a = 4r \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,r = \frac{{\sqrt 2 \times 361}}{4} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 127\,pm \cr} $$

Given, ionic radius of cation $$\left( {{A^ + }} \right) = 0.98 \times {10^{ - 10}}m$$
Ionic radius of anion $$\left( {{B^ - }} \right) = 1.81 \times {10^{ - 10}}m$$
∴ Coordination number of each ion in $$AB$$ = ?
Now, we have
$$\eqalign{ & {\text{Radius ratio}} = \frac{{{\text{Radius of cation}}}}{{{\text{Radius of anion}}}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{0.98 \times {{10}^{ - 10}}m}}{{1.81 \times {{10}^{ - 10}}m}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 0.541 \cr} $$
If radius ratio range is in between $$0.441 - 0.732,\,ion$$    would have octahedral structure with coordination number $$‘six’.$$

Correction factor for attractive force for n moles of real gas is given by the term mentioned in (B).

$$\eqalign{ & {p_{{O_2}}} = {x_{{O_2}}} \times {P_{{\text{total}}}} \cr & {\text{Given}}:{P_{{\text{total}}}} = 1\,atm,{x_{{O_2}}} = \frac{4}{5} \cr} $$
$${p_{{O_2}}} = \frac{4}{5} \times 1 = 0.8\,atm$$       $$\left( {\because \,\,1\,atm = 1.01325 \times {{10}^5}Pa\,\,{\text{or}}\,\,N\,{m^{ - 2}}} \right)$$
$$\eqalign{ & \,\,\,\,\,\,\, = 0.8 \times {10^5}\,N\,{m^{ - 2}} \cr & \,\,\,\,\,\,\, = 8 \times {10^4}N\,{m^{ - 2}} \cr} $$