For ideal gas $$PV=$$  constant at all pressures

Volume decreases when pressure is increased at constant temperature. When temperature is decreased at constant pressure, the volume decreases.

$$\eqalign{ & \frac{{{P_1}{V_1}}}{{{T_1}}} = \frac{{{P_2}{V_2}}}{{{T_2}}} \cr & \Rightarrow \frac{{1 \times 0.4}}{{273}} = \frac{{1 \times {V_2}}}{{546}} \cr & {V_2} = \frac{{1 \times 0.4 \times 546}}{{273 \times 1}} = 0.8\,L \cr} $$

$$\eqalign{ & {\text{Given,}}\,{P_1} = 15\,atm,\,{P_2} = 60\,atm \cr & {V_1} = 76\,c{m^3},\,{V_2} = 20.5\,c{m^3}. \cr} $$
If the gas is an ideal gas, then according to Boyle's law, it must follow the equation,
$$\eqalign{ & {P_1}{V_1} = {P_2}{V_2} \cr & {P_1} \times {V_1} = 15 \times 76 = 1140 \cr & {P_2} \times {V_2} = 60 \times 20.5 = 1230 \cr & \therefore \,\,{P_1}{V_1} \ne {P_2}{V_2} \cr} $$
∴ The gas behaves non-ideally.
The given information is not sufficient to comment on other statements.

$$\eqalign{ & {P_1}{V_1} = {P_2}{V_2}\,\,{\text{or}}\,\,760 \times 40 = 800 \times {V_2} \cr & {V_2} = \frac{{760 \times 40}}{{800}} = 38\,mL \cr} $$

Liquids tend to minimise their surface due to surface tension.

$${\text{Pressure correction}} = \frac{{a{n^2}}}{{{V^2}}}\,\,{\text{or}}\,\, \propto \frac{{{n^2}}}{{{V^2}}}$$

$$KE = \frac{3}{2}RT$$   ( for one mole of a gas )
The temperature is constant and kinetic energy is independent on molecular weights. So,
$$\overline {KE} co = {\overline {KE} _{{N_2}}}$$

$$\eqalign{ & {V_c} = \frac{{60}}{{0.80}} \cr & = 75c{m^3}\,mo{l^{ - 1}}; \cr & b = \frac{{{V_c}}}{3} = 25c{m^3}\,mo{l^{ - 1}};\, = 0.25\,L\,mo{l^{ - 1}} \cr & \therefore \,\,{T_c} = \frac{{8a}}{{27\,Rb}} \cr & \frac{{4 \times {{10}^5}}}{{821}} \cr & = \frac{{2.45}}{{0.425}} \cr & = 5.76\,g{L^{ - 1}} \cr & \Rightarrow a = 3.375 \cr} $$

Due to increase in the temperature, the kinetic energy of the gas molecules increases resulting in an increase in average molecular speed. The molecules are bombarded to the walls of the container with a greater velocity resulting in an increase in pressure.