Equal moles of $$CO$$  and $${{N_2}}$$  
$${n_{CO}} = {n_{{N_2}}}$$
then, according to ideal gas equation, pressure of both gases $$CO$$  and $${N_2}$$  becomes equal
$$\therefore \,\,{p_{CO}} = {p_{{N_2}}}$$
Given, $${p_{CO}} + {p_{{N_2}}} = $$   Total pressure of mixture.
$$\eqalign{ & {\text{or}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2{p_{{N_2}}} = 1\,atm \cr & {\text{or}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{p_{{N_2}}} = 0.5\,atm \cr} $$

Upon increase of temperature the internal energy of water or any system increases resulting in decrease in intermolecular force and hence decrease in surface tension. Surface tension decreases with increase in mobility due to increase in temperature.

$$\eqalign{ & {P_{{O_2}}} = \frac{3}{{10}} \times {P_T}; \cr & {\text{After removing }}2{\text{ }}mole{\text{ of }}{O_2}, \cr & P{'_{{O_2}}} = \frac{1}{{10}} \times {P_T} \cr & {\text{Decrease in partial pressure of}}\,{O_2} \cr & = \frac{{\frac{{3{P_T}}}{{10}} - \frac{{{P_T}}}{8}}}{{\frac{{3{P_T}}}{{10}}}} \times 100 \cr & = 58.33\% \cr} $$

$${\text{For}}\,\,T = 223\,K,$$    $$P = 81.06\,MPa = 81.06 \times {10^3}\,k\,Pa,$$       $$Z = 1.95,{\text{and}}\,\,V = 1.0\,d{m^3},{\text{we have}}$$
$$\eqalign{ & n = \frac{{PV}}{{ZRT}} \cr & \,\,\,\,\, = \frac{{81.06 \times {{10}^3} \times 1}}{{1.95 \times 8.314 \times 223}} \cr & \,\,\,\,\, = 22.42\,mol \cr} $$
$${\text{Now, at}}\,\,T = 373\,K,$$     $$P = 20.265\,MPa = 20.265 \times {10^3}\,k\,Pa,$$        $$Z = 1.10,$$   $${\text{the volume occupied will be}}$$
$$\eqalign{ & {\text{V = }}\frac{{ZnRT}}{P} \cr & \,\,\,\,\, = \frac{{1.10 \times 22.42 \times 8.314 \times 373}}{{20.265 \times {{10}^3}}} \cr} $$
$$\therefore V = 3.774\,d{m^3}$$

Different particles of a gas have different speeds due to constant collisions and different kinetic energies.

Particles of a gas are always in constant and random motion. If the particles were at rest and occupy fixed positions, then a gas would have a fixed shape which is not observed.

$$\eqalign{ & {V_1} = 2\,L,\,{T_2} = \left( {26.1 + 273} \right)\,K = 299.1\,K, \cr & {V_2} = ? \cr & {T_1} = \left( {23.4 + 273} \right)\,K = 296.4\,K \cr & {\text{From Charle's law,}} \cr & \frac{{{V_1}}}{{{T_1}}} = \frac{{{V_2}}}{{{T_2}}} \Rightarrow {V_2} = \frac{{{V_1}{T_2}}}{{{T_1}}} \cr & \Rightarrow {V_2} = \frac{{2L \times 299.1K}}{{296.4K}} = 2L \times 1.009 \cr & = 2.018\,L \cr} $$

In case of body centred cubic $$(bcc)$$  crystal,
$$a\sqrt 3 = 4r$$
Given, edge length, $$a = 351\,pm$$
Hence, atomic radius of lithium,
$$\eqalign{ & r = \frac{{a\sqrt 3 }}{4} \cr & \,\,\,\, = \frac{{351 \times 1.732}}{4} \cr & \,\,\,\, = 151.98\,pm \cr} $$

$$\eqalign{ & {\text{Initial volume,}}\,{V_1} = 500\,mL \cr & {\text{Initial temperature,}} \cr & {T_1} = {27^ \circ }C \cr & \,\,\,\,\,\,\, = 27 + 273 \cr & \,\,\,\,\,\,\, = 300K \cr & {\text{Final temperature,}} \cr & {T_2} = - 5 + 273 \cr & \,\,\,\,\,\,\, = 268\,K \cr & {V_2} = ? \cr & \frac{{{V_1}}}{{{T_1}}} = \frac{{{V_2}}}{{{T_2}}} \cr & {V_2}\, = \frac{{{V_1}{T_2}}}{{{T_1}}} \cr & \,\,\,\,\,\,\, = \frac{{500 \times 268}}{{300}} \cr & \,\,\,\,\,\,\, = 446.66\,mL \cr} $$

$$m = 88\,g,M = 44\,\,g\,\,mo{l^{ - 1}},$$      $$T = 273 + 30 = 303\,K$$
$$\eqalign{ & P = 1\,bar,V = ? \cr & PV = nRT\,\,{\text{or}}\,\,V = \frac{{nRT}}{P} = \frac{m}{M}\frac{{RT}}{P} \cr & V = \frac{{88 \times 0.0831 \times 303}}{{44 \times 1}} = 50.36\,L \cr} $$