Root mean square velocity at $${T_1}$$  temperature,
$$\eqalign{ & {U_1} = \sqrt {\frac{{3R{T_1}}}{M}} \cr & \,\,\,\,\,\,\,\, = \sqrt {\frac{{3R\left( {27 + 273} \right)}}{M}} \,\,\,\,...{\text{(i)}} \cr} $$
Root mean square velocity at $${T_2}$$  temperature,
$$\eqalign{ & {U_2} = \sqrt {\frac{{3R{T_2}}}{M}} \cr & \,\,\,\,\,\,\,\, = \sqrt {\frac{{3R\left( {927 + 273} \right)}}{M}} \,\,...{\text{(ii)}} \cr & {\text{Eq}}{\text{. (i) divided by Eq}}{\text{. (ii)}} \cr & \frac{{{U_1}}}{{{U_2}}} = \sqrt {\frac{{27 + 273}}{{927 + 273}}} \cr & \,\,\,\,\,\,\,\,\, = \sqrt {\frac{{300}}{{1200}}} \cr & \,\,\,\,\,\,\,\, = \frac{1}{2} \cr & {U_2} = 2{U_1} \cr} $$

The graph is a straight line parallel to the axis of pressure with intercept
$$ = P{V_m} = RT = $$    $$0.082\,atm\,L{K^{ - 1}}mo{l^{ - 1}} \times 300K$$
$$ = 24.6\,atm\,L\,mo{l^{ - 1}}.$$

$$\eqalign{ & {\text{Volume of alveoli}}\, = \frac{4}{3}\pi {R^3} \cr & = \frac{4}{3} \times 3.14 \times {\left( {5 \times {{10}^{ - 3}}} \right)^3}\,mL \cr & = 5.23 \times {10^{ - 7}}mL = 5.23 \times {10^{ - 7}}L \cr & {n_2} = \frac{{14}}{{100}} \times {n_{air}} = 0.14 \times \frac{{PV}}{{RT}} \cr & = 0.14 \times 1 \times \frac{{5.23 \times {{10}^{ - 10}}}}{{0.0821 \times 310}} \cr & = 2.87 \times {10^{ - 12}} \cr & {\text{Molecules}} = 2.87 \times 6.02 \times {10^{23}} \times {10^{ - 12}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1.7 \times {10^{12}} \cr} $$

Vapour pressure of all liquids are different at their freezing points.

$$\left( {p + \frac{a}{{{V^2}}}} \right)\,\,\left( {V - b} \right) = RT;\,{\text{Here}}\left( {p + \frac{a}{{{V^2}}}} \right)$$
represents the intermolecular forces.

$$\eqalign{ & \frac{{{P_1}{V_1}}}{{{T_1}}} = \frac{{{P_2}{V_2}}}{{{T_2}}} \cr & {V_2} = \frac{{745 \times 1.41 \times {{10}^4} \times 225}}{{294 \times 63.1}} \cr & \,\,\,\,\,\,\, = 1.274 \times {10^5}L \cr} $$

From ideal gas equation, $$PV = nRT$$
$$PV = \left( {\frac{m}{M}} \right)RT\,\,{\text{or}}\,\,M = m\frac{{RT}}{{PV}}$$
Let the molecular $$wt.$$  of $$A$$  and $$B$$  be $${M_A}$$  and $${M_B}$$  respectively.
$$\eqalign{ & {\text{Then}}\,{M_A} = 2 \times \frac{{RT}}{{1 \times V}};\,\,{M_B} = \frac{{3 \times RT}}{{0.5 \times V}} \cr & \therefore \,\,\frac{{{M_A}}}{{{M_B}}} = \frac{{2RT}}{V} \times \frac{{0.5V}}{{3RT}} = \frac{{2 \times 0.5}}{3} = \frac{1}{3} \cr} $$
Therefore, the ratio $${M_A}:{M_B} = 1:3$$

Glycerol in water has highest surface tension due to $$H$$ - bonding.

NOTE : The value of $$'a'$$  indicates the magnitude of attractive forces between gas molecules.
Value of $$\,'a' \propto \,$$  size of molecule.
∴ inert gas will have minimum value of $$\,'a'$$  followed by $${H_2}O,\,{C_6}{H_6}$$   and $${C_6}{H_5}C{H_3}$$

$$P = \frac{n}{V}RT$$
$$n\,\,{\text{for}}\,\,C{H_4} = \frac{{3.2}}{{16}} = 0.2;$$     $$n\,\,{\text{for}}\,\,C{O_2} = \frac{{4.4}}{{44}} = 0.1$$
$${n_{{\text{total}}}} = 0.2 + 0.1 = 0.3;$$     $$P = \frac{{0.3 \times 0.0821 \times 300}}{9} = 0.82\,\,atm$$