States of Matter Solid, Liquid and Gas MCQ Questions & Answers in Physical Chemistry | Chemistry

(i) Order of intermolecular attraction is $$C{O_2} > C{H_4} > {O_2} > {H_2}$$
In $$C{O_2},$$  intermolecular forces increases with number of electrons in a molecule.
(ii) Size increases in order of $${H_2} < {O_2} < C{H_4} < C{O_2}$$
Hence, $${H_2}$$  will have lowest value of $$b.$$

$$H-F$$   has dipole-dipole interaction, London forces and hydrogen bonding due to highest electronegativity of $$F.$$  Hence, boiling point of $$H - F$$   is highest.

From the ideal gas equation :
$$PV = nRT\,\,or\,\,n = \frac{{PV}}{{RT}} = \frac{{3170 \times {{10}^{ - 3}}}}{{8.314 \times 300}} = 1.27 \times {10^{ - 3}}$$

According to ideal gas equation, at two condition
$$\eqalign{ & {\text{At}}\,300K;\,\,{P_0}V = {n_0}R{T_0} \cr & 1 \times V = 1.08 \times R \times 300\,\,\,\,\,...\left( {\text{i}} \right) \cr & {\text{At}}\,600K;\,\,{P_1}{V_1} = {n_1}R{T_1} \cr & {P_1} \times V = \left( {0.86 + 0.43} \right) \times R \times 600\,\,\,\,\,...\left( {{\text{ii}}} \right) \cr & {\text{Divide (ii) by (i),}} \cr & \frac{{{P_1}}}{1} = \frac{{1.29 \times 600}}{{1.08 \times 300}}; \cr & {P_1} = \frac{{1.29 \times 2}}{{1.08}} = 2.38\,\,atm. \simeq 2.4\,\,atm. \cr} $$

Gay Lussac's law states that at constant volume, pressure of a fixed amount of a gas varies directly with the temperature.
$$\therefore \,\,P \propto T$$   ( at constant volume )
Thus, if temperature is increased at constant volume pressure increases.

$$\eqalign{ & r.m.s\,\,{\text{velocity}}\,\,{V_{rms}} = \sqrt {\frac{{3RT}}{M}} \cr & i.e.,\,\,\frac{{{V_1}}}{{{V_2}}} = \sqrt {\frac{{{T_1}}}{{{T_2}}}} \cr & \frac{{5 \times {{10}^4}}}{{10 \times {{10}^4}}} = \frac{1}{2} = \sqrt {\frac{{{T_1}}}{{{T_2}}}} \cr & \therefore \,\,{T_2} = 4{T_1} \cr} $$

The higher the value of $$'a’,$$  more the value of $${T_c},$$  easy is the liquefaction

Molecular weight of $$C{l_2}$$  is more than molecular weight of $${N_2}.$$

For a given mass of an ideal gas, the volume and amount (moles) of the gas are directly proportional if the temperature and pressure are constant. i.e
$$V \propto n$$
Hence in the given case.
Initial moles and final moles are equal $${\left( {{n_T}} \right)_i} = {\left( {{n_T}} \right)_f}$$
$$\eqalign{ & \frac{{{P_i}V}}{{R{T_1}}} + \frac{{{P_i}V}}{{R{T_1}}} = \frac{{{P_f}V}}{{R{T_1}}} + \frac{{{P_f}V}}{{R{T_2}}} \cr & 2\frac{{{P_i}}}{{{T_1}}} = \frac{{{P_f}}}{{{T_1}}} + \frac{{{P_f}}}{{{T_2}}} \Rightarrow {P_f} = \frac{{2{P_i}{T_2}}}{{{T_1} + {T_2}}} \cr} $$

Critical temperature $$ = \frac{{8a}}{{27Rb}}$$
Value of $$\frac{a}{b}$$ is highest for $$Kr.$$  Therefore, $$Kr$$  has greatest value of critical temperature.