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

$$PV = nRT\left( {{\text{number}}\,{\text{of}}\,{\text{moles}} = \frac{n}{V}} \right)\,\,\therefore \,\,\frac{n}{V} = \frac{P}{{RT}}.$$

Root mean square speed is given by the expression
$$\eqalign{ & {U_{rms}} = \sqrt {\frac{{3RT}}{M}} \,\,\,\,\left[ {\because \,M = mN} \right] \cr & \therefore \,\,{U_{rms}} \propto \sqrt {\frac{1}{m}} \cr & \Rightarrow \,\,{U_{rms}} \propto {\left( m \right)^{ - \frac{1}{2}}} \cr} $$

$$\eqalign{ & {\text{The expression of root mean square speed is}} \cr & {U_{rms}} = \sqrt {\frac{{3RT}}{M}} \cr & {\text{Hence,}} \cr & \frac{{{U_{rms}}\left( {{H_2}} \right)}}{{{U_{rms}}\left( {{O_2}} \right)}} = {\left[ {\frac{{\frac{{3R\left( {50K} \right)}}{{2g\,mo{l^{ - 1}}}}}}{{\frac{{3R\left( {800K} \right)}}{{32g\,mo{l^{ - 1}}}}}}} \right]^{\frac{1}{2}}} = 1 \cr} $$

Intermolecular interactions increases from gas to solid. Thermal energy converts solids to liquids and liquids to gases.

$${\text{Density of }}CsBr = \frac{{Z \times M}}{{{a^3} \times {N_0}}}$$
$$Z \to $$  number of atoms in the bec unit cell = 2
$$M \to $$  molar mass of $$CsBr$$  = 133 + 80 =213
$$a \to $$  edge length of unit cell = 436.6 $$pm$$
$$ = 436.6 \times {10^{ - 10}}cm$$
$$\therefore {\text{Density}} = $$   $$\frac{{2 \times 213}}{{{{\left( {436.6 \times {{10}^{ - 10}}} \right)}^3} \times 6.023 \times {{10}^{23}}}}$$
$$\eqalign{ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 8.49 \times {10^{ - 7}} \times {10^7}g/c{m^3} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 8.50\,g/c{m^3} \cr} $$
$${\text{For a unit cell = }}\frac{{8.50}}{2} = 4.25\,g/c{m^3}$$

At high altitude, atmospheric pressure is low. Therefore,liquids at high altitude boil at lower temperatures in comparison to that at sea level hence pressure cooker is used for cooking food which otherwise takes more time.

$$\eqalign{ & {V_{rms}} = \sqrt {\frac{{3\,RT}}{M}} \cr & {V_{rms\left( {{O_2}} \right)}} = {V_{rms\left( {He} \right)}} \cr & \sqrt {\frac{{3\,R{T_{{O_2}}}}}{{{M_{{O_2}}}}}} = \sqrt {\frac{{3\,R{T_{He}}}}{{{M_{He}}}}} \cr & {\text{or}}\,\,\,\frac{{{T_{{O_2}}}}}{{{M_{{O_2}}}}} = \frac{{{T_{He}}}}{{{M_{He}}}} \cr & \therefore \,\,{T_{{O_2}}} = \frac{{300 \times 32}}{4} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 2400K \cr} $$

$$\eqalign{ & \left( {p + \frac{a}{{{V^2}}}} \right)\left( {V - b} \right) = RT\,{\text{at}}\,{\text{high}}\,{\text{pressure}}\,\frac{a}{{{V^2}}}\,{\text{can}}\,{\text{be}}\,{\text{neglected}} \cr & PV - Pb = RT\,\,{\text{and}}\,PV = RT + Pb \cr & \frac{{PV}}{{RT}} = 1 + \frac{{Pb}}{{RT}} \cr & Z = 1 + \frac{{Pb}}{{RT}};\,\,\,Z > 1\,\,{\text{at}}\,{\text{high}}\,{\text{pressure}} \cr} $$

$$\eqalign{ & {\text{According to Graham}}'{\text{s}}\,{\text{Law Diffusion :}} \cr & \frac{{{r_1}}}{{{r_2}}} = \sqrt {\frac{{{d_2}}}{{{d_1}}}} \cr & {\text{Since rate of diffusion}} \cr & = \frac{{Vol.{\text{ of gas diffused}}\left( V \right)}}{{{\text{Time taken for diffusion}}\,\left( t \right)}} \cr & \therefore \,\,\frac{{{r_1}}}{{{r_2}}} = \frac{{\frac{{{V_1}}}{{{t_1}}}}}{{\frac{{{V_2}}}{{{t_2}}}}} \cr & {\text{or}}\,\,\frac{{{r_1}}}{{{r_2}}} = \frac{{\frac{{{V_1}}}{{{t_1}}}}}{{\frac{{{V_2}}}{{{t_2}}}}} \cr & = \sqrt {\frac{{{d_2}}}{{{d_1}}}} \cr & = \frac{{\frac{{20}}{{60}}}}{{\frac{{{V_2}}}{{30}}}} \cr & = \sqrt {\frac{{\frac{{16}}{2}}}{{\frac{{32}}{2}}}} \cr & = \sqrt {\frac{1}{2}} \cr & \because \,\,Mol.\,wt = 2 \times V.D \cr & \therefore \,\,V.D = \frac{{Mol.\,wt}}{2} \cr & {\text{On calculating}}, \cr & {V_2} = 14.1 \cr} $$

$$\eqalign{ & {\text{Partial pressure of hydrogen gas :}} \cr & {V_1} = 0.5\,L\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{V_2} = 1.0\,L \cr & {P_1} = 0.8\,bar\,\,\,\,\,\,\,\,\,\,{P_2} = ? \cr} $$
$${\text{Applying Boyle's law}}$$     $${\text{(At constant }}T{\text{ and }}n{\text{)}}$$
$$\eqalign{ & {P_1}{V_1} = {P_2}{V_2}\,\,\,{\text{or}}\,\,\,{P_2} = \frac{{{P_1}{V_1}}}{{{V_2}}} \cr & \therefore \,\,{P_2} = \frac{{\left( {0.8\,bar} \right) \times \left( {0.5\,L} \right)}}{{\left( {1.0\,L} \right)}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = 0.40\,bar \cr & {\text{Partial pressure of oxygen gas :}} \cr & {V_1} = 2.0\,L,\,\,\,\,\,\,\,\,\,\,\,{V_2} = 1.0\,L \cr & {P_1} = 0.7\,bar\,\,\,\,\,\,\,\,{P_2} = ? \cr & {\text{or}}\,\,\,{P_2} = \frac{{{P_1}{V_1}}}{{{V_2}}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{\left( {0.7\,bar} \right) \times \left( {2.0\,L} \right)}}{{\left( {1.0\,L} \right)}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1.40\,bar \cr & {\text{Pressure of the gas mixture,}} \cr & {P_{mix}} = {P_{{H_2}}} + {P_{{O_2}}} \cr & \,\,\,\,\,\,\,\,\,\,\,\, = 0.40 + 1.40 \cr & \,\,\,\,\,\,\,\,\,\,\,\, = 1.80\,bar \cr} $$