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

$$\eqalign{ & P,V\,{\text{and }}R\,{\text{are constant}} \cr & \therefore \,\,{n_1}{T_1} = {n_2}{T_2},\,100 \times 300 = {n_2} \times 500,:{n_2} = 60 \cr & {\text{Air escaped is}}\,\,40\% . \cr} $$

$$\eqalign{ & {\text{For centred unit cell,}} \cr & \,\,\,2\left( {{r^ + } + {r^ - }} \right) = a \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{r^ + } = radii{\text{ of cation}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{r^ - } = radii{\text{ of anion}} \cr & {\text{2}}\left( {100 + {r^ - }} \right) = 508 \cr & \,\,\,\,\,\,\,\,100 + {r^ - } = \frac{{508}}{2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{r^ - } = \frac{{508}}{2} - 100 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 254 - 100 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 154\,pm \cr} $$

$$\eqalign{ & {\text{Volume of atoms in a unit cell}} \cr & \left( V \right) = \frac{4}{3}\pi {r^3} \cr & {\text{For primitive cell,}} \cr & \,\,r = \frac{a}{2} \cr & V = \frac{4}{3}\pi {\left( {\frac{a}{2}} \right)^3} \cr & \,\,\,\,\,\, = \frac{{\pi {a^3}}}{6} \cr & {\text{Volume of the unit cell}}\left( V \right) = {a^3} \cr} $$
Thus, total volume occupied by the atoms

$$\eqalign{ & = \frac{{{\text{Volume of the atoms in unit cell}}}}{{{\text{Volume of unit cell}}}} \cr & = \frac{{\pi {a^3}}}{6} \times \frac{1}{{{a^3}}} \cr & = \frac{\pi }{6} \cr & = 0.52 \cr & = 1.00 - 0.52 \cr & = 0.48 \cr} $$

$$\eqalign{ & {\text{Number of moles of}}\,{O_2} \cr & = \frac{{70.6\,g}}{{32\,g\,mo{l^{ - 1}}}} \cr & = 2.21\,mole \cr & {\text{Number of moles of }}Ne \cr & = \frac{{167.5\,g}}{{20\,g\,mo{l^{ - 1}}}} \cr & = 8.375\,mole \cr & {\text{Mole fraction of}}\,{O_2} \cr & = \frac{{2.21}}{{2.21 + 8.375}} \cr & = 0.21\,mole \cr & {\text{Mole fraction of}}\,Ne \cr & = 1 - 0.21 \cr & = 0.79 \cr & {\text{Partial pressure of a gas}} \cr & {\text{ = Mole fraction}} \times {\text{total pressure}} \cr & {\text{Partial pressure of}}\,{O_2} \cr & = 0.21 \times 25 \cr & = 5.25\,bar \cr & {\text{Partial pressure of}}\,Ne \cr & = 0.79 \times 25 \cr & = 19.75\,bar \cr} $$

Such type of phenomenon is only exhibited by liquid crystals.

$$\eqalign{ & \left( {P + \frac{{a{n^2}}}{{{V^2}}}} \right)\left( {V - nb} \right) = nRT \cr & \left( {P + \frac{a}{{{V^2}}}} \right)\left( {V - b} \right) = RT\,\,\,\,\,\,\left( {\because \,\,n = 1} \right) \cr & {\text{If }}b{\text{ is negligible}}\,\,P = \frac{{RT}}{V} - \frac{a}{{{V^2}}} \cr & {\text{The equation is quadratic in }}V,{\text{thus}} \cr & V = \frac{{ + RT \pm \sqrt {{R^2}{T^2} - 4aP} }}{{2P}} \cr} $$
Since $$V$$ has one value at given $$P$$  and $$T,$$  thus numerical value of discriminant $$= 0$$
$$\eqalign{ & {R^2}{T^2} = 4aP \cr & P = \frac{{{R^2}{T^2}}}{{4a}} = \frac{{{{\left( {0.0821} \right)}^2}{{\left( {300} \right)}^2}}}{{4 \times 3.592}} \cr & \Rightarrow P = 42.22\,atm \cr} $$

$${\text{Given that,}}V = 2\,L,$$     $${T_1} = 50 + 273 = 323\,K,$$     $${P_1} = 3\,atm$$
$${\text{On heating}}\,\,V = 2L,$$     $${T_2} = 200 + 273 = 473,{P_2} = ?$$
$${\text{Using }}P - T{\text{ law,}}\,\,\frac{{{P_1}}}{{{T_1}}} = \frac{{{P_2}}}{{{T_2}}}$$
$$\eqalign{ & {P_2} = \frac{{3 \times 473}}{{323}} \cr & \,\,\,\,\,\,\,\, = 4.39\,atm \cr} $$
$$\eqalign{ & {\text{and}}\,\,n = \frac{{PV}}{{RT}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{3 \times 2}}{{0.0821 \times 323}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 0.226 \cr} $$
Now valve is opened till the pressure is maintained at $$1\,atm.$$
Thus, at constant $$V$$ and $$T,P \propto n$$
$$\eqalign{ & \therefore {P_2} \propto n\,\,\,\,\therefore 1 \propto {n_{{\text{left}}}} \cr & \therefore \frac{{{P_2}}}{1} = \frac{n}{{{n_{{\text{left}}}}}}\,\,\,\therefore \,\,{n_{{\text{left}}}} = \frac{{0.226}}{{4.39}} = 0.052 \cr} $$
$$\therefore {\text{Moles escaped out }}$$    $$ = 0.226 - 0.052 = 0.174$$
$$\therefore {\text{Fraction of moles escaped out}}$$       $$ = \frac{{0.174}}{{0.226}} = 0.77$$

It will spread as a thin layer

$$\eqalign{ & {\text{Let }}w{\text{ }}g{\text{ of each gas is taken}}{\text{.}} \cr & {\text{Mole fraction of}} \cr & X = \frac{{\frac{w}{4}}}{{\frac{w}{4} + \frac{w}{{40}}}} \cr & \,\,\,\,\,\,\, = \frac{{10}}{{11}} \cr & {\text{Partial pressure of}} \cr & X = {P_{{\text{total}}}} \times {\text{Mole}}\,\,{\text{fraction}} \cr & \,\,\,\,\,\, = 1.1 \times \frac{{10}}{{11}} \cr & \,\,\,\,\,\, = 1\,atm \cr} $$

$${\text{We know that from ideal equation,}}$$       $$V \propto \frac{T}{P}$$
$${\text{Given}}\,\,{T_1} = 15 + 273 = 288\,K,$$       $${P_1} = 1.5\,bar$$
$${T_2} = 25 + 273 = 298\,K,{P_2} = 1\,bar$$
$${V_1} \propto \frac{{288}}{{1.5}}\,\,{\text{i}}{\text{.e}}{\text{.,}}{V_1} \propto 192;$$     $${V_2} \propto \frac{{298}}{1};\frac{{{V_2}}}{{{V_1}}} = \frac{{298}}{{192}} = 1.55$$