Sol. It is the point at which solid, liquid and vapour of a substance are in equilibrium with each other. For example, triple point of water is the temperature at which ice, water and its vapour coexist.

Sol. HF molecules are associated with intermolecular $H-$ bonding, therefore, it is liquid whereas $H C l$ is gas because less Vander Waal’s forces of attraction.

Sol. $(i)<100^{\circ} C(\text { ii })>100^{\circ} C$

Sol. Heavier air will come down and lighter air goes up. Air at lower level is denser since it is compressed by mass of air above it.

Sol. Solid $\mathrm{CO}_{2}$ It is because solid $\mathrm{CO}_{2}$ is directly converted into gaseous state (sublimes) and does not change into liquid, so it is called dry ice.

Sol. Balloon will expand because rate of diffusion of $H_{2}$ is greater than that of $D_{2}$

Sol. A straight line parallel to pressure axis.

Sol. Charle’s Law.

Sol. $\frac{P_{1} V_{1}}{T_{1}}=\frac{P_{2} V_{2}}{T_{2}}$ or $P_{2}=\frac{P_{1} V_{1} T_{2}}{T_{1} V_{2}}$

But $d \propto \frac{1}{V} .$ Hence $P_{2}=\frac{P_{1} T_{2}}{T_{1}}\left(\frac{d_{2}}{d_{1}}\right)$

$=\frac{760 \times 263}{273}\left(\frac{1}{10}\right)=73.2 \mathrm{mm}$

Sol. Number of years $=\frac{6.023 \times 10^{23}}{10^{10} \times 365 \times 24 \times 60 \times 60}=1,908,00$ years.

Sol.

Sol. $\frac{p V^{2} T^{2}}{n}=\frac{\left(N m^{-2}\right)\left(m^{3}\right)^{2}(K)^{2}}{m o l}$

$=N m^{4} K^{2} m o l^{-1}$

Sol. $R=8.314 \mathrm{JK}^{-1} \mathrm{mol}^{-1}$

Sol. $^{c} a^{\prime}$ is a measure of the magnitude of the intermolecular forces of attraction while $b$ is a measure of the effective size of the gas molecules.

Sol. The air pressure decreases with increases in altitude. That is why jet aeroplane flying at high altitude need pressurization of the cabin so that partial pressure of oxygen is sufficient for breathing.

Sol. Both have same molar mass $\left(=44 g m o l^{-1}\right)$. According to Graham’s law of diffusion, rates of diffusion of different gases are inversely proportional to the square root of their molar masses under same conditions of temperature and pressure.

Sol. At $30^{\circ} \mathrm{C},$ kinetic energy depends only on absolute temperature and not on the identity of a gas.

Sol. (i) Inter molecular forces between molecules are negligible.

(ii) Molecules of a gas have negligible volumes.

Sol. (i) High compressibility is due to large empty spaces between the molecules.

(ii) Due to absence of attractive forces between molecules, the molecules of gases can easily separate from one another.

Sol. Higher the critical temperature, more easily the gas can be liquefied i.e. greater are the intermolecular forces of attraction. Therefore, $\mathrm{CO}_{2}$ has stronger intermolecular forces than $\mathrm{CH}_{4}$

Sol. $(\mathrm{i}) \mathrm{pCH}_{4}=\frac{n R T}{\mathrm{V}}=\frac{3.2}{16} \times \frac{0.0821 \mathrm{L} \operatorname{atm} \mathrm{K}^{-1} \mathrm{mol}^{-1} \times 300 \mathrm{K}}{9 \mathrm{L}}$$p C H_{4}=\frac{0.2 \times 24.63}{9}=\frac{4.926}{9}=0.547 \mathrm{atm}$

$p C H_{4}=\frac{0.2 \times 24.63}{9}=\frac{4.926}{9}=0.547 \mathrm{atm}$

(ii) Boron and silicon are example of Covalent solids.

Or

(i) $28 g$ of Nitrogen gas contains $2 \times 7 \times 6.023 \times 10^{23}$ electrons. $1+1$

$\begin{array}{llll}{1.4} & {g} & {\text { of }} & {\text { Nitrogen }} & {\text { gas }} & {\text { contains }}\end{array}$

$\frac{2 \times 7 \times 6.023 \times 10^{23}}{28} \times 1.4$

$=\frac{2 \times 7 \times 6.023 \times 10^{23}}{20}=\frac{84.32 \times 10^{23}}{20}$

$=4.2161 \times 10^{23}$ electrons

(ii) $N H_{3}$ will diffuse faster

$\frac{r_{N H_{3}}}{r_{H C l}}=\sqrt{\frac{36.5}{17}}=\sqrt{2.14}=1.46$ times faster.

(iii) Urea is a crystalline solid therefore it has sharp melting point whereas glass does not have sharp melting point because it is amorphous, i.e., does not have regular three dimensional structure.

Sol.

Sol. Pressure of gas $=$ Atmospheric pressure $+$ Difference between levels of $\mathrm{Hg}$

$=743 \mathrm{mm}+(43.7 \mathrm{cm}-15.6 \mathrm{cm})=74.3 \mathrm{cm}+28.1 \mathrm{cm}$

$P=102.4 \mathrm{cm} P$ in bar $=\frac{102.4}{76}=1.347 \mathrm{bar}$

Sol. $P V=n R T, \quad P=1$ atm, $\quad V=500 \mathrm{cm}^{3}, \quad n=?$

$R=82.1 \mathrm{atm} \mathrm{cm}^{3} \mathrm{K}^{-1} \mathrm{mol}^{-1}, T=300 \mathrm{K}$

$n=\frac{1 a t m, \times 500 \mathrm{cm}^{3}}{\left(82.1 \mathrm{atm} \mathrm{cm}^{3} \mathrm{K}^{-1} \mathrm{mol}^{-1}\right) \times(300 \mathrm{K})}=0.02 \mathrm{mol}$

Sol.

$p V=n R T$

1 bar $\times \frac{34.05}{1000} L=\frac{0.0625}{M} \times 0.083 \times 819 K$

$M=\frac{0.0625 \times 0.083 \times 819 K \times 1000}{34.05}=\frac{4248.5625}{34.05}$

$M=124.77 \mathrm{g} \mathrm{mol}^{-1}$

Or

Charle’s plotted the volume against temperature in $^{\circ} C$. These plots when extraplotted intersect the temperature axis at the same point $-273^{\circ} \mathrm{C} .$ He concluded that all gases at this temperature could have zero volume and below this temperature volume would be negative. It shows $-273^{\circ} \mathrm{C}$ is lowest temperature attainable.

Sol. Since temperature and amount of gas remains constant, therefore, Boyle’s law is applicable.

Sol. According to Boyle’s law, at constant temperature,

Since balloon bursts at 0.2 bar pressure, the volume of the balloon should be less than 11.35 L.

Sol. Suppose volume of vessel $=V c m^{3}$

i.e., volume of air in the flask at $27^{\circ} \mathrm{C}=\mathrm{Vcm}^{3}$

$\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}, \quad$ i.e., $\quad \frac{V}{300}=\frac{V_{2}}{750} \quad$ or $\quad V_{2}=2.5 \mathrm{V}$

$\therefore$ Volume expelled $=2.5 \mathrm{V}-\mathrm{V}=1.5 \mathrm{V}$

Fraction of air expelled $=\frac{1.5 \mathrm{V}}{2.5 \mathrm{V}}=\frac{3}{5}$

Sol. Pressure $=$ height $\times$ density $\times g$

Case (i). Pressure

$=76 \mathrm{cm} \times 0.99 \mathrm{g} / \mathrm{cm}^{3} \times 981 \mathrm{cm} / \mathrm{s}^{2}$

$=7.38 \times 10^{4}$ dynes $\mathrm{cm}^{-2}$

$=0.073$ atm $\left(1 \mathrm{atm}=1.013 \times 10^{6} \text { dynes } \mathrm{cm}^{-2}\right)$

Case (ii). Pressure

$=76 \mathrm{cm} \times 13.6 \mathrm{g} / \mathrm{cm}^{3} \times 981 \mathrm{cm} / \mathrm{s}^{2}$

$=1.013 \times 10^{6}$ dynes $\mathrm{cm}^{-2}=1 \mathrm{atm}$

Sol.

Sol. According to Charles’ law

$\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}$

$V_{1}=2 L$

$V_{2}=?$

$T_{1}=273+23.4=296.4 \mathrm{K} \quad T_{2}=273+26.1=299.1$

$\therefore V_{2}=\frac{V_{1} T_{2}}{T_{1}}=\frac{2 L \times 299.1 K}{296.4 K}=2.018 L$

Sol. Since the amount of gas and the pressure remains constant, Charles’ law is applicable. i.e.

$\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}$

$V_{1}=800 \mathrm{mL}$

$V_{2}=?$

$T_{1}=273+27=300 K ; \quad T_{2}=273+47=320 K$

$\frac{800 m L}{300 K}=\frac{V_{2}}{320 K}$

or $\quad V_{2}=\frac{(800 \mathrm{mL})}{(300 \mathrm{K})} \times(320 \mathrm{K})=853.3 \mathrm{mL}$

Increase in volume of air $=853.3-800=53.3 \mathrm{mL}$

Sol. since gas is confined in a cylinder, its volume will remain constan

Sol.

Sol. According to ideal gas equation:

$p V=n R T$

$p=3.32$ bar, $V=5 d m^{3}, n=4.0 \mathrm{mol}$

$R=0.083$ bar $d m^{3} K^{-1} \mathrm{mol}^{-1}$

$T=\frac{p V}{n R}=\frac{3.32 \text { bar } \times 5 d m^{3}}{4.0 \mathrm{mol} \times 0.083 \text { bar } d m^{3} K^{-1} \mathrm{mol}^{-1}}=50 \mathrm{K}$

Sol. According to ideal gas equation,

Sol. Weight of liquid $=148-50=98 g$

Density of liquid $=0.98 g m l^{-1}$

$\therefore$ Volume of liquid $=\frac{98}{0.98}=100 \mathrm{ml}$

Weight of gas $=50.5-50.0=0.5 g$

Volume $=\frac{100}{1000} L, P=1$ atm, $T=300 K$

$p V=n R T$ or $p V=\frac{W}{M} R T$

$\frac{1 \times 100}{1000}=\frac{0.5}{M} \times 0.082 \times 300$

$\therefore M=123.15 g \mathrm{mol}^{-1}$

Sol. Calculation of density of $N_{2}$ at 5 bar and $0^{\circ} \mathrm{C}$

$d=\frac{P M}{R T}=\frac{5(\text { bar }) \times 28\left(g m o l^{-1}\right)}{0.0831\left(L \text { bar } K^{-1} \mathrm{mol}^{-1}\right) \times 273.15(K)}$

$=6.168 g L^{-1}$

Calculation of molar mass of gaseous oxide

$M=\frac{d R T}{P}$

$=\frac{6.168\left(g L^{-1}\right) \times 0.0831\left(L \text { bar } K^{-1} \mathrm{mol}^{-1}\right) \times 273.15(K)}{2}$

$=70.0 \mathrm{g} \mathrm{mol}^{-1}$

Sol.

Sol. $p=\frac{n}{V} R T=\frac{w}{M} \frac{R T}{V}$

$p_{C H_{4}}=\left(\frac{3.2}{16} m o l\right) \frac{0.0821 d m^{3} a t m K^{-1} m o l^{-1} \times 300 K}{9 d m^{3}}$

$=0.55 \mathrm{atm}$

$p_{\mathrm{CO}_{2}}=\left(\frac{4.4}{44} \mathrm{mol}\right) \frac{0.0821 \mathrm{dm}^{3} \mathrm{atm} \mathrm{K}^{-1} \mathrm{mol}^{-1} \times 300 \mathrm{K}}{9 \mathrm{dm}^{3}}$

$=0.27 \mathrm{atm}$

$p_{\text {Total }}=0.55+0.27=0.82 \mathrm{atm}$

Sol. $p V=n R T$

$p \times 1=\frac{8}{32} \times 0.083 \times 300=\frac{24.9}{4}=6.225$ bar

$p \times 1=\frac{4}{2} \times 0.083 \times 300=49.80 \mathrm{bar}$

Total Pressure $=6.225+49.80=56.025$ bar

Sol.

Sol. Pay load is the difference between the mass of displaced air and the mass of the balloon.

Volume of ballon $=\frac{4}{3} \pi r^{3}$

Radius of balloon, $r=10 \mathrm{m}$

$V=\frac{4}{3} \times 3.14 \times(10)^{3}=4186.7 \mathrm{m}^{3}$

Mass of displaced air $=4186.7 \mathrm{m}^{3} \times 1.2 \mathrm{kg} \mathrm{m}^{-3}$

$=5024.04 \mathrm{kg}$

Moles of gas present $=\frac{p V}{R T}$

$=\frac{1.66 \times 4186.7 \times 10^{3}}{0.083 \times 300}=279.11 \times 10^{3} \mathrm{mol}$

Mass of helium present $=279.11 \times 10^{3} \times 4$

$=1116.44 \times 10^{3} g=1116.4 \mathrm{kg}$

Mass of filled balloon $=100+1116.4=1216.4 \mathrm{kg}$ Pay load $=$ mass of displaced air – Mass of balloon

$=5024.4-1216.44=3807.6 \mathrm{kg}$

Sol. Volume of balloon,

$V=\frac{4}{3} \pi r^{3} r=10 m$

$\therefore V=\frac{4}{3} \times 3.14 \times(10)^{3}=4187 \mathrm{m}^{3}$

Mass of displaced air $=4187 \mathrm{m}^{3} \times 1.2 \mathrm{kg} \mathrm{m}^{-3}$

$=5024.44 \mathrm{kg}$

Moles of gas present, $n=\frac{P V}{R T}=\frac{1 \times 4187 \times 10^{3}}{0,082 \times 298}$

$=171.3 \times 10^{3} \mathrm{mol}$

Mass of He present $=171.3 \times 10^{3} \times 4$

$=685.3 \times 10^{3} g$

$=685.3 \mathrm{kg}$

Mass of filled balloon $=100+685.3=785.3 \mathrm{kg}$

Pay load $=$ Mass of displaced air – Mass of balloon $=5024.4-785.3=4239.1 \mathrm{kg}$

Sol. $\left(p+\frac{a n^{2}}{V^{2}}\right)(V-n b)=n R T$

$n=2 \mathrm{mol}, V=4 L, p=11 \mathrm{atm}$

$T=300 K, b=0.05 L m o l^{-1}$

Substituting the values

$\left(11+\frac{a \times 4}{16}\right)(4-2 \times 0.05)=2 \times 0.082 \times 300$

$\frac{176+4 a}{16} \times 3.9=49.2$

$(176+4 a) 3.9=49.2 \times 16$

$15.6 a=787.2-686.4$

$a=6.4616 \mathrm{atm} \mathrm{L}^{2} \mathrm{mol}^{-2}$

Sol. Mole of phosphorus vapour $(n)=\frac{P V}{R T}$

$=\frac{1(\text {bar}) \times 34.05 \times 10^{-3}(L)}{0.0831\left(\text {bar } L K^{-1} \mathrm{mol}^{-1}\right) \times 819.15(K)}=5.0 \times 10^{-4}$

Let molar mass of phosphorus be $M g m o l^{-1}$

$\therefore$ Mole of phosphorus vapour $=\frac{0.0625}{M}$

Now, $\frac{0.0625}{M}=5.0 \times 10^{-4}$ or $M=\frac{0.0625}{5.0 \times 10^{-4}}$

$=125 g \mathrm{mol}^{-1}$

Sol. The chemical reaction taking place is

Sol. For a real gas, the plot of $P V_{m}$ vs $P$ can be of the type $A$ or $B$ but at the point of intercept, $P=0$ and at any low pressure, vander Waal’s equation reduce to ideal gas equation.

$\mathrm{PV}=\mathrm{nRT}$ or $\mathrm{PV}_{\mathrm{m}}=\mathrm{RT}$

Hence, $y$ -intercept of graph will be $=R T$

Sol. (i) For statement of the Graham’s law. Molar mass of:

$\mathrm{CO}_{2}=44 \mathrm{u} ; \mathrm{SO}_{2}=64 \mathrm{u}$ and $\mathrm{NO}_{2}=46 \mathrm{u}$

As $r_{d i f f} \propto \frac{1}{\sqrt{M}},$ therefore, larger the molar mass lesser

will be the rate of diffusion under similar condition. Thus, increasing order of rates of diffusion is

$r_{S O_{2}}<r_{N O_{2}}<r_{C O_{2}}$

(ii) $V=\frac{n R T}{P}=\frac{0.30(\text { mol }) \times 0.0821\left(L \text { atm } K^{-1} \mathrm{mol}^{-1}\right) \times 333(\mathrm{K})}{0.821(\mathrm{atm})}$

$=9.99 L \approx 10.8 L$

Sol. $\frac{r_{N H_{3}}}{r_{H C l}}=\frac{l_{1}}{200-l_{1}}=\sqrt{\frac{M_{H C l}}{M_{N H_{3}}}}=\sqrt{\frac{36.5}{17}}$

$\frac{l_{1}}{200-l_{1}}=\sqrt{2.147}=1.465$

$l_{1}=293-1.465 l_{1} ; 2.465 l_{1}=293$

$l_{1}=\frac{293}{2.465}=118.88 \mathrm{cm}$

$200-l_{1}=200-118.88=81.12 \mathrm{cm}$ from $\mathrm{HCl}$ end.

Sol. $\frac{r_{A}}{r_{B}}=\sqrt{\frac{M_{B}}{M_{A}}}$ or $\frac{t_{B}}{t_{A}}=\sqrt{\frac{M_{B}}{M_{A}}}$

$t_{A}=20$ sec, $t_{B}=10$ sec, $M_{A}=80, M_{B}=?$

$\frac{10}{20}=\sqrt{\frac{M_{B}}{80}}$

$M_{B}=\frac{80}{4}=20 g \mathrm{mol}^{-1}$

Sol. As the mixture $H_{2}$ and $O_{2}$ contains $20 \%$ by weight of

hydrogen, therefore, if $H_{2}=20 g,$ then $O_{2}=80 g$

$n_{H_{2}}=\frac{20}{2}=10$ moles, $n_{O_{2}}=\frac{80}{32}=2.5$ moles

$p_{H_{2}}=\frac{n_{H_{2}}}{n_{H_{2}}+n_{O_{2}}} \times P_{\text {total }}=\frac{10}{10+2.5} \times 1$ bar $=0.8$ bar.

Sol. Partial pressure of oxygen gas,

$p=\frac{n R T}{V}$

$n=\frac{8}{32}$ mol, $\quad \mathrm{V}=1 \mathrm{dm}^{3}, T=300 \mathrm{K}$

$p_{\left(O_{2}\right)}=\frac{8 \times 0.083 \times 300}{32 \times 1}=6.225$ bar

Partial pressure of hydrogen gas

$p=\frac{n R T}{V}$

$n=\frac{4}{2}=2 m o l$

$p\left(H_{2}\right)=\frac{2 \times 0.083 \times 300}{1}=49.8 \mathrm{bar}$

Total pressure $=p_{\left(O_{2}\right)}+p_{\left(H_{2}\right)}$

$=6.225+49.8=56.025$ bar

Sol. Let the partial pressure of hydrogen be $P_{H_{2}}$ and the partial

pressure of oxygen be $P_{O_{2}}$

The number of mole of hydrogen $\left(n_{1}\right)=\frac{4}{2}=2$ mole

The number of mole of oxygen $\left(n_{2}\right)=\frac{8}{32}=0.25$ mole

Now, applying ideal gas equation for each gas $p_{H_{2}} \times V=n_{1} R T$

$p_{H_{2}}=\frac{n_{1} R T}{V}=\frac{2 \times 0.083 \times 300}{1}=49.8$ bar

Similarly, $p_{O_{2}} V=n_{2} R T$

$p_{O_{2}}=\frac{n_{2} R T}{V}=\frac{0.25 \times 0.083 \times 300}{1}=6.225 \mathrm{bar}$

Total pressure of gaseous mixture

$=p_{H_{2}}+p_{O_{2}}=49.8+6.225=56.025$ bar.

Sol. (i) Average kinetic energy is given as $E_{k}=\frac{3 n R T}{2}$

Here, $n=\frac{32}{16}=2 ; R=8.314 \mathrm{JK}^{-1} \mathrm{mol}^{-1} ; T=300 \mathrm{K}$

$\therefore E_{k}=\frac{3 \times 2 \times 8.314 \times 300}{2}=7482.6 \mathrm{J}$

(ii) Root mean square speed is given as

$u_{r m s}=\sqrt{\frac{3 R T}{M}}$

Here use $R=8.314 \times 10^{7}$ ergs $K^{-1} \mathrm{mol}^{-1}$ to get speed in

$\mathrm{cm} \mathrm{s}^{-1}$

$u_{m s}=\sqrt{\frac{3 \times 8.314 \times 10^{7} \times 300}{16}}=68385.85 \mathrm{cm} s^{-1}=68.38 \mathrm{ms}^{-1}$

(iii) Most probable speed

$(\alpha)=\frac{u_{r m s}}{1.224}=\frac{683.8}{1.224}=558.7 \mathrm{ms}^{-1}$

Sol. Volume occupied by $1 \mathrm{mol}$ of $\mathrm{O}_{3}$ at $20^{\circ} \mathrm{C}$ and $82 \mathrm{mm}$ pressure is calculated by applying general gas equation,

$\frac{P_{1} V_{1}}{T_{1}}=\frac{P_{2} V_{2}}{T_{2}}$

$\therefore \quad V_{2}=\frac{P_{1} V_{1} T_{2}}{T_{1} P_{2}}=\frac{76 \times 22400 \times 293}{273 \times 82}=22281.92 \mathrm{cm}^{3}$

Now, $u=\sqrt{\frac{3 P V}{M}}$

Here, we use $P$ in dyne/cm$^{2}, P=82 \times 13.6 \times 981$ dyne/cm$^{2}$

$\therefore u=\sqrt{\frac{3 \times 82 \times 13.6 \times 981 \times 22281.92}{48}}=3.90 \times 10^{4} \mathrm{cm} \mathrm{s}^{-1}$

Sol. (i) The kinetic theory of gases assumes that pressure of gas is due to collision of gas molecules with the walls of the container. The more will be frequency of collision, more will be pressure. The reduction in volume of gas increases no. of molecules per unit volume to which pressure is directly proportional. Therefore, the volume of the gas is reduced if pressure is increased or pressure is inversely proportional to volume.

$\frac{1}{2} m u^{2}=\frac{3}{2} k T$

$p=\frac{1}{3} \frac{N}{V} m u^{2}$ or $p=\frac{2}{3} \frac{N}{V} \times \frac{1}{2} m u^{2}=\frac{2}{3} \frac{N}{V} \times \frac{3}{2} k T$

$\Rightarrow p V=N k T \Rightarrow P \alpha \frac{1}{V}$

It can be seen that at a constant temperature for a fixed number of gas molecules, the pressure is inversely proportional to volume.

(ii) $\frac{T_{1}}{T_{2}}=\frac{n_{1}}{n_{2}} \quad \begin{array}{cc}{T_{1}=27^{\circ} C+273} & {=300 K} \\ {T_{2}=477+273=} & {750 K}\end{array}$

$\frac{300 K}{750 K}=\frac{n_{1}}{n_{2}}, \quad \frac{n_{1}}{n_{2}}=\frac{2}{5} b v$

fraction of air escaped $=1-\frac{2}{5}=\frac{3}{5}$

Sol. Average velocity $=\sqrt{\frac{8 R T}{\pi M}}$

Root mean square velocity

$=\sqrt{\frac{3 R T}{M}}=\sqrt{\frac{3 R \times 300 K}{M}}$

For equal values, $\sqrt{\frac{8 R T}{\pi M}}=\sqrt{\frac{3 R \times 300}{M}}$

or $\quad \frac{8 R T}{\pi M}=\frac{3 R \times 300}{M}$ or $\frac{8 T}{\pi}=900$

or $\quad T=353.57 K=80.57^{\circ} \mathrm{C}$

Sol. Boyle’s law is applicable as the amount and temperature are unaltered

$p_{1} V_{1}=p_{2} V_{2}$

or $p_{1} / p_{2}=V_{2} / V_{1}$ ‘Substituting the values

0.720 bar/ 0.900 bar $=V_{2} / 200 \mathrm{mL}$

$V_{2}=\frac{720}{900} \times 200 m L=160 m L$

Boyle’s law is manifested in the working of many devices used in daily life such as cycle pump, aneroid barometer and tyre pressure gauge etc.

Sol. Number of moles of nitrogen

$=2.8 g / 28 g \mathrm{mol}^{-1}=0.1 \mathrm{mol}$

Number of nitrogen molecules $=0.1 \mathrm{mol} \times 6.022 \times 10^{23}$

$m o l^{-1}=6.022 \times 10^{22}$

Sol. $p V_{m}=R T$ or $p M / d=R T$ or $M=d R T / p$

$=\frac{1.29 \mathrm{kg} m^{-3} \times 8.314 \mathrm{NmK}^{-1} \mathrm{mol}^{-1} \times 273.15 \mathrm{K}}{1.0 \times 10^{5} \mathrm{Nm}^{-2}(\mathrm{or} P a)}$

$=\frac{1.29 \times 8.314 \times 273.15 k g m o l^{-1}}{1 \times 10^{5}}$

$=0.0293 \mathrm{kg} \mathrm{mol}^{-1}$ or molar mass is $29.3 \mathrm{g} \mathrm{mol}^{-1}$

Sol. $r_{N H_{3}} / r_{H C 1}=\left(M_{H C 1} / M_{N H_{3}}\right)^{1 / 2}$

$=(36.5 / 17)^{1 / 2}=1.46$ or $r_{N H_{3}}=1.46 r_{H C 1}$

Thus ammonia will diffuse 1.46 times faster than hydrogen chloride gas.

Sol. Partial pressure of sulphur dioxide,

$p_{S O_{2}}=n R T / V$

$=\frac{0.25 \mathrm{mol} \times 8.314 \mathrm{JK}^{-1} \mathrm{mol}^{-1} \times 300 \mathrm{K}}{2.5 \times 10^{-3} \mathrm{m}^{3}}$

$=2.49 \times 10^{5} \mathrm{Nm}^{-2}=2.49 \times 10^{5} \mathrm{Pa}$

Similarly $p_{N_{2}}=2.49 \times 10^{5} \mathrm{Pa}$

Following Dalton’s law $p_{\text {Total }}=p_{N_{2}}+p_{S O_{2}}$

$=2.49 \times 10^{5} P a+2.49 \times 10^{5} P a=4.98 \times 10^{5} P a$

Sol. As the vessel is open, pressure and volume remain constant. Thus, if $n_{1}$ moles are present at $T_{1}$ and $n_{2}$ moles are present at $T_{2},$ we can write

$P V=n_{1} R T_{1} ; P V=n_{2} R T_{2}$

Hence, $n_{1} R T_{1}=n_{2} R T_{2}$ or $n_{1} T_{1}=n_{2} T_{2}$

or, $\quad \frac{n_{1}}{n_{2}}=\frac{T_{2}}{T_{1}}$

Suppose the no. of moles of air originally present $=n$

After heating, no. of moles of air expelled $=\frac{3}{5} n$

$\therefore$ No. of moles left after heating $=n-\frac{3}{5} n=\frac{2}{5} n$

Thus, $n_{1}=n, T_{1}=300 K ; n_{2}=\frac{2}{5} n, T_{2}=?$

$\frac{n}{\frac{2}{5} n}=\frac{T_{2}}{300}$ or $\frac{5}{2}=\frac{T_{2}}{300}$ or $, T_{2}=750 \mathrm{K}$

Alternatively, suppose the volume of the vessel $=V$

i.e. Volume of air initially at $27^{\circ} C=V$

Volume of air expelled $=\frac{3}{5} V$

$\therefore$ Volume of air left at $27^{\circ} \mathrm{C}=\frac{2}{5} \mathrm{V}$

However, on heating to $T^{\circ} K,$ it would become $=V$ As pressure remains constant, (vessel being open),

$\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}$ i.e. $\frac{2 / 5 \mathrm{V}}{300 \mathrm{K}}=\frac{\mathrm{V}}{T_{2}}$ or $T_{2}=750 \mathrm{K}$

Sol. Weight of $2.24 \mathrm{L}$ of $\mathrm{H}_{2}$ at $\mathrm{S} . T . P .=0.2 \mathrm{g}$

(Mol. mass of $\left.H_{2}=2\right)=0.1 \mathrm{mol}$

Weight of $1.12 \mathrm{L}$ of $D_{2}$ at $S . T . P .=0.2 \mathrm{g}$

(Mol. mass of $\left.D_{2}=4\right)=0.05 \mathrm{mol}$

As number of moles of two gases are different but V and T are same, therefore, their partial pressures will be different, i.e., in the ratio of their number of moles. Thus,

$\frac{P_{H_{2}}}{P_{D_{2}}}=\frac{n_{H_{2}}}{n_{D_{2}}}=\frac{0.1}{0.5}=2$

Now, $D_{2}$ present in the first bulb $=0.1 g$ (Given)

$D_{2}$ diffused into the second bulb

$=0.2-0.1=0.1 g=0.56 L$ at $S . T . P .$

Now, $\frac{r_{H_{2}}}{r_{D_{2}}}=\frac{P_{H_{2}}}{P_{D_{2}}} \times \sqrt{\frac{M_{D_{2}}}{M_{H_{2}}}}$

Or, $\frac{v_{H_{2}}}{t} \times \frac{t}{v_{D_{2}}}=\frac{P_{H_{2}}}{P_{D_{2}}} \times \sqrt{\frac{M_{D_{2}}}{M_{H_{2}}}}$

$\frac{v_{H_{2}}}{t} \times \frac{t}{0.56 L}=2 \times \sqrt{\frac{4}{2}}$

or $\quad v_{H_{2}}=1.584 L=0.14 g$ of $H_{2}$

$\therefore$ Weight of the gases in 2 nd bulb

$=0.10 g\left(D_{2}\right)+0.14 g\left(H_{2}\right)=0.24 g$

Hence, in the 2 nd bulb,

$\%$ of $D_{2}$ by weight $=\frac{0.10}{0.24} \times 100=41.67 \%$

$\%$ of $H_{2}$ by weight $=100-41.67=58.33 \%$

Sol. Weight of $L P G$ originally present $=29.0-14.8$ $=14.2 \mathrm{kg}$

Pressure $=2.5$ atm Weight of LPG present after use $=23.2-14.8$ $=8.4 \mathrm{kg}$

since volume of the cylinder is constant, applying

since volume of the cylinder is constant, applying

$p V=n R T$

$\frac{p_{1}}{p_{2}}=\frac{n_{1}}{n_{2}}=\frac{W_{1} / M}{W_{2} / M}=\frac{W_{1}}{W_{2}}$

$\frac{2.5}{p_{2}}=\frac{14.2}{8.4}$

$p_{2}=\frac{2.5 \times 8.4}{14.2}=1.48 \mathrm{atm}$

or Weight of used gas $=14.2-8.4=5.8 \mathrm{kg}$

Moles of gas $=\frac{5.8 \times 10^{3}}{58}=100$ moles

Normal conditions

$p=1$ atm;

$T=273+27=300 K$

Volume of 100 moles of $L P G$ at 1 atm and $300 \mathrm{K}$

$V=\frac{n R T}{p}=\frac{100 \times 0.082 \times 300}{1}=2460$ litre

$V=2.460 \mathrm{m}^{3}$

Sol. Compressibility factor

$Z=\frac{p V}{R T} ; \quad 0.5=\frac{100 \times V}{0.082 \times 273}$

$\therefore V=\frac{0.5 \times 0.082 \times 273}{100}=0.1119 L$

If volume of molecules is negligible i.e. $b$ is negligible vander Waals’ equation:

$\left(p+\frac{a}{V^{2}}\right)(\mathrm{V})=R T$

or $p V=R T-a / V$ or $a=R T V-P V^{2}$

$=(0.082 \times 273 \times 0.1119)-(100 \times 0.1119 \times 0.1119)$

$=1.253 \mathrm{atm} L^{2} \mathrm{mol}^{-2}$