 NCERT Solutions for Class 11 Chemistry chapter 5 States of Matter PDF – eSaral
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### Download NCERT Solutions for Class 11 Chemistry chapter 5 States of Matter PDF

Question 1: What will be the minimum pressure required to compress $500 \mathrm{dm}^{3}$ of air at 1 bar to $200 \mathrm{dm}^{3}$ at $30^{\circ} \mathrm{C} ?$

Solution. Given,

Initial pressure, $p_{1}=1$ bar Initial volume, $V_{1}=500 \mathrm{dm}^{3}$

Final volume, $V_{2}=200 \mathrm{dm}^{3}$

Since the temperature remains constant, the final pressure $\left(p_{2}\right)$ can be calculated using Boyle’s law.

According to Boyle’s law,

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

$\Rightarrow p_{2}=\frac{p_{1} V_{1}}{V_{2}}$

$=\frac{1 \times 500}{200} \mathrm{bar}$

$=2.5 \mathrm{bar}$Therefore, the minimum pressure required is $2.5$ bar.

Question 2: A vessel of $120 \mathrm{~mL}$ capacity contains a certain amount of gas at $35^{\circ} \mathrm{C}$ and $1.2$ bar pressure. The gas is transferred to another vessel of volume $180 \mathrm{~mL}$ at $35^{\circ} \mathrm{C}$. What would be its pressure?

Solution. Given,

Initial pressure, $p_{1}=1.2$ bar

Initial volume, $V_{1}=120 \mathrm{~mL}$

Final volume, $V_{2}=180 \mathrm{~mL}$

Since the temperature remains constant, the final pressure $\left(p_{2}\right)$ can be calculated using Boyle’s law.

According to Boyle’s law,

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

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

$=\frac{1.2 \times 120}{180} \mathrm{bar}$

$=0.8$ bar

Therefore, the pressure would be $0.8$ bar.

Question 3: Using the equation of state $p V=n R T ;$ show that at a given temperature density of a gas is proportional to gas pressurep.

Solution. The equation of state is given by,

$p V=n \mathrm{R} T \ldots \ldots \ldots$ (i)

Where,

$p \rightarrow$ Pressure of gas

$V \rightarrow$ Volume of gas

$n \rightarrow$ Number of moles of gas

$\mathrm{R} \rightarrow$ Gas constant

$T \rightarrow$ Temperature of gas

From equation (i) we have,

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

Replacing $n$ with $\frac{m}{M}$, we have

$\frac{m}{M V}=\frac{p}{\mathrm{R} T} \ldots \ldots \ldots . .(\mathrm{ii})$

Where,

$m \rightarrow$ Mass of gas

$M \rightarrow$ Molar mass of gas

But, $\frac{m}{V}=d(d=$ density of gas)

Thus, from equation (ii), we have

Thus, from equation (ii), we have

\begin{equation}

\frac{d}{M}=\frac{p}{\mathrm{R} T}

\end{equation}

$\Rightarrow d=\left(\frac{M}{\mathrm{R} T}\right) p$

Molar mass $(M)$ of a gas is always constant and therefore, at constant temperature $(T), \frac{M}{R T}=$ constant.

$d=($ constant $) p$

$\Rightarrow d \propto p$

Hence, at a given temperature, the density $(d)$ of gas is proportional to its pressure $(p)$

Question $4$ : At $0^{\circ} \mathrm{C}$, the density of a certain oxide of a gas at 2 bar is same as that of dinitrogen at 5 bar. What is the molecular mass of the oxide?

Solution. Density (d) of the substance at temperature ( $T$ ) can be given by the expression,

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

Now, density of oxide $\left(d_{1}\right)$ is given by,

$d_{1}=\frac{M_{1} p_{1}}{\mathrm{R} T}$

Where, $M_{1}$ and $p_{1}$ are the mass and pressure of the oxide respectively.

Density of dinitrogen gas $\left(d_{2}\right)$ is given by,

$d_{2}=\frac{M_{2} p_{2}}{\mathrm{R} T}$

Where, $M_{2}$ and $p_{2}$ are the mass and pressure of the oxide respectively.

According to the given question,

$d_{1}=d_{2}$

$\therefore M_{1} p_{1}=M_{2} p_{2}$

Given,

$p_{1}=2$ bar

$p_{2}=5 \mathrm{bar}$

Molecular mass of nitrogen, $M_{2}=28 \mathrm{~g} / \mathrm{mol}$

Now, $M_{1}=\frac{M_{2} p_{2}}{p_{1}}$

$=\frac{28 \times 5}{2}$

$=70 \mathrm{~g} / \mathrm{mol}$

Hence, the molecular mass of the oxide is $70 \mathrm{~g} / \mathrm{mol}$.

Question $5$ : Pressure of $1 \mathrm{~g}$ of an ideal gas $\mathrm{A}$ at $27^{\circ} \mathrm{C}$ is found to be 2 bar. When $2 \mathrm{~g}$ of another ideal gas $\mathrm{B}$ is introduced in the same flask at same temperature the pressure becomes 3 bar. Find a relationship between their molecular masses.

Solution. For ideal gas $A$, the ideal gas equation is given by,

$p_{\mathrm{A}} V=n_{\mathrm{A}} \mathrm{R} T$$\ldots \ldots .$ (i)

Where, $p_{\mathrm{A}}$ and $n_{\mathrm{A}}$ represent the pressure and number of moles of gas $\mathrm{A}$.

For ideal gas $\mathrm{B}$, the ideal gas equation is given by,

$p_{B} V=n_{B} \mathrm{R} T \ldots \ldots \ldots$ (ii)

Where, $p_{\mathrm{B}}$ and $n_{\mathrm{B}}$ represent the pressure and number of moles of gas $\mathrm{B}$.

$[V$ and $T$ are constants for gases $A$ and $B]$

From equation (i), we have

$p_{A} V=\frac{m_{A}}{\mathrm{M}_{A}} \mathrm{R} T \Rightarrow \frac{p_{A} \mathrm{M}_{A}}{m_{A}}=\frac{\mathrm{R} T}{V} \ldots \ldots \ldots(\mathrm{iii})$

From equation (ii), we have

$p_{\mathrm{B}} V=\frac{m_{\mathrm{B}}}{\mathrm{M}_{\mathrm{B}}} \mathrm{R} T \Rightarrow \frac{p_{\mathrm{B}} \mathrm{M}_{\mathrm{B}}}{m_{\mathrm{B}}}=\frac{\mathrm{R} T}{V} \ldots \ldots \ldots$ (iv)

Where, $M_{A}$ and $M_{B}$ are the molecular masses of gases $A$ and $B$ respectively.

Now, from equations (iii) and (iv), we have

$\frac{p_{A} \mathrm{M}_{A}}{m_{A}}=\frac{p_{B} \mathrm{M}_{B}}{m_{B}} \ldots \ldots \ldots . .(v)$

Given,

$m_{A}=1 \mathrm{~g}$

$p_{A}=2$ bar

$m_{B}=2 \mathrm{~g}$

$p_{B}=(3-2)=1$ bar

(Since total pressure is 3 bar)

Substituting these values in equation (v), we have

$\frac{2 \times \mathrm{M}_{A}}{1}=\frac{1 \times \mathrm{M}_{B}}{2}$

$\Rightarrow 4 \mathrm{M}_{A}=\mathrm{M}_{B}$

Thus, a relationship between the molecular masses of $A$ and $B$ is given by

$4 \mathrm{M}_{A}=\mathrm{M}_{B} .$

Question 6: The drain cleaner, Drainex contains small bits of aluminum which react with caustic soda to produce dihydrogen. What volume of dihydrogen at $20^{\circ} \mathrm{C}$ and one bar will be released when $0.15 \mathrm{~g}$ of aluminum reacts?

Solution. The reaction of aluminium with caustic soda can be represented as:

At STP $(273.15 \mathrm{~K}$ and $1 \mathrm{~atm}), 54 \mathrm{~g}(2 \times 27 \mathrm{~g})$ of Al gives $3 \times 22400 \mathrm{~mL}$ of $\mathrm{H}_{2}$

$\therefore 0.15 \mathrm{~g} \mathrm{Al}$ gives $\frac{3 \times 22400 \times 0.15}{54} \mathrm{~mL}$ of $\mathrm{H}_{2}$ i.e., $186.67 \mathrm{~mL}$ of $\mathrm{H}_{2}$.

At STP,

$p_{1}=1$ atm

$V_{1}=186.67 \mathrm{~mL}$

$T_{1}=273.15 \mathrm{~K}$

Let the volume of dihydrogen be $V_{2}$ at $p_{2}=0.987 \mathrm{~atm}$ (since 1 bar $\left.=0.987 \mathrm{~atm}\right)$ and $T_{2}=20^{\circ} \mathrm{C}=(273.15+20) \mathrm{K}=293.15 \mathrm{~K}$.

Now,

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

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

$=\frac{1 \times 186.67 \times 293.15}{0.987 \times 273.15}$

$=202.98 \mathrm{~mL}$

$=203 \mathrm{~mL}$

Therefore, $203 \mathrm{~mL}$ of dihydrogen will be released.

Question 7: What will be the pressure exerted by a mixture of $3.2 \mathrm{~g}$ of methane and $4.4 \mathrm{~g}$ of carbon dioxide contained in a $9 \mathrm{dm}^{3}$ flask at $27^{\circ} \mathrm{C}$ ?

Solution. It is known that,

$p=\frac{m}{M} \frac{\mathrm{R} T}{V}$

For methane $\left(\mathrm{CH}_{4}\right)$,

$p_{\mathrm{CH}_{+}}=\frac{3.2}{16} \times \frac{8.314 \times 300}{9 \times 10^{-3}}\left[\begin{array}{l}\text { Since } 9 \mathrm{dm}^{3}=9 \times 10^{-3} \mathrm{~m}^{3} \\ 27^{\circ} \mathrm{C}=300 \mathrm{~K}\end{array}\right]$

$=5.543 \times 10^{4} \mathrm{~Pa}$

For carbon dioxide $\left(\mathrm{CO}_{2}\right)$

$p_{\mathrm{CO}_{2}}=\frac{4.4}{44} \times \frac{8.314 \times 300}{9 \times 10^{-3}}$

$=2.771 \times 10^{4} \mathrm{~Pa}$

Total pressure exerted by the mixture can be obtained as:

$p=p_{\mathrm{CH}_{4}}+p_{\mathrm{CO}_{2}}$

$=\left(5.543 \times 10^{4}+2.771 \times 10^{4}\right) \mathrm{Pa}$

$=8.314 \times 10^{4} \mathrm{~Pa}$

Hence, the total pressure exerted by the mixture is $8.314 \times 10^{4} \mathrm{~Pa}$.

Question 8: What will be the pressure of the gaseous mixture when $0.5 \mathrm{~L}$ of $\mathrm{H}_{2}$ at $0.8$ bar and $2.0 \mathrm{~L}$ of dioxygen at $0.7$ bar are introduced in a $1 \mathrm{~L}$ vessel at $27^{\circ} \mathrm{C} ?$

Solution. Let the partial pressure of $\mathrm{H}_{2}$ in the vessel be $p_{\mathrm{H}_{2}}$.

Now,

$p_{1}=0.8$ bar $\quad p_{2}=p_{H_{2}}=?$

$V_{1}=0.5 \mathrm{~L} \quad V_{2}=1 \mathrm{~L}$

It is known that,

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

$\Rightarrow p_{2}=\frac{p_{1} V_{1}}{V_{2}}$

$\Rightarrow p_{H_{2}}=\frac{0.8 \times 0.5}{1}$

$=0.4$ bar

Now, let the partial pressure of $\mathrm{O}_{2}$ in the vessel be $p_{\mathrm{O}_{2}}$

Now,

$p_{1}=0.7$ bar $\quad p_{2}=p_{\mathrm{O}_{2}}=?$

$V_{1}=2.0 \mathrm{~L} \quad \mathrm{~V}_{2}=1 \mathrm{~L}$

$\mathrm{p}_{1} \mathrm{~V}_{1}=\mathrm{p}_{2} \mathrm{~V}_{2}$

$\Rightarrow \mathrm{p}_{2}=\frac{\mathrm{p}_{1} \mathrm{~V}_{1}}{\mathrm{~V}_{2}}$

$\Rightarrow \mathrm{P}_{\mathrm{O}_{2}}=\frac{0.7 \times 2.0}{1}=1.4 \mathrm{bar}$

Total pressure of the gas mixture in the vessel can be obtained as:

$p_{\text {total }}=p_{\mathrm{H}_{2}}+p_{\mathrm{O}_{2}}$

$=0.4+1.4$

$=1.8$ bar

Hence, the total pressure of the gaseous mixture in the vessel is $1.8$ bar.

Question 9: Density of a gas is found to be $5.46 \mathrm{~g} / \mathrm{dm}^{3}$ at $27^{\circ} \mathrm{C}$ at 2 bar pressure. What will be its density at STP?

Solutioin. Given,

$d_{1}=5.46 \mathrm{~g} / \mathrm{dm}^{3}$

$p_{1}=2$ bar

$T_{1}=27^{\circ} \mathrm{C}=(27+273) \mathrm{K}=300 \mathrm{~K}$

$p_{2}=1 \mathrm{bar}$

$\mathrm{T}_{2}=273 \mathrm{~K}$

$d_{2}=?$

The density $\left(d_{2}\right)$ of the gas at STP can be calculated using the equation,

$d=\frac{M p}{\mathrm{R} T}$

$\therefore \frac{d_{1}}{d_{2}}=\frac{\frac{M p_{1}}{R T_{1}}}{\frac{M p_{2}}{R T_{2}}}$

$\Rightarrow \frac{d_{1}}{d_{2}}=\frac{p_{1} T_{2}}{p_{2} T_{1}}$

$\Rightarrow d_{2}=\frac{p_{2} T_{1} d_{1}}{p_{1} T_{2}}$

$=\frac{1 \times 300 \times 5.46}{2 \times 273}$

$=3 \mathrm{~g} \mathrm{dm}^{-3}$

Hence, the density of the gas at STP will be $3 \mathrm{~g} \mathrm{dm}^{-3}$.

Question 10: $34.05 \mathrm{~mL}$ of phosphorus vapour weighs $0.0625 \mathrm{~g}$ at $546^{\circ} \mathrm{C}$ and $0.1$ bar pressure. What is the molar mass of phosphorus?

Solution. Given,

$p=0.1$ bar

$V=34.05 \mathrm{~mL}=34.05 \times 10^{-3} \mathrm{~L}=34.05 \times 10^{-3} \mathrm{dm}^{3}$

$\mathrm{R}=0.083 \mathrm{bar} \mathrm{dm}^{3} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$

$T=546^{\circ} \mathrm{C}=(546+273) \mathrm{K}=819 \mathrm{~K}$

The number of moles $(n)$ can be calculated using the ideal gas equation as:

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

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

$=\frac{0.1 \times 34.05 \times 10^{-3}}{0.083 \times 819}$

$=5.01 \times 10^{-5} \mathrm{~mol}$

Therefore, molar mass of phosphorus

$=\frac{0.0625}{5.01 \times 10^{-5}}=1247.5 \mathrm{~g} \mathrm{~mol}^{-1}$

Hence, the molar mass of phosphorus is $1247.5 \mathrm{~g} \mathrm{~mol}^{-1}$.

Question 11: A student forgot to add the reaction mixture to the round bottomed flask at $27^{\circ} \mathrm{C}$ but instead he/she placed the flask on the flame. After a lapse of time, he realized his mistake, and using a pyrometer he found the temperature of the flask was $477^{\circ} \mathrm{C}$. What fraction of air would have been expelled out?

Solution. Let the volume of the round bottomed flask be $V$.

Then, the volume of air inside the flask at $27^{\circ} \mathrm{C}$ is $V$.

Now,

$V_{1}=V$

$T_{1}=27^{\circ} \mathrm{C}=300 \mathrm{~K}$

$V_{2}=?$

$T_{2}=477^{\circ} \mathrm{C}=750 \mathrm{~K}$

According to Charles’s law,

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

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

$=\frac{750 \mathrm{~V}}{300}$

$=2.5 \mathrm{~V}$

Therefore, volume of air expelled out $=2.5 \mathrm{~V}-\mathrm{V}=1.5 \mathrm{~V}$

Hence, fraction of air expelled out $=\frac{1.5 V}{2.5 V}=\frac{3}{5}$

Question 12: Calculate the temperature of $4.0$ mol of a gas occupying $5 \mathrm{dm}^{3}$ at $3.32$ bar.

Solution. Given,

$n=4.0 \mathrm{~mol}$

$V=5 \mathrm{dm}^{3}$

$p=3.32$ bar

$\mathrm{R}=0.083 \mathrm{bar} \mathrm{dm}^{3} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$

The temperature (T) can be calculated using the ideal gas equation as:

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

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

$=\frac{3.32 \times 5}{4 \times 0.083}$

$=50 \mathrm{~K}$

Hence, the required temperature is 50 K.

Question 13: Calculate the total number of electrons present in $1.4 \mathrm{~g}$ of dinitrogen gas.

Solution. Molar mass of dinitrogen $\left(\mathrm{N}_{2}\right)=28 \mathrm{~g} \mathrm{~mol}^{-1}$

Thus, $1.4 \mathrm{~g}$ of $\mathrm{N}_{2}=\frac{1.4}{28}=0.05 \mathrm{~mol}$

$=0.05 \times 6.02 \times 10^{23}$ number of molecules

$=3.01 \times 10^{23}$ number of molecules

Now, 1 molecule of $\mathrm{N}_{2}$ contains 14 electrons.

Therefore, $3.01 \times 10^{23}$ molecules of $\mathrm{N}_{2}$ contains $=1.4 \times 3.01 \times 10^{23}$

$=4.214 \times 10^{23}$ electrons

Question $14$ How much time would it take to distribute one Avogadro number of wheat grains, if $10^{10}$ grains are distributed each second?

Solution. Avogadro number $=6.02 \times 10^{23}$

Thus, time required

$=\frac{6.02 \times 10^{23}}{10^{10}} \mathrm{~s}$

$=6.02 \times 10^{23} \mathrm{~s}$

$=\frac{6.02 \times 10^{23}}{60 \times 60 \times 24 \times 365}$ years

$=1.909 \times 10^{6}$ years

Hence, the time taken would be $1.909 \times 10^{6}$ years $.$

Question $15:$ Calculate the total pressure in a mixture of $8 \mathrm{~g}$ of dioxygen and $4 \mathrm{~g}$ of dihydrogen confined in a vessel of $1 \mathrm{dm}^{3}$ at $27^{\circ} \mathrm{C} . \mathrm{R}=0.083$ bar $\mathrm{dm}^{3} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$.

Solution. Given,

Mass of dioxygen $\left(\mathrm{O}_{2}\right)=8 \mathrm{~g}$

Thus, number of moles of $\mathrm{O}_{2}=\frac{8}{32}=0.25$ mole

Mass of dihydrogen $\left(\mathrm{H}_{2}\right)=4 \mathrm{~g}$

Thus, number of moles of $\mathrm{H}_{2}=\frac{4}{2}=2$ mole

Therefore, total number of moles in the mixture $=0.25+2=2.25$ mole

Given,

$V=1 \mathrm{dm}^{3}$

$n=2.25 \mathrm{~mol}$

$n=2.25 \mathrm{~mol}$

$\mathrm{R}=0.083$ bar $\mathrm{dm}^{3} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$

$T=27^{\circ} \mathrm{C}=300 \mathrm{~K}$

Total pressure $(p)$ can be calculated as:

$p V=n R T$

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

$=\frac{225 \times 0.083 \times 300}{1}$

$=56.025$ bar

Hence, the total pressure of the mixture is $56.025$ bar.

Question 16: Pay load is defined as the difference between the mass of displaced air and the mass of the balloon. Calculate the pay load when a balloon of radius $10 \mathrm{~m}$, mass $100 \mathrm{~kg}$ is filled with helium at $1.66$ bar at $27^{\circ} \mathrm{C}$. (Density of air $=1.2 \mathrm{~kg} \mathrm{~m}^{-3}$ and $\mathrm{R}=0.083 \mathrm{bar} \mathrm{dm}^{3} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$ ).

Splution. Given,

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

$\therefore$ Volume of the balloon $=\frac{4}{3} \pi r^{3}$

$=\frac{4}{3} \times \frac{22}{7} \times 10^{3}$

$=4190.5 \mathrm{~m}^{3}($ approx $)$

Thus, the volume of the displaced air is $4190.5 \mathrm{~m}^{3}$.

Given,

Density of air $=1.2 \mathrm{~kg} \mathrm{~m}^{-3}$

Then, mass of displaced air $=4190.5 \times 1.2 \mathrm{~kg}$

$=5028.6 \mathrm{~kg}$

Now, mass of helium $(m)$ inside the balloon is given by,

$m=\frac{M p V}{\mathrm{R} T}$

Here,

$M=4 \times 10^{-3} \mathrm{~kg} \mathrm{~mol}^{-1}$

$p=1.66 \mathrm{bar}$

$V=$ Volume of the balloon

$=4190.5 \mathrm{~m}^{3}$

$\mathrm{R}=0.083 \mathrm{bar} \mathrm{dm}^{3} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$

$T=27^{\circ} \mathrm{C}=300 \mathrm{~K}$

Then, $m=\frac{4 \times 10^{-3} \times 1.66 \times 4190.5 \times 10^{3}}{0.083 \times 300}$

$=1117.5 \mathrm{~kg}($ approx $)$

Now, total mass of the balloon filled with helium $=(100+1117.5) \mathrm{kg}$

$=3811.1 \mathrm{~kg}$

Hence, the pay load of the balloon is $3811.1 \mathrm{~kg}$.

Question 17: Calculate the volume occupied by $8.8 \mathrm{~g}$ of $\mathrm{CO}_{2}$ at $31.1^{\circ} \mathrm{C}$ and 1 bar pressure.

$\mathrm{R}=0.083 \mathrm{bar} \mathrm{L} \mathrm{K}^{-1} \mathrm{~mol}^{-1}$

Solution. It is known that,

$p V=\frac{m}{M} \mathrm{R} T$

$\Rightarrow V=\frac{m \mathrm{R} T}{M p}$

Here,

$m=8.8 \mathrm{~g}$

$\mathrm{R}=0.083 \mathrm{bar} \mathrm{LK}^{-1} \mathrm{~mol}^{-1}$

$T=31.1^{\circ} \mathrm{C}=304.1 \mathrm{~K}$

$M=44 \mathrm{~g}$

$p=1$ bar

Thus, volume $(V)=\frac{8.8 \times 0.083 \times 304.1}{44 \times 1}$

$=5.04806 \mathrm{~L}$

$=5.05 \mathrm{~L}$

Hence, the volume occupied is $5.05 \mathrm{~L}$.

Question 18: $2.9 \mathrm{~g}$ of a gas at $95^{\circ} \mathrm{C}$ occupied the same volume as $0.184 \mathrm{~g}$ of dihydrogen at $17^{\circ} \mathrm{C}$, at the same pressure. What is the molar mass of the gas?

Solution. Volume ( $V$ occupied by dihydrogen is given by,

$V=\frac{m}{M} \frac{\mathrm{R} T}{p}$

$=\frac{0.184}{2} \times \frac{\mathrm{R} \times 290}{p}$

Let $M$ be the molar mass of the unknown gas. Volume $(V)$ occupied by the unknown gas can be calculated as:

$V=\frac{m}{M} \frac{\mathrm{R} T}{p}$

$=\frac{2.9}{M} \times \frac{\mathrm{R} \times 368}{p}$

According to the question,

$\frac{0.184}{2} \times \frac{\mathrm{R} \times 290}{p}=\frac{2.9}{M} \times \frac{\mathrm{R} \times 368}{p}$

$\Rightarrow \frac{0.184 \times 290}{2}=\frac{2.9 \times 368}{M}$

$\Rightarrow M=\frac{2.9 \times 368 \times 2}{0.184 \times 290}$

$=40 \mathrm{~g} \mathrm{~mol}^{-1}$

Hence, the molar mass of the gas is $40 \mathrm{~g} \mathrm{~mol}^{-1}$.

Question 19: A mixture of dihydrogen and dioxygen at one bar pressure contains $20 \%$ by weight of dihydrogen. Calculate the partial pressure of dihydrogen. Solution. Let the weight of dihydrogen be $20 \mathrm{~g}$ and the weight of dioxygen be $80 \mathrm{~g}$.

Then, the number of moles of dihydrogen, $n_{\mathrm{H}_{2}}=\frac{20}{2}=10$ moles and the number of moles of dioxygen, $n_{0_{2}}=\frac{80}{32}=2.5$ moles

Given,

Total pressure of the mixture, $p_{\text {total }}=1$ bar

Then, partial pressure of dihydrogen,

$p_{\mathrm{H}_{2}}=\frac{n_{\mathrm{H}_{2}}}{n_{\mathrm{H}_{2}}+n_{\mathrm{O}_{2}}} \times P_{\text {total }}$

$=\frac{10}{10+2.5} \times 1$

$=0.8$ bar

Hence, the partial pressure of dihydrogen is $0.8 \mathrm{bar}$.

Question 20: What would be the SI unit for the quantity $p V^{2} T^{2} / n ?$

Solution. The SI unit for pressure, $p$ is $\mathrm{Nm}^{-2}$.

The $S I$ unit for volume, $V$ is $\mathrm{m}^{3}$.

The SI unit for temperature, $T$ is $\mathrm{K}$.

The SI unit for the number of moles, $n$ is mol.

Therefore, the SI unit for quantity $\frac{p V^{2} T^{2}}{n}$ is given by,

$=\frac{\left(\mathrm{Nm}^{-2}\right)\left(m^{3}\right)^{2}(\mathrm{~K})^{2}}{\mathrm{~mol}}$

$=\mathrm{Nm}^{4} \mathrm{~K}^{2} \mathrm{~mol}^{-1}$

Question 21: In terms of Charles’ law explain why $-273^{\circ} \mathrm{C}$ is the lowest possible temperature.

Solution. Charles’ law states that at constant pressure, the volume of a fixed mass of gas is directly proportional to its absolute temperature. It was found that for all gases (at any given pressure), the plots of volume vs. temperature (in $\left.{ }^{\circ} \mathrm{C}\right)$ is a straight line. If this line is extended to zero volume, then it intersects the temperature-axis at $-273^{\circ} \mathrm{C}$. In other words, the volume of any gas at $-273^{\circ} \mathrm{C}$ is zero. This is because all gases get liquefied before reaching a temperature of $-273^{\circ} \mathrm{C}$. Hence i can be concluded that $-273^{\circ} \mathrm{C}$ is the lowest possible temperature

Question $22$ Critical temperature for carbon dioxide and methane are $31.1^{\circ} \mathrm{C}$ and $-81.9{ }^{\circ} \mathrm{C}$ respectively. Which of these has stronger intermolecular forces and why?

Solution. Higher is the critical temperature of a gas, easier is its liquefaction. This means that the intermolecular forces of attraction between the molecules of a gas are directly proportional to its critical temperature. Hence, intermolecular forces of attraction are stronger in the case of $\mathrm{CO}_{2}$.

Question 23: Explain the physical significance of Van der Waals parameters.

Solution. Physical significance of ‘ $\mathrm{a}$ ‘:

‘a’ is a measure of the magnitude of intermolecular attractive forces within a gas.

Physical significance of ‘b’:

$\mathrm{b}^{\prime}$ is a measure of the volume of a gas molecule.\