States of Matter(Deleted)
5.1 What will be the minimum pressure required to compress $500 \mathrm{dm}^{3}$ of air at 1 bar to $200 \mathrm{dm}^{3}$ at $30^{\circ} \mathrm{C}$ ?
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Answer
Given,
Initial pressure, $p_1=1$ bar
Initial volume, $V_1=500 dm^{3}$
Final volume, $V_2=200 dm^{3}$
Since the temperature remains constant, the final pressure $(p_2)$ can be calculated using Boyle’s law.
According to Boyle’s law,
$ \begin{aligned} p_1 V_1 & =p_2 V_2 \\ \Rightarrow p_2 & =\frac{p_1 V_1}{V_2} \\ & =\frac{1 \times 500}{200} bar \\ & =2.5 bar \end{aligned} $
Therefore, the minimum pressure required is 2.5 bar.
5.2 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?
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Answer
Given,
Initial pressure, $p_1=1.2$ bar
Initial volume, $V_1=120 mL$
Final volume, $V_2=180 mL$
Since the temperature remains constant, the final pressure $(p_2)$ can be calculated using Boyle’s law.
According to Boyle’s law,
$ \begin{aligned} p_1 V_1 & =p_2 V_2 \\ p_2 & =\frac{p_1 V_1}{V_2} \\ & =\frac{1.2 \times 120}{180} bar \\ & =0.8 bar \end{aligned} $
Therefore, the pressure would be 0.8 bar.
5.3 Using the equation of state $p V=n R T$, show that at a given temperature density of a gas is proportional to gas pressure $p$.
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Answer
The equation of state is given by,
$p V=n R T$
Where,
p $\rightarrow$ Pressure of gas
V $\rightarrow$ Volume of gas
n $\rightarrow$ Number of moles of gas
R $\rightarrow$ Gas constant
T $\rightarrow$ Temerature of gas
From equation (i) we have,
$\frac{n}{V}=\frac{p}{R T}$
Replacing $n$ with $\frac{m}{M}$, we have
$\frac{m}{M V}=\frac{p}{R T}$
Where,
m $ \rightarrow$ mass of gas
M $\rightarrow$ Molar mass of gas
But, $\frac{m}{V}=d _{(d=\text{ density of gas })}$
Thus, from equation (ii), we have $\frac{d}{M}=\frac{p}{R T}$
$\Rightarrow d=(\frac{M}{R T}) p$
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 ( $p$ )
5.4 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?
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Answer
Density (d) of the substance at temperature ( $T$ ) can be given by the expression,
$d=\frac{M p}{R T}$
Now, density of oxide $(d_1)$ is given by,
$d_1=\frac{M_1 p_1}{R T}$
Where, $M_1$ and $p_1$ are the mass and pressure of the oxide respectively.
Density of dinitrogen gas $(d_2)$ is given by,
$d_2=\frac{M_2 p_2}{R T}$
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
$p_2=5$ bar
Molecular mass of nitrogen, $M_2=28 g / mol$
Now, $M_1=\frac{M_2 p_2}{p_1}$
$ \begin{aligned} & =\frac{28 \times 5}{2} \\ & =70 g / mol \end{aligned} $
Hence, the molecular mass of the oxide is $70 g / mol$.
5.5 Pressure of $1 \mathrm{~g}$ of an ideal gas $\mathrm{A}$ at $27^{\circ} \mathrm{C}$ is found to be $2 \mathrm{bar}$. 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.
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Answer
For ideal gas $A$, the ideal gas equation is given by,
$p_A V=n_A R T$.
Where, $p_A$ and $n_A$ represent the pressure and number of moles of gas $A$.
For ideal gas $B$, the ideal gas equation is given by,
$p_B V=n_B R T$
Where, $p_B$ and $n_B$ represent the pressure and number of moles of gas $B$.
[ $V$ and $T$ are constants for gases $A$ and $B$ ]
From equation (i), we have
$$ p_A V=\frac{m_A}{M_A} R T \rightarrow \frac{p_A M_A}{m_A}=\frac{R T}{V} \ldots \ldots \tag{iii} $$
From equation (ii), we have
$$ p_B V=\frac{m_B}{M_B} R T \Rightarrow \frac{p_B M_B}{m_B}=\frac{R T}{V} \ldots \ldots \tag{iv} $$
Where, $M_A$ and $M_B$ are the molecular masses of gases $A$ and $B$ respectively.
Now, from equations (iii) and (iv), we have
$$ \frac{p_A M_A}{m_A}=\frac{p_B M_B}{m_B} \tag{v} $$
Given,
$ \begin{aligned} & m_A=1 g \\ & p_A=2 bar \\ & m_B=2 g \\ & p_B=(3-2)=1 bar \end{aligned} $
(Since total pressure is 3 bar)
Substituting these values in equation (v), we have
$ \begin{aligned} & \frac{2 \times M_A}{1}=\frac{1 \times M_B}{2} \\ & \Rightarrow 4 M_A=M_B \end{aligned} $
Thus, a relationship between the molecular masses of $A$ and $B$ is given by
$ 4 M_A=M_B . $
5.6 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?
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Answer
The reaction of aluminium with caustic soda can be represented as:
$2 Al+2 NaOH+2 H_2 O \longrightarrow 2 NaAlO_2+3 H_2$
$2 \times 27 g$ $3 \times 22400 mL$
At STP $(273.15 K$ and $1 atm), 54 g(2 \times 27 g)$ of Al gives $3 \times 22400 mL$ of $H_2$
$\therefore 0.15 g$ Al gives $\frac{3 \times 22400 \times 0.15}{54} mL$ of $H_2$ i.e., $186.67 mL$ of $H_2$.
At STP,
$p_1=1 atm$
$V_1=186.67 mL$
$T_1=273.15 K$
Let the volume of dihydrogen be ${ }^{V}$ at $p_2=0.987$ atm (since 1 bar $.=0.987 atm)$ and $T_2=20^{\circ} C=(273.15+20) K=$ 293.15 K.
Now,
$ \begin{aligned} \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 mL \\ & =203 mL \end{aligned} $
Therefore, $203 mL$ of dihydrogen will be released.
5.7 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 $27{ }^{\circ} \mathrm{C}$ ?
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Answer
It is known that,
$ p=\frac{m}{M} \frac{R T}{V} $
For methane $(CH_4)$,
$ \begin{aligned} p _{CH_4} & =\frac{3.2}{16} \times \frac{8.314 \times 300}{9 \times 10^{-3}} \begin{bmatrix} \text{ Since } 9 dm^{3}=9 \times 10^{-3} m^{3} \\ 27^{\circ} C=300 K \end{bmatrix} \\ & =5.543 \times 10^{4} Pa \end{aligned} $
For carbon dioxide $(CO_2)$,
$ \begin{aligned} p _{CO_2} & =\frac{4.4}{44} \times \frac{8.314 \times 300}{9 \times 10^{-3}} \\ & =2.771 \times 10^{4} Pa \end{aligned} $
Total pressure exerted by the mixture can be obtained as:
$ \begin{aligned} p & =p _{CH_4}+p _{CO_2} \\ & =(5.543 \times 10^{4}+2.771 \times 10^{4}) Pa \\ & =8.314 \times 10^{4} Pa \end{aligned} $
Hence, the total pressure exerted by the mixture is $8.314 \times 10^{4} Pa$.
5.8 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 $0.7 \mathrm{bar}$ are introduced in a $1 \mathrm{~L}$ vessel at $27^{\circ} \mathrm{C}$ ?
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Answer
Let the partial pressure of $H_2$ in the vessel be $p _{H_2}$.
Now,
$ \begin{matrix} p_1=0.8 bar & p_2=p _{H_2}=? \\ V_1=0.5 L & V_2=1 L \end{matrix} $
It is known that,
$ \begin{aligned} 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 \end{aligned} $
Now, let the partial pressure of $O_2$ in the vessel be $p _{O_2}$.
Now,
$p_1=0.7$ bar $\quad p_2=p _{O_2}=$ ?
$V_1=2.0 L \quad V_2=1 L$
$p 1 V 1=p 2$
5.9 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?
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Answer
Given,
$ \begin{aligned} & d_1=5.46 g / dm^{3} \\ & p_1=2 bar \\ & T_1=27^{\circ} C=(27+273) K=300 K \\ & p_2=1 bar \\ & T_2=273 K \\ & d_2=? \end{aligned} $
The density $(d_2)$ of the gas at STP can be calculated using the equation,
$ \begin{aligned} & d=\frac{M p}{R T} \\ & \begin{aligned} \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 g dm^{-3} \end{aligned} \end{aligned} $
Hence, the density of the gas at STP will be $3 g dm^{-3}$.
5.10 $34 .05 \mathrm{~mL}$ of phosphorus vapour weighs $0.0625 \mathrm{~g}$ at $546{ }^{\circ} \mathrm{C}$ and $0.1 \mathrm{bar}$ pressure. What is the molar mass of phosphorus?
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Answer
Given,
$p=0.1$ bar
$V=34.05 mL=34.05 \times 10^{- 3} L=34.05 \times 10^{- 3} dm^{3}$
$R=0.083$ bar $dm^{3} K^{{-1}} mol^{{-1}}$
$T=546^{\circ} C=(546+273) K=819 K$
The number of moles ( $n$ ) can be calculated using the ideal gas equation as:
$ \begin{aligned} p V & =n 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} mol \end{aligned} $
Therefore, molar mass of phosphorus $=\frac{0.0625}{5.01 \times 10^{-5}}=1247.5 g mol^{-1}$
Hence, the molar mass of phosphorus is $1247.5 g mol^{-1}$.
5.11 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?
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Answer
Let the volume of the round bottomed flask be $V$.
Then, the volume of air inside the flask at $27^{\circ} C$ is $V$.
Now,
$V_1=V$
$T_1=27^{\circ} C=300 K$
$V_2=$ ?
$T_2=477^{\circ} C=750 K$
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 V}{300} $
$ =2.5 V $
Therefore, volume of air expelled out $=2.5 V - V=1.5 V$
Hence, fraction of air expelled out
$ =\frac{1.5 V}{2.5 V}=\frac{3}{5} $
5.12 Calculate the temperature of $4.0 \mathrm{~mol}$ of a gas occupying $5 \mathrm{dm}^{3}$ at 3.32 bar. ( $\mathrm{R}=0.083$ bar $\mathrm{dm}^{3} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$ ).
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Answer
Given,
$n=4.0 mol$ $V=5 dm^{3}$
$p=3.32$ bar
$R=0.083$ bar $dm^{3} K^{{-1}} mol^{-1}$
The temperature $(T)$ can be calculated using the ideal gas equation as:
$ \begin{aligned} p V & =n R T \\ \Rightarrow T & =\frac{p V}{n R} \\ & =\frac{3.32 \times 5}{4 \times 0.083} \\ & =50 K \end{aligned} $
Hence, the required temperature is $50 K$.
5.13 Calculate the total number of electrons present in $1.4 \mathrm{~g}$ of dinitrogen gas.
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Answer
Molar mass of dinitrogen $(N_2)=28 g mol^{{-1}}$
Thus, $1.4 g$ of
$ N_2=\frac{1.4}{28}=0.05 mol $
$=0.05 \times 6.02 \times 10^{23}$ number of molecules
$=3.01 \times 10^{23}$ number of molecules
Now, 1 molecule of $N_2$ contains 14 electrons.
Therefore, $3.01 \times 10^{23}$ molecules of $N_2$ contains $=1.4 \times 3.01 \times 10^{23}$
$=4.214 \times 10^{23}$ electrons
5.14 How much time would it take to distribute one Avogadro number of wheat grains, if $10^{10}$ grains are distributed each second?
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Answer
Avogadro number $=6.02 \times 10^{23}$
Thus, time required $=\frac{6.02 \times 10^{23}}{10^{10}} s$
$=6.02 \times 10^{23} s$
$=\frac{6.02 \times 10^{23}}{60 \times 60 \times 24 \times 365}$ years
$=1.909 \times 10^{6}$ years
Hence, the time taken would be $1.909 \times 10^{6}$ years
5.15 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 $\mathrm{dm}^{3} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$.
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Answer
Given,
Mass of dioxygen $(O_2)=8 g$
Thus, number of moles of
$ O_2=\frac{8}{32}=0.25 mole $
Mass of dihydrogen $(H_2)=4 g$
Thus, number of moles of
$ H_2=\frac{4}{2}=2 mole $
Therefore, total number of moles in the mixture $=0.25+2=2.25$ mole
Given,
$V=1 dm^{3}$
$n=2.25 mol$
$dm^3 $ at $27^oC.R = 0.083 \text{ bar d}m^3K^{-1}mol^{-1}$
$T=27^{\circ} C=300 K$
Total pressure $(p)$ can be calculated as:
$p V=n R T$
$\Rightarrow p=\frac{n R T}{V}$
$ \begin{aligned} & =\frac{225 \times 0.083 \times 300}{1} \\ & =56.025 bar \end{aligned} $
Hence, the total pressure of the mixture is 56.025 bar.
5.16 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 $\mathrm{R}=0.083$ bar $\mathrm{dm}^{3} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$ ).
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Answer
Given,
Radius of the balloon, $r=10 m$
$\therefore$ Volume of the balloon $=\frac{4}{3} \pi r^{3}$
$=\frac{4}{3} \times \frac{22}{7} \times 10^{3}$
$=4190.5 m^{3}$ (approx)
Thus, the volume of the displaced air is $4190.5 m^{3}$.
Given,
Density of air $=1.2 kg m^{- 3}$
Then, mass of displaced air $=4190.5 \times 1.2 kg$
$=5028.6 kg$
Now, mass of helium $(m)$ inside the balloon is given by,
$m=\frac{M p V}{R T}$
Here,
$M=4 \times 10^{-3} kg mol^{-1}$
$p=1.66$ bar
$V=$ Volume of the balloon
$ =4190.5 m^{3} $
$R=0.083$ bar dm $^{3} K^{-1} mol^{-1}$
$T=27^{\circ} C=300 K$
Then, $m=\frac{4 \times 10^{-3} \times 1.66 \times 4190.5 \times 10^{3}}{0.083 \times 300}$
$ =1117.5 kg \text{ (approx) } $
Now, total mass of the balloon filled with helium $=(100+1117.5) kg$
$=1217.5 kg$
Hence, pay load = (5028.6 - 1217.5) kg
$=3811.1 kg$
Hence, the pay load of the balloon is $3811.1 kg$.
5.17 Calculate the volume occupied by $8.8 \mathrm{~g}$ of $\mathrm{CO}_{2}$ at $31.1^{\circ} \mathrm{C}$ and 1 bar pressure. $\mathrm{R}=0.083$ bar $\mathrm{L} \mathrm{K}^{-1} \mathrm{~mol}^{-1}$.
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Answer
It is known that,
$ \begin{aligned} & p V=\frac{m}{M} R T \\ & \Rightarrow V=\frac{m R T}{M p} \end{aligned} $
Here,
$m=8.8 g$
$R=0.083$ bar $LK^{-1} mol^{{-1}}$
$T=31.1^{\circ} C=304.1 K$
$M=44 g$
$p=1$ bar
Thus, volume $(V)=\frac{8.8 \times 0.083 \times 304.1}{44 \times 1}$
$ \begin{aligned} & =5.04806 L \\ & =5.05 L \end{aligned} $
Hence, the volume occupied is $5.05 L$.
5.18 $2 .9 \mathrm{~g}$ 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?
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Answer
Volume ( $V$ ) occupied by dihydrogen is given by,
$ \begin{aligned} V & =\frac{m}{M} \frac{R T}{p} \\ & =\frac{0.184}{2} \times \frac{R \times 290}{p} \end{aligned} $
Let $M$ be the molar mass of the unknown gas. Volume $(V)$ occupied by the unknown gas can be calculated as:
$ \begin{aligned} V & =\frac{m}{M} \frac{R T}{p} \\ & =\frac{2.9}{M} \times \frac{R \times 368}{p} \end{aligned} $
According to the question,
$ \begin{aligned} & \frac{0.184}{2} \times \frac{R \times 290}{p}=\frac{2.9}{M} \times \frac{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 g mol^{-1} \end{aligned} $
Hence, the molar mass of the gas is $40 g mol^{-1}$.
5.19 A mixture of dihydrogen and dioxygen at one bar pressure contains $20 %$ by weight of dihydrogen. Calculate the partial pressure of dihydrogen.
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Answer
Let the weight of dihydrogen be $20 g$ and the weight of dioxygen be $80 g$.
Then, the number of moles of dihydrogen,
$ n _{H_2}=\frac{20}{2}=10 \text{ moles } \text{ and the number of moles of } $
dioxygen,
$ n _{O_2}=\frac{80}{32}=2.5 moles $
Given,
Total pressure of the mixture, $p _{\text{total }}=1$ bar
Then, partial pressure of dihydrogen,
$ \begin{aligned} p _{H_2} & =\frac{n _{H_2}}{n _{H_2}+n _{O_2}} \times P _{\text{total }} \\ & =\frac{10}{10+2.5} \times 1 \\ & =0.8 bar \end{aligned} $
Hence, the partial pressure of dihydrogen is 0.8 bar .
5.20 What would be the SI unit for the quantity $p V^{2} T^{2} / n$ ?
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Answer
The $SI$ unit for pressure, $p$ is $Nm^{- 2}$.
The $SI$ unit for volume, $V$ is $m^{3}$
The $SI$ unit for temperature, $T$ is $K$.
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,
$=\frac{(Nm^{-2})(m^{3})^{2}(K)^{2}}{mol}$
$=Nm^{4} K^{2} mol^{-1}$
5.21 In terms of Charles’ law explain why $-273^{\circ} \mathrm{C}$ is the lowest possible temperature.
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Answer
Charles’ law states that at constant pressure, the volume of a fixed mass of gas is directly proportional to its absolute temperature.
![](https://temp-public-img-folder.s3.amazonaws.com/images/c11_ncert_solution_images/2024_03_14_2813f9ba0f068094bb9cg-131.jpg)
It was found that for all gases (at any given pressure), the plots of volume vs. temperature (in ${ }^{\circ} C$ ) is a straight line. If this line is extended to zero volume, then it intersects the temperature-axis at $-273^{\circ} C$. In other words, the volume of any gas at $-273^{\circ} C$ is zero. This is because all gases get liquefied before reaching a temperature of $-273^{\circ} C$. Hence, it can be concluded that $-273^{\circ} C$ is the lowest possible temperature.
5.22 Critical temperature for carbon dioxide and methane are $31 .{ }^{\circ} \mathrm{C}$ and $-81.9{ }^{\circ} \mathrm{C}$ respectively. Which of these has stronger intermolecular forces and why?
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Answer
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 $CO_2$.
5.23 Explain the physical significance of van der Waals parameters.
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Answer
Physical significance of ’ $a$ ‘:
’ $a$ ’ is a measure of the magnitude of intermolecular attractive forces within a gas.
Physical significance of ’ $b$ ‘:
‘b’ is a measure of the volume of a gas molecule.