Text Book Back Questions and Answers - Chapter 6 Gaseous State 11th Chemistry Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

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On April 24, 2024, 11:35 AM

Gaseous State
Textual Evaluation Solved
I. Choose the correct answer from the following:
Question 1.
Gases deviate from ideal behavior at high pressure. Which of the following statement (s) is correct for non - ideality?
(a) at high pressure the collision between the gas molecule become enormous
(b) at high pressure the gas molecules move only in one direction
(c) at high pressure, the volume of gas become insignificant
(d) at high pressure the inter molecular interactions become significant

Answer:
(d) at high pressure the inter molecular interactions become significant
Question 2.
Rate of diffusion of a gas is
(a) directly proportional to its density
(b) directly proportional to its molecular weight
(c) directly proportional to its square root of its molecular weight
(d) inversely proportional to the square root of its molecular weight

Answer:
(d) inversely proportional to the square root of its molecular weight
Question 3.
Which of the following is the correct expression for the equation of state of van der Waals gas?
(a) $\left(\mathrm{P}+\frac{\mathrm{a}}{\mathrm{n}^2 \mathrm{~V}^2}\right)(\mathrm{V}-\mathrm{nb})=\mathrm{nRT}$
(b) $\left(\mathrm{P}+\frac{\mathrm{na}}{\mathrm{n}^2 \mathrm{~V}^2}\right)(\mathrm{V}-\mathrm{nb})=\mathrm{nRT}$

(c) $\left(\mathrm{P}+\frac{\mathrm{an^{2 }}}{\mathrm{V}^2}\right)(\mathrm{V}-\mathrm{nb})=\mathrm{nRT}$
(d) $\left(\mathrm{P}+\frac{\mathrm{n}^2 \mathrm{a}^2}{\mathrm{~V}^2}\right)(\mathrm{V}-\mathrm{nb})=\mathrm{nRT}$
Answer:
(c) $\left(\mathrm{P}+\frac{\mathrm{an}^2}{\mathrm{~V}^2}\right)(\mathrm{V}-\mathrm{nb})=\mathrm{nRT}$
Question 4.
When an ideal gas undergoes unrestrained expansion, no cooling occurs because the molecules

(a) are above inversion temperature
(b) exert no attractive forces on each other
(c) do work equal to the loss in kinetic energy
(d) colide without loss of energy
Answer:
(b) exert no attractive forces on each other
Question 5.
Equal weights of methane and oxygen is mixed in an empty container at $298 \mathrm{~K}$. The fraction of total pressure exerted by oxygen
(a) $1 / 3$
(b) $1 / 2$
(c) $2 / 3$
(d) $1 / 3 \times 273 \times 298$
Answer:
(a) $1 / 3$
Hint:
mass of methane $=$ mass of oxygen $=\mathrm{a}$
number of moles of methane $=\frac{a}{16}$
number of moles of Oxygen $=\frac{a}{32}$
$\frac{\frac{a}{32}}{\frac{a}{16}+\frac{a}{32}}=\frac{\frac{a}{32}}{\frac{3 a}{32}}=\frac{1}{3}$ mole fraction of Oxygen $=$ Partial pressure of oxygen $=$ mole fraction $\mathrm{x}$ Total Pressure $=\frac{1}{3} \mathrm{P}$
Question 6.
The temperatures at which real gases obey the ideal gas laws over a wide range of pressure is called
(a) Critical temperature
(b) Boyle temperature
(c) Inversion temperature

(d) Reduced temperature
Answer:
(b) Boyle temperature
Hint:
The temperature at which real gases obey the ideal gas laws over a wide range of pressure is called Boyle temperature
Question 7.
In a closed room of $1000 \mathrm{~m}^3$ a perfume bottle is opened up. The room develops a smell. This is due to which property of gases?
(a) Viscosity
(b) Density
(c) Diffusion
(d) None
Answer:
(c) Diffusion
Question 8.
A bottle of ammonia and a bottle of $\mathrm{HCl}$ connected through a long tube are opened simultaneously at both ends. The white ammonium chloride ring first formed will be
(a) At the center of the tube
(b) Near the hydrogen chloride bottle

(c) Near the ammonia bottle
(d) Throughout the length of the tube

Answer:
(b) Near the hydrogen chloride bottle
Hint:
Rate of diffusion $\alpha 1 / \sqrt{\mathrm{m}}$
$\mathrm{m}_{\mathrm{NH}_3}=17$
$\mathrm{m}_{\mathrm{HCl}}=36.5$
$\gamma_{\mathrm{NH}_3}>\gamma_{\mathrm{HCl}}$
Hence white fumes first formed near hydrogen chloride.
Question 9.
The value of universal gas constant depends upon
(a) Temperature of the gas
(b) Volume of the gas
(c) Number of moles of the gas
(d) units of Pressure and volume
Answer:
(d) units of Pressure and volume
Question 10.
The value of the gas constant $R$ is
(a) $0.082 \mathrm{dm}^3 \mathrm{~atm}$.
(b) $0.987 \mathrm{cal} \mathrm{mol}^{-1} \mathrm{~K}^{-1}$
(c) $8.3 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$
(d) $8 \mathrm{erg} \mathrm{mol}^{-1} \mathrm{~K}^{-1}$
Answer:
(c) $8.3 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$
Question 11.
Use of hot air balloon in sports at meteorological observation is an application of
(a) Boyle's law

(b) Newton's law
(c) Kelvin's law
(d) Brown's law
Answer:
(a) Boyle's law
Question 12.
The table indicates the value of van der Waals constant ' $a$ ' in $\left(\mathrm{dm}^3\right)^2 \mathrm{~atm} . \mathrm{mol}^{-2}$

The gas which can be most easily liquefied is
(a) $\mathrm{O}_2$
(b) $\mathrm{N}_2$
(c) $\mathrm{NH}_3$
(d) $\mathrm{CH}_4$
Answer:
(c) $\mathrm{NH}_3$
Hint:
Higher the value of 'a', greater the intermolecular force of attraction, easier the liquefaction. Option (c) is correct
Question 13.
Consider the following statements.
(i) Atmospheric pressure is less at the top of a mountain than at sea level
(ii) Gases are much more compressible than solids or liquids
(iii) When the atmospheric pressure increases the height of the mercury column rises Select the correct statement.
(a) (i) and (ii)
(b) (ii) and (iii)
(c) (i) and (iii)

(d) (i), (ii) and (iii)
Answer:
(d) (i), (ii) and (iii)
Question 14.
Compressibility factor for $\mathrm{CO}_2$ at $400 \mathrm{~K}$ and 71.0 bar is 0.8697 . The molar volume of $\mathrm{CO}_2$ under these conditions is ..........
(a) $22.04 \mathrm{dm}^3$
(b) $2.24 \mathrm{dm}^3$
(c) $0.41 \mathrm{dm}^3$
(d) $19.5 \mathrm{dm}^3$
Answer:
(c) $0.41 \mathrm{dm}^3$
Compressibility factor $(\mathrm{z})=\frac{P v}{n R T}$
$
\begin{aligned}
& \mathrm{V}=\frac{z x n R T}{p} \\
& \mathrm{~V}=\frac{\mathrm{Z} \times \mathrm{nRT}}{\mathrm{P}}=\frac{0.8697 \times 1 \times 8.314 \times 10^{-2} \text { bar dm}^3 \mathrm{~K}^{-1} \mathrm{~mol}^{-1} \times 400 \mathrm{~K}}{71 \text { bar }} \\
& \mathrm{V}=0.41 \mathrm{dm}^3
\end{aligned}
$
Question 15.
If temperature and volume of an ideal gas is increased to twice its values, the initial pressure $P$ becomes
(a) $4 P$
(b) $2 \mathrm{P}$
(c) $\mathrm{P}$
(d) $3 \mathrm{P}$
Answer:
(c) $\mathrm{P}$
Hint:

$
\begin{aligned}
& \mathrm{T}_1 \quad \mathrm{~T}_2=2 \mathrm{~T}_1 \\
& \mathrm{~V}_1 \quad \mathrm{~V}_2=2 \mathrm{~V}_1 \\
& \mathrm{P}_1 \quad \mathrm{P}_2=? \\
& \frac{\mathrm{P}_1 \mathrm{~V}_1}{\mathrm{~T}_1}=\frac{\mathrm{P}_2 \mathrm{~V}_2}{\mathrm{~T}_2} \\
& \mathrm{P}_2=\frac{\mathrm{P}_1 \mathrm{~V}_1}{\mathrm{~T}_1} \times \frac{\mathrm{T}_2}{\mathrm{~V}_2}=\frac{\mathrm{P}_1 \mathrm{~V}_1}{\mathrm{~T}_1} \times \frac{2 \mathrm{~T}_1}{2 \mathrm{~V}_1} \\
& \mathrm{P}_2=\mathrm{P}_2 \text { Option (c) }
\end{aligned}
$
Question 16.
At identical temperature and pressure, the rate of diffusion of hydrogen gas is $3 \sqrt{3}$ times that of a hydrocarbon having molecular formula What is the value of $n$ ?
(a) 8
(b) 4
(c) 3
(d) 1
Answer:
(b) 4
Hint:
$
\begin{aligned}
& \frac{\gamma_{\mathrm{H}_2}}{\gamma_{\mathrm{C}_{\mathrm{n}} \mathrm{H}_{2 \mathrm{n}+2}}}=\sqrt{\frac{\mathrm{m}_{\mathrm{C}_{\mathrm{n}} \mathrm{H}_{2 \mathrm{n}+2}}}{\mathrm{~m}_{\mathrm{H}_2}}} \\
& 3 \sqrt{3}=\sqrt{\frac{m_{\mathrm{C}_{\mathrm{n}} \mathrm{H}_{2 n+2}}}{2}} \\
&
\end{aligned}
$
Squaring on both sides and rearranging
$
\begin{aligned}
& 27 \times 2=\mathrm{m}_{\mathrm{C}_{\mathrm{n}}} \mathrm{H}_{2 \mathrm{n}-2} \\
& 54=\mathrm{n}(12)+(2 \mathrm{n}-2)(1)
\end{aligned}
$

$
\begin{aligned}
& 54=12 \mathrm{n}+2 \mathrm{n}-2 \\
& 54=14 \mathrm{n}-2 \\
& \mathrm{n}=(54+2) / 14=56 / 14=4
\end{aligned}
$
Question 17.
Equal moles of hydrogen and oxygen gases are placed in a container, with a pin-hole through which both can escape what fraction of oxygen escapes in the time required for one-half of the hydrogen to escape. (NEET phase 1)
(a) $3 / 8$
(b) $1 / 2$
(c) $1 / 8$
(d) $1 / 4$
Answer:
(c) $1 / 8$
Hint:
$
\frac{\gamma_{\mathrm{O}_2}}{\gamma_{\mathrm{H}_2}}=\sqrt{\frac{\mathrm{m}_{\mathrm{H}_2}}{\mathrm{~m}_{\mathrm{O}_2}}}=\sqrt{\frac{2}{32}}=\frac{1}{4}
$

$
\gamma_{\mathrm{O}_2}=\frac{1}{4} \gamma_{\mathrm{H}_2}
$
The fraction of oxygen that escapes in the time required for one half of the hydrogen to escape is $1 / 8$
Question 18
The variation of volume $\mathrm{V}$, with temperature $\mathrm{T}$, keeping pressure constant is called the coefficient of thermal expansion ie $\alpha=\frac{1}{V}\left(\frac{\partial V}{\partial T}\right)_P$ For an ideal gas $\alpha$ is equal to ..........
(a) $\mathrm{T}$
(b) $1 / T$
(c) $\mathrm{P}$
(d) none of these
Answer:
(b) $1 / \mathrm{T}$
Hint:
$
\begin{aligned}
& \frac{1}{V}\left(\frac{\partial V}{\partial T}\right)_P[\text { For ani deal gas } P V=n R T] \\
& =\frac{1}{V}\left(\frac{\partial\left(\frac{n R T}{P}\right)}{\partial T}\right)_P=\frac{n R}{P V}\left(\frac{\partial T}{\partial T}\right)=\frac{n R}{n R T}=\frac{1}{T}
\end{aligned}
$
Question 19.
Four gases $P, Q, R$ and $S$ have almost same values of ' $b$ ' but their ' $a$ ' values ( $a, b$ are Van der Waals Constants) are in the order $\mathrm{Q}<\mathrm{R}<\mathrm{S}<\mathrm{P}$. At a particular temperature, among the four gases the most easily liquefiable one is ...........
(a) $\mathrm{P}$
(b) $\mathrm{Q}$
(c) $\mathrm{R}$
(d) $\mathrm{S}$

Answer:
(a) $\mathrm{P}$
Hint:
Greater the ' $a$ ' value, casier the liquefaction
Question 20.
Maximum deviation from ideal gas is expected from 
(a) $\mathrm{CH}_{4(\mathrm{~g})}$
(b) $\mathrm{NH}_{3(\mathrm{~g})}$
(c) $\mathrm{H}_2$ (g)
(d) $\mathrm{N}_2(\mathrm{~g})$
Answer:
(b) $\mathrm{NH}_{3(\mathrm{~g})}$
Question 21.
The units of Van der Waals constants 'b' and ' $a$ ' respectively
(a) $\mathrm{mol} \mathrm{L}^{-1}$ and $\mathrm{L} \mathrm{atm}^2 \mathrm{~mol}^{-1}$
(b) $\mathrm{mol} \mathrm{L}$ and $\mathrm{L}$ atm $\mathrm{mol}^2$
(c) $\mathrm{mol}^{-1} \mathrm{~L}$ and $\mathrm{L}^2 \mathrm{~atm} \mathrm{~mol}^{-1}$
(d) none of these
Answer:

(c) $\mathrm{mol}^{-1} \mathrm{~L}$ and $\mathrm{L}^2$ atm $\mathrm{mol}^{-1}$
Hint:
$
\begin{aligned}
& \mathrm{an}^2 / \mathrm{V}^2 \mathrm{~atm} \\
& \mathrm{a}=\mathrm{atm} \mathrm{L}^2 / \mathrm{mol}^2=\mathrm{L}^2 \mathrm{~mol}^{-2} \text { atm } \\
& \mathrm{nb}=\mathrm{L} \\
& \mathrm{b}=\mathrm{L} / \mathrm{mol}=\mathrm{L} \mathrm{mol}^{-1}
\end{aligned}
$
Question 22.
Assertion : Critical temperature of $\mathrm{CO}_2$ is $304 \mathrm{~K}$, it can be liquefied above $304 \mathrm{~K}$.
Reason : For a given mass of gas, volume is to directly proportional to pressure at constant temperature.
(a) both assertion and reason arc true and reason is the correct explanation of assertion
(b) both assertion and reason are true but reason is not the correct explanation of assertion
(c) assertion is true but reason is false
(d) both assertion and reason are false
Answer:
(d) both assertion and reason are false
Hint:
Correct Statement: Critical temperature of $\mathrm{CO}_2$ is $304 \mathrm{~K}$. It means that $\mathrm{CO}_2$ cannot be liquefied above 304 $\mathrm{K}$, whatever the pressure may applied. Pressure is inversely proportional to volume.
Question 23.
What is the density of $\mathrm{N}$, gas at $227^{\circ} \mathrm{C}$ and $5.00 \mathrm{~atm}$ pressure? $\left(\mathrm{R}=0.082 \mathrm{~L} \mathrm{~atm} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right)$
(a) $1.40 \mathrm{~g} / \mathrm{L}$
(b) $2.81 \mathrm{~g} / \mathrm{L}$
(c) $3.41 \mathrm{~g} / \mathrm{L}$
(d) $0.29 \mathrm{~g} / \mathrm{L}$
Answer:
(c) $3.41 \mathrm{~g} / \mathrm{L}$
Hint:
Density $=\frac{\text { Mass }}{\text { Volume }}$

$
\begin{aligned}
& =\frac{\mathrm{m}}{\left(\frac{\mathrm{nRT}}{\mathrm{P}}\right)}=\left(\frac{\mathrm{m}}{\mathrm{n}}\right) \frac{\mathrm{P}}{\mathrm{RT}} \\
& =\text { Molar mass } \times \frac{\mathrm{P}}{\mathrm{RT}} \\
& =14 \mathrm{~g} \mathrm{~mol}^{-1} \\
& =\frac{5 \mathrm{~atm}}{0.082 \mathrm{~L} \mathrm{~atm} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} \times 500 \mathrm{~K}} \\
& =3.41 \mathrm{~g} \mathrm{~L}^{-1}
\end{aligned}
$
Question 24 .
Which of the following diagrams correctly describes the behaviour of a fixed mass of an ideal gas? ( $\mathrm{T}$ is measured in $\mathrm{K}$ )

(d) All of these
Answer:
(c)

For a fixed mass of an ideal gas $\mathrm{V} \alpha \mathrm{T}$
$\mathrm{P} \propto 1 / \mathrm{V}$
and $\mathrm{PV}=$ Constant
Question 25.
$25 \mathrm{~g}$ of each of the following gases are taken at $27^{\circ} \mathrm{C}$ and $600 \mathrm{~mm} \mathrm{Hg}$ pressure. Which of these will have the least volume?
(a) $\mathrm{HBr}$
(b) $\mathrm{HCl}$
(c) $\mathrm{HF}$
(d) $\mathrm{HI}$
Answer:
(d) $\mathrm{HI}$
Hint:
At a given temperature and pressure
Volume $\alpha$ number of moles
Volume $\alpha$ Mass / Molar mass
Volume $\alpha 28$ / Molar mass
i.e. if molar mass is more, volume is less. Hence $\mathrm{Hl}$ has the least volume.
II. Answer these questions briefly.

Question 26.
State Boyle's law.
Answer:
Boyle's law states that at a given temperature. the volume occupied by a fixed mass of a gas is inversely proportional to its pressure.
$\mathrm{V} \propto \frac{1}{P}$
where $T$ and $n$ are fixed or $\mathrm{PV}=$ Constant $=\mathrm{k}$
Question 27
A balloon filled with air at room temperature and cooled to a much lower temperalure can be used as a model for Charles' law.
Answer:
Charles' law:
- $\mathrm{VT}$ at constant $\mathrm{P}$ and $\mathrm{n}$ (or) $\frac{V}{T}=$ Constant
- A balloon filled with air at room temperature and cooled to a much lower temperature. the size of the balloon is reduced. Because if the temperature of the gas decreases, the volume also decreases in a direct proportion.
- When temperature is reduced, the gas molecules inside in over slower due to decreased temperature and hence the volume decreases.
Question 28.
Name two items that can serve as a model for Gay Lussac's law and explain.
Answer:
Gay Lussac's law:
1. $\mathrm{P} \alpha \mathrm{T}$ at constant volume (or) $=\frac{V}{T}$
2. Example - 1:
You fill the car type completely full of air on the hottest day of summer. The type cannot change it shape and volume. But when winter comes, the pressure inside the lyre is reduced and the shape is also reduced. This confirms that pressure and temperature are direct related to each other.
3. Example -2 :
The egg in the bottle experiment.

A glass bottle is taken, inside the bottle put some pieces of cotton with fire. Then place a boiled egg (shell removed) at the top of the bottle. The temperature inside the bottle increases from the fire, rising (he pressure. By scaling the bottle with egg, the fire goes on, dropping the temperature and pressure. This
causes the egg to be sucked into the bottle.
$\mathrm{P} \alpha \mathrm{T}$ is proved (or) $=\frac{P_1}{V_1}=\frac{P_2}{V_2}$
Question 29.
Give the mathematical expression that relates gas volume and moles. Describe in words what the mathematical expression means.
Answer:
1. The mathematical relationship betwêen the volume of a gas and the number of moles is $\mathrm{V} \alpha \mathrm{n}$
2. $\frac{V_1}{n_1}=\frac{V_2}{n_2}=$ Constant
Where $V_1$ and $n_1$ are the volume and number of moles of a gas and $V_2$ and $n_2$ are the values of volume and number of moles of same gas at a different set of conditions.
3. If the volume of the gas increase then the number of moles of the gas also increases.
4. At a certain temperature and pressure, the volume of a gas is dirctly proportional to the number of the moles of the gas.

Question 30 .
What are ideal gases? In what way real gases differ from ideal gases.
Answer:
1. Ideal gases are the gases that obey gas laws or gas equation $P V=n R$.
2. Real gases do not obey gas equation. $P V=n R T$.
3. The deviation of real gases from ideal behaviour is measure in terms of a ratio of $P V$ to $n R T$. This is termed as compression factor $(\mathrm{Z}) . \mathrm{Z}=\frac{P V}{n R T}$
4. For ideal gases $Z=1$.
5. For real gases $Z>1$ or $Z<1$. For example, at high pressure real gases have $Z>1$ and at intermediate pressure $Z<1$.
6. Above the Boyle point $Z>1$ for real gases and below the Boyle point, the real gases first show a decrease for $Z$, reaches a minimum and then increases with the increase in pressure.
7. So, it is clear that at low pressure and high temperature, the real gases behave as ideal gases.
Question 31.
Can a Van der Waals gas with a $=0$ be liquefied? Explain.
Answer:
- $a=0$ for a Van der Waals gas i.e. for a real gas. Van der Waals constant a $=0$. It cannot be liquefied.
- If a $=0$, there is a very less interaction between the molecules of gas.
- ' $a$ ' is the measure of strength of Van der Waals force of attraction between the molecules of the gas.
- If a is equal to zero, the Van der Waals force of attraction is very less and the gas cannot be liquefied.
Question 32.
Suppose there is a tiny sticky area on the wan of a container of gas. Molecules hitting this area stick there permanently. Is the pressure greater or less than on the ordinary area of walls?
Answer:
- Molecules hitting the tiny sticky area on the wall of the container of gas moves faster as they get closer to adhesive surface, but this effect is not permanent.
- The pressure on the sticky wall is greater than on the ordinary area of walls.

Question 33.
Explain the following observations
(a) Aerated water bottles are kept under water during summer
(b) Liquid ammonia bottle is cooled before opening the seal
(c) The type of an automobile is inflated to slightly lesser pressure in summer than in winter
(d) The size of a weather balloon becomes larger and larger as it ascends up into larger altitude

Answer:
(a) In aerated water bottles, $\mathrm{CO}_2$ gas is passed through the aqueous solution under pressure because the solubility of the gas in water is not very high. In summer, the solubility of the gas in water is likely to decrease because of the rise in temperature. Thus, in summer, more of gas will be present above the liquid surface in the glass bottle.

In case, the pressure of the gas becomes too high, the glass will not be able to withstand the pressure and the bottle may explode. To avoid this, the bottles are kept under water. As a result, the temperature is likely to decrease and the solubility of $\mathrm{CO}_2$ is likely to increase in aqueous solution resulting in decreased pressure.
(b) Liquid ammonia bottle contains the gas under very high pressure. If the bottle is opened as such, then the sudden decrease in pressure will lead to a large increase in volume of the gas. As a result, the gas will come out of the bottle all of a sudden with force. This will lead to the breakage of the bottle and also causes accident.
However, if the bottle is cooled under tap water for sometime, there will be a decrease in the volume of a gas to a large extent. if the seal is opened now, the gas will corne out of the bottle at a slower rate, reduces the chances of accident.
(c) The pressure of air is directly proportional to the temperature. Since the temperature is higher in summer than in higher, the pressure of the air in the tube of the lyre is likely to be quite high as compared to winter. It is quite likely that the tube may burst under high pressure in summer, Therefore, it is advisable to inflate the types to lesser pressure in summer than in winter.
(d) The volume of the gas is inversely proportional to pressure at a given temperature according to Boyle's law. As the weather balloon ascends, the pressure tends to decrease. As a result, the volume of the gas inside the balloon or the size of the balloon is likely to increase.

Question 34 .
Give suitable explanation for the following facts about gases.
(a) Gases don't settle at the bottom of a container
(b) Gases diffuse through all the space available to them and
(c) Explain with an increase in temperature
Answer:
(a) Gases by definition are the least dense state of matter. They have negligible intermolecular forces of attraction. So they are all free to roam separately. So the least dense gas particles will not sink at the bottom of a container.
(b) When a sample of a gas introduced to one part of a closed container, its molecules very quickly disperse throughout the container, this process by which molecules disperse in space in response to differences in concentration is called diffusion. For e.g., you can smell perfume in a room, because it difluses into the air totally inside the room.
(c) Diffusion is faster at higher temperature because the gas molecules have greater kinetic energy. Since heat increase the motion, then diffusion happens faster.
Question 35 .
Suggest why there is no hydrogen $\left(\mathrm{H}_2\right)$ in our atmosphere. Why does the moon have no atmosphere?
Answer:
1. Hydrogen is the lightest element thus when produced in free state, it rises above all the other gases to the top of the atmosphere, where it is open to cosmic storms and solar flares. There it literally leaks from the atmosphere to the empty space. Hydrogen easily gains velocity required to escape Earth's magnetic field. Hydrogen is very reactive in nature. So it would have reacted with $\mathrm{O}_2$, in its way to produce $\mathrm{H}_2 \mathrm{O}$. So majority portion of $\mathrm{H}_2$ reacts and very less amount of it present in the upper level of atmosphere and gains velocity to escape the atmosphere.
2. Moon has no atmosphere because the value of acceleration due to gravity ' $\mathrm{g}$ ' on the surface of the moon is small. Therefore, the value of escape velocity on the surface of the moon is very small. The molecule of the atmospheric gases on the surface of the moon have thermal velocities greater than the escape velocity. That's why all the molecules of gases have escaped and there is no atmosphere in the moon. The moon has insufficient gravity to retain an atmosphere. So we conclude that moon has no atmosphere.
Question 36.
Explain whether a gas approaches ideal behaviour or deviates from ideal behaviour if-
(a) it is compressed to a smaller volume at constant temperature
(b) the temperature is raised while keeping the volume constant
(c) more gas is introduced into the same volume and at the same temperature
Answer:
(a) it a gas is compressed to a smaller volume at constant temperature, pressure is increased. At high pressure with a smaller volume, the gas deviates from ideal behaviour.
(b) If a gas temperature is raised keeping the volume constant, the pressure of the gas will increase. At high pressure, the gas deviates from ideal behaviour.
(c) if more gas is introduced into the same volume and at the same temperature, the number of moles are increasing. if the volume remains same, the increased number of moles collide with each other and kinetic energy increases and pressure decreases. At increased pressure, the gas deviates from ideal behaviour.
Question 37.
Which of the following gases would you expect to deviate from jdeal behaviour under conditions of low temperature $\mathrm{F} \mathrm{Cl}_2$, or $\mathrm{Br}_2$ ? Explain.
Answer:
1. Bromine deviates $\left(\mathrm{Br}_2\right)$ from the ideal gas maximum than $\mathrm{Cl}_2$ and $\mathrm{F}_2$. Because $\mathrm{Br}_2$ has biggest size (atomic weight 79.9) provides maximum attraction between bromine molecules which is directly proportional to the size of the molecule and the boiling point of the liquid made from those molecules.

2. $\mathrm{Br}_2$ deviates from ideal behaviour because it has largest atomic radii compared to $\mathrm{Cl}_2$ and $\mathrm{F}_2$. So it contains more electrons than other two, and the Vander Waals forces are stronger in $\mathrm{Br}_2$ than in $\mathrm{Cl}_2$ and $\mathrm{F}_2$. So $\mathrm{Br}_2$ deviates from ideal behaviour.
Question 38 .
Distinguish between diffusion and effusion.
Answer:
Diffusion:
- Diffusion is the spreading of molecules of a substance throughout a space or a second substance.
- Diffusion refers to the ability of the gases to mix with each other.
- E.g.. Spreading of something such as brown tea liquid spreading through the water in a tea cup.
Effusion:
- Effusion is the escape of gas molecules through a very small hole in a membrane into an evacuated area.
- Effusion is a ability of a gas to travel through a small pin-hole.
- E.g., pouring out something like the soap studs bubbling out from a bucket of water.
Question 39.
Aerosol cans carry clear warning of heating of the can. Why?

Answer:
Aerosol cans carry clear warning of heating of the can. As the temperature rises, pressure in the can will increase and ambient temperatures about $120^{\circ} \mathrm{F}$ may lead to explosions. So aerosol cans should always be stored in dry areas where they will not be exposed to excessive temperatures. You should never throw an aerosol can onto a fire or leave it in the direct sunlight. even it is empty. This is because the pressure will build up so much that the can will burst. It is due to 2 reasons.
- The gas pressure increases.
- More of the liquefied propellant turns into a gas.
Question 40 .
When the driver of an automobile applies brake, the passengers are pushed toward the front of the car but a helium balloon is pushed toward back of the car. Upon forward acceleration the passengers are pushed toward the front of the car. Why?
Answer:
1. When the driver of an automobile applies brake, the passengers are pushed toward the front of the car due to inertia of the body, but a helium balloon pushed toward back of the car. Helium balloon responds to the air around it. Helium molecules are lighter than air of our atmosphere, and so they move toward back by gravity as a result of the accelerating frame.
2. Upon forward acceleration, the passengers arc pushed toward the front of the car, because the body in motion tends to stay in motion until acted upon by an outside force. Helium balloon is going to move opposite to this pseudo gravitational force.
Question 41.
Would it be easier to drink water with a straw on the top of Mount Everest?
Answer:
It is difficult to drink water with a straw on the top of Mount Everest. This is because the reduced atmospheric pressure is less effective in pushing water into the straw at the top of the mountain because gravity falls off gradually with height. The air pressure falls off, there isn't enough atmospheric pressure to push the water up in the straw all the way to the mouth.
Question 42.
Write the Van der Waals equation for a real gas. Explain the correction term for pressure and volume.
Answer:

Van der Waals equation of state for real gases is -
$
\left(P+\frac{a n^2}{V^2}\right)(V-n b)=\mathrm{nRT}
$
Correction term for pressure:
$\frac{\mathrm{an}^2}{\mathrm{~V}^2}$ is the pressure correction. It represents the intermolecular interaction that causes the non ideal
behaviour.
Correction term for Volume:
$\mathrm{V}-\mathrm{nb}$ is the volume correction. it is the effective volume occupied by real gas.
Question 43.
Derive the values of critical constants from the Van der Waals constants.
Answer:
Derivation of critical constants from the Van der Waals constants:
Van der Waals equation is,
$\left(P+\frac{a n^2}{V^2}\right)(V-n b)=\mathrm{nRT}$ for 1 mole
From this equation, the values of critical constant $\mathrm{P}_{\mathrm{C}} \mathrm{V}_{\mathrm{C}}$ and $\mathrm{T}_{\mathrm{C}}$ arc derived in terms of $a$ and $b$ the Van der Waals constants.
$
\left(P+\frac{a n^2}{V^2}\right)(V-b)=\mathrm{RT}
$

On expanding the equestion (1)
$
\mathrm{P} \mathrm{V}+\frac{a}{V}-\mathrm{pb}-\frac{\mathrm{ab}}{\mathrm{V}^2}-\mathrm{RT}=0
$
Multiplying eqestion (2) by $\frac{V^2}{P}$,
$
\begin{aligned}
& \frac{V^2}{P}\left(P V+\frac{a}{V}-P b-\frac{a b}{V^2}-R T\right)=0 \\
& V^3+\frac{a V}{P}-b V^2-\frac{a b}{P}-\frac{R T V^2}{P}=0
\end{aligned}
$
equation (3) is rearranged in the powers of $\mathrm{V}$
$
\mathrm{V}^3-\left[\frac{\mathrm{RT}}{\mathrm{P}}+\mathrm{b}\right] \mathrm{V}^2+\frac{a V}{P}-
$
The above equation (4) is an cubic equation of $\mathrm{V}$, which can have three roots. At the critical point. all the three values of $\mathrm{V}$ are equal to the critical volume $\mathrm{V}_{\mathrm{C}}$.
$
\begin{aligned}
& \text { i.e. } V=V_C \\
& \mathrm{~V}-\mathrm{V}_{\mathrm{C}}=\mathrm{O} \ldots \ldots \ldots(5) \\
& \left(\mathrm{V}-\mathrm{V}_{\mathrm{C}}\right)^3=\mathrm{O} \ldots \ldots \ldots(6) \\
& \left(\mathrm{V}^3-3 \mathrm{~V}_{\mathrm{C}} \mathrm{V}^2+3 \mathrm{~V}_{\mathrm{C}}{ }^3 \mathrm{~V}-\mathrm{V}_{\mathrm{C}}{ }^3=0\right.
\end{aligned}
$
As the equation (4) is identical with equation (7), comparing the ' $V$ ' ternis in (4) and (7),
$
\begin{aligned}
-3 \mathrm{~V}_{\mathrm{C}} \mathrm{V}^2 & =-\left[\frac{\mathrm{RT}_{\mathrm{C}}}{\mathrm{P}_{\mathrm{C}}}+\mathrm{b}\right] \mathrm{V}^2 \\
3 \mathrm{~V}_{\mathrm{C}} & =\mathrm{b}+\frac{\mathrm{RT}_{\mathrm{C}}}{\mathrm{P}_{\mathrm{C}}} \\
3 \mathrm{~V}_{\mathrm{C}}{ }^2 & =\frac{\mathrm{a}}{\mathrm{P}_{\mathrm{C}}} \\
\mathrm{V}_{\mathrm{C}}{ }^3 & =\frac{\mathrm{ab}}{\mathrm{P}_{\mathrm{C}}}
\end{aligned}
$

Divide equation (11) by (10)
$
\begin{aligned}
& \frac{\mathrm{V}_{\mathrm{C}}^3}{3 \mathrm{~V}_{\mathrm{C}}^2}=\frac{a b / \mathrm{P}_{\mathrm{C}}}{\mathrm{a} / \mathrm{P}_{\mathrm{C}}} \\
& \frac{\mathrm{V}_{\mathrm{C}}}{3}=\mathrm{b} \\
& \therefore \mathrm{V}_{\mathrm{C}}=3 \mathrm{~b} \ldots \ldots
\end{aligned}
$
When equation (12) is substituted in (10)
$
\begin{aligned}
& 3 \mathrm{~V}_{\mathrm{C}}^2=\frac{\mathrm{a}}{\mathrm{P}_{\mathrm{C}}} \\
& \mathrm{P}_{\mathrm{C}}=\frac{\mathrm{a}}{3 \mathrm{~V}_{\mathrm{C}}^2}=\frac{\mathrm{a}}{3(3 \mathrm{~b})^2}=\frac{\mathrm{a}}{3 \times 9 \mathrm{~b}^2}=\frac{\mathrm{a}}{27 \mathrm{~b}} \\
& \therefore \mathrm{P}_{\mathrm{C}}=\frac{\mathrm{a}}{27 \mathrm{~b}^2} \ldots \ldots \ldots \ldots(13)
\end{aligned}
$
substituting the values of $\mathrm{Vc}$ and $\mathrm{Pc}$ in equation (9)

$
\begin{aligned}
& 3 \mathrm{~V}_{\mathrm{C}}=\mathrm{b}+\frac{\mathrm{RT}_{\mathrm{C}}}{\mathrm{P}_{\mathrm{C}}} \\
& 3 \times 3 \mathrm{~b}=\mathrm{b}+\frac{\mathrm{RT}_{\mathrm{C}}}{\mathrm{a} / 27 \mathrm{~b}^2} \\
& 9 \mathrm{~b}-\mathrm{b}=\frac{\mathrm{RT}_{\mathrm{C}}}{\mathrm{a}} \times 27 \mathrm{~b}^2 \\
& 8 \mathrm{~b}=\frac{\mathrm{T}_{\mathrm{C}} \cdot \mathrm{R} 27 \mathrm{~b}^2}{\mathrm{a}} \\
& \therefore \mathrm{T}_{\mathrm{C}}=\frac{8 \mathrm{ab}}{27 \mathrm{Rb}^2}=\frac{8 \mathrm{a}}{27 \mathrm{Rb}} \\
& \mathrm{T}_{\mathrm{C}}=\frac{8 \mathrm{a}}{27 \mathrm{Rb}}
\end{aligned}
$
Critical constant $\mathrm{a}$ and $\mathrm{b}$ can be calculated using Van der Waals Constant as follows:
$
\begin{gathered}
\mathrm{a}=3 \mathrm{~V}_{\mathrm{C}}{ }^2 \mathrm{P}_{\mathrm{C}} \\
\mathrm{b}=\frac{\mathrm{V}_{\mathrm{C}}}{3}
\end{gathered}
$
Question 44
Why do astronauts have to wear protective suits when they are on the surface of moon?
Answer:
In space, there is no pressure, if we do wear a pressurised suit, our body will die. In space, we have to wear a pressurised suit, otherwise our body will continue to push out and blow up like a balloon. It would look cool, but we will be dead. So the astronauts in space must wear a pressurised suit (protective suits).
Question 45.
When ammonia combines with $\mathrm{HCl}, \mathrm{NH}_4 \mathrm{Cl}$ is formed as white dense fumes. Why do more funies appear near $\mathrm{HCl}$ ?
Answer:
1. When ammonia combines with $\mathrm{HCl}, \mathrm{NH}_4 \mathrm{Cl}$ is formed as white dense fumes. The reaction takes place in neutralization between a weak base and a strong acid.
2. The property of the gas is diffusion.
3. Diffusion of gases Ammonia and hydrogen chloride. Concentrated ammonia solution is placed on a pad in one end of a tube and concentrated $\mathrm{HCl}$ on the pad at the other. After about a minute, the gases diffuses far enough to meet and a ring of solid ammonium chloride is formed near the $\mathrm{HCl}$ end.
Question 46.
A sample of gas at $15^{\circ} \mathrm{C}$ at 1 atm has a volume of $2.58 \mathrm{dm}^3$. Vhen the temperature is raised to $38^{\circ} \mathrm{C}$ at $\mathrm{I}$ atm does the volume of the gas increase? if so, calculate the final volume.
Answer:
$
\begin{aligned}
& \mathrm{T}_1=15^{\circ} \mathrm{C}+273 \quad \mathrm{~T}_2=38^{\circ} \mathrm{C}+273 \\
& \mathrm{~T}_1=288 \mathrm{~K} \quad \mathrm{~T}_2=311 \mathrm{~K} \\
& \mathrm{~V}_1=2.58 \mathrm{dm}^3 \quad \mathrm{~V}_2=\text { ? } \\
& (\mathrm{P}=1 \text { atom constant }) \\
& \frac{V_1}{T_1}=\frac{V_2}{T_2} \\
& \mathrm{~V}_2=\left(\frac{\mathrm{V}_1}{\mathrm{~T}_1}\right) \times \mathrm{T}_2 \\
& =\frac{2.58 \mathrm{dm}^3}{288 \mathrm{~K}} \times 311 \mathrm{~K} \\
& \mathrm{~V}_2=2.78 \mathrm{dm}^3 \text { i.e. volume increased from } 2.58 \mathrm{dm}^3 \text { to } 2.78 \mathrm{dm}^3 \text {. } \\
&
\end{aligned}
$

Question 47.
A sample of gas has a volume of $8.5 \mathrm{dm}^3$ at an unknown temperature. When the sample is submerged in ice water at $0^{\circ} \mathrm{C}$, its volume gets reduced to $6.37 \mathrm{dm}^3$. What $i$ s its initial temperature?
Answer:
$
\begin{array}{ll}
\mathrm{V}_1=8.5 \mathrm{dm}^3 & \mathrm{~V}_2=6.37 \mathrm{dm}^3 \\
\mathrm{~T}_1=? & \mathrm{~T}_2=0^{\circ} \mathrm{C}=273 \mathrm{~K} \\
\frac{\mathrm{V}_1}{\mathrm{~T}_1}=\frac{\mathrm{V}_2}{\mathrm{~T}_2} \\
\mathrm{~V}_1 \times\left(\frac{\mathrm{T}_2}{\mathrm{~V}_2}\right)=\mathrm{T}_1 \\
\mathrm{~T}_1=8.5 \mathrm{dm} \times \frac{273 \mathrm{~K}}{6.37 \mathrm{dm}^3} \\
\mathrm{~T}_1=364.28 \mathrm{~K}
\end{array}
$
Question 48.
Of two samples of nitrogen gas, sample A contains 1.5 moles of nitrogen in a vessel of volume of 37.6
$\mathrm{dm}^3$ at $298 \mathrm{~K}$, and the sample $\mathrm{B}$ is in a vessel of volume $16.5 \mathrm{dm}^3$ at $298 \mathrm{~K}$. Calculate the number of moles in sample $B$.

Answer:
$
\begin{aligned}
& \mathrm{n}_{\mathrm{A}}=1.5 \mathrm{~mol} \mathrm{n}_{\mathrm{B}}=? \\
& \mathrm{~V}_{\mathrm{A}}=37.6 \mathrm{dm}^3 \mathrm{~V}_{\mathrm{B}}=16.5 \mathrm{dm}^3 \\
& (\mathrm{~T}=298 \mathrm{~K} \text { constant }) \\
& \frac{\mathrm{V}_{\mathrm{A}}}{\mathrm{n}_{\mathrm{A}}}=\frac{\mathrm{V}_{\mathrm{B}}}{\mathrm{n}_{\mathrm{B}}} \\
& \mathrm{n}_{\mathrm{A}}=\left(\frac{\mathrm{n}_{\mathrm{A}}}{\mathrm{V}_{\mathrm{A}}}\right) \mathrm{V}_{\mathrm{B}} \\
& =\frac{1.5 \mathrm{~mol}}{37.6 \mathrm{dm}^3} \times 16.5 \mathrm{dm}^3
\end{aligned}
$
Question 49.
Sulphur hexafluoride is a colourless, odourless gas; calculate the pressure exerted by 1.82 moles of the gas in a steel vessel of volume $5.43 \mathrm{dm}^3$ at $69.5^{\circ} \mathrm{C}$, assuming jdeal gas behaviour.
Answer:
$
\begin{aligned}
& \mathrm{n}=1.82 \mathrm{~mole} \\
& \mathrm{~V}=5.43 \mathrm{dm}^3 \\
& \mathrm{~T}=69.5+273=342.5 \\
& \mathrm{P}=? \\
& \mathrm{PV}=\mathrm{nRT} \\
& \mathrm{P}=\frac{n R T}{V} \\
& \mathrm{P}=\frac{1.82 \mathrm{~mol} \times 0.821 \mathrm{dm}^3 \mathrm{~atm} \mathrm{~mol}^{-1} \mathrm{~K}^{-1} \times 342.5 \mathrm{~K}}{5.43 \mathrm{dm}^3} \\
& \mathrm{P}=94.25 \mathrm{~atm} .
\end{aligned}
$
Question 50 .
Argon is an inert gas used in light bulbs to retard the vapourlzation of the tungsten filament. A certain light bulb containing argon at $1.2 \mathrm{~atm}$ and $18^{\circ} \mathrm{C}$ is heated to $85^{\circ} \mathrm{C}$ at constant volume. Calculate its final pressure in atm.

Answer:
$
\begin{aligned}
\mathrm{P}_1 & =1.2 \mathrm{~atm} \\
\mathrm{~T}_1 & =18^{\circ} \mathrm{C}+273=291 \mathrm{~K} \\
\mathrm{~T}_2 & =85^{\circ} \mathrm{C}+273=358 \mathrm{~K} \\
\mathrm{P}_2 & =? \\
\frac{P_1}{T_1} & =\frac{P_2}{T_2} \\
\mathrm{P}_2 & =\left(\frac{\mathrm{P}_1}{\mathrm{~T}_1}\right) \times \mathrm{T}_2 \\
& =\frac{1.2 \mathrm{~atm}}{291 \mathrm{~K}} \times 358 \mathrm{~K}
\end{aligned}
$

Question 51.
A small bubble rises from the bottom of a lake, where the temperature and pressure are $6^{\circ} \mathrm{C}$ and 4 atm. to the water surface, where the temperature is $25^{\circ} \mathrm{C}$ and pressure is $1 \mathrm{~atm}$. Calculate the final volume in (mL) of the bubble, if its initial volume is $1.5 \mathrm{~mL}$.
Answer:
$
\begin{aligned}
& \mathrm{T}_1=6^{\circ} \mathrm{C}+273=279 \mathrm{~K} \\
& \mathrm{P}_1=4 \mathrm{~atm} \mathrm{~V}_1=1.5 \mathrm{~m} \\
& \mathrm{~T}_2=25^{\circ} \mathrm{C}+273=298 \mathrm{~K} \\
& \mathrm{P}_2=1 \mathrm{~atm} \mathrm{~V}_1=? \\
& \frac{P_1 V_1}{T_1}=\frac{P_2 V_2}{T_2} \\
& \mathrm{~V}_2=\frac{\mathrm{P}_1 \mathrm{~V}_1}{\mathrm{~T}_1} \times \frac{\mathrm{T}_2}{\mathrm{P}_2} \\
& =\frac{4 \mathrm{~atm} \times 1.5 \mathrm{ml} \times 298 \mathrm{~K}}{279 \mathrm{~K} \times 1 \mathrm{~atm}} \\
& \mathrm{~V}_2=6.41 \mathrm{~mol}
\end{aligned}
$
Question 52.
Hydrochloric acid is treated with a metal to produce hydrogen gas. Suppose a student carries out this reaction and collects a volume of $154.4 \times 10^{-3} \mathrm{dm}^3$ of a gas at a pressure of $742 \mathrm{~mm}$ of $\mathrm{Hg}$ at a temperature of $298 \mathrm{~K}$. What mass of hydrogen gas (in $\mathrm{mg}$ ) did the student collect?
Answer:
$
\begin{aligned}
& =0.006 \mathrm{~mol} \\
& \mathrm{n}=\frac{P V}{R T} \\
& \mathrm{n}=\frac{\text { Mass }}{\text { MolarMass }} \\
& \text { Mass }=\mathrm{n} \times \text { Molar mass } \\
& =0.006 \times 2.016 \\
& =0.0121 \mathrm{~g}=12.1 \mathrm{mg} \text {. } \\
&
\end{aligned}
$
Question 53 .
It takes $192 \mathrm{sec}$ for an unknown gas to diffuse through a porous wall and $84 \mathrm{sec}$ for $\mathrm{N} 2$ gas to effuse at the same temperature and pressure. What is the molar mass of the unknown gas?

Answer:
$
\begin{aligned}
& \frac{\gamma_{\text {unknown }}}{\gamma_{\mathrm{N}_2}}=\frac{\mathrm{t}_{\mathrm{N}_2}}{\mathrm{t}_{\text {unknown }}}=\sqrt{\frac{\mathrm{m}_{\mathrm{N}_2}}{\mathrm{~m}_{\text {unknown }}}} \\
& \frac{84 \mathrm{sec}}{192 \mathrm{sec}}=\sqrt{\frac{14 \mathrm{~g} \mathrm{~mol}^{-1}}{\mathrm{~m}_{\text {unknown }}}} \\
& =\frac{14 \mathrm{~g} \mathrm{~mol}^{-1}}{\mathrm{~m}_{\text {unknown }}}
\end{aligned}
$
$
\begin{aligned}
m_{\text {unknown }} & =14 \mathrm{~g} \mathrm{~mol}^{-1} \times\left(\frac{192 \mathrm{sec}}{84 \mathrm{sec}}\right)^2 \\
\mathrm{~m}_{\text {unknown }} & =73.14 \mathrm{~g} \mathrm{~mol}^{-1}
\end{aligned}
$
Question 54
A tank contains a mixture of $52.5 \mathrm{~g}$ of oxygen and $65.1 \mathrm{~g}$ of $\mathrm{CO} 2$ at $300 \mathrm{~K}$ the total pressure in the tank is $9.21 \mathrm{~atm}$. Calculate the partial pressure (in atm.) of each gas in the mixture.
Answer:
$
\begin{aligned}
& \mathrm{m}_{\mathrm{O}_2}=52.5 \mathrm{~g} \\
& \mathrm{P}_{\mathrm{O}_2}=\text { ? } \\
& \mathrm{m}_{\mathrm{CO}_2}=65.1 \mathrm{~g} \\
& \mathrm{P}_{\mathrm{CO}_2}=\text { ? } \\
& \mathrm{T}=300 \mathrm{~K} \mathrm{P}=9.21 \mathrm{~atm} \\
& \mathrm{P}_{\mathrm{O}_2}=\mathrm{X}_{\mathrm{O}_2} \mathrm{x} \text { total pressure } \\
& \mathrm{X}_{\mathrm{O}_2}=\frac{\mathrm{n}_{\mathrm{O}_2}}{\mathrm{n}_{\mathrm{O}_2}+\mathrm{n}_{\mathrm{CO}_2}} \\
& \mathrm{n}_{\mathrm{O}_2}=\frac{\text { Mass of } \mathrm{O}_2}{\text { Molar mass of } \mathrm{O}_2} \\
& =\frac{52.5 \mathrm{~g}}{32 \mathrm{~g} \mathrm{~mol}^{-1}}=1.64 \mathrm{~mol} \\
& \mathrm{n}_{\mathrm{CO}_2}=\frac{\text { Mass of } \mathrm{CO}_2}{\text { Molar mass of } \mathrm{CO}_2} \\
& =\frac{65.1 \mathrm{~g}}{44 \mathrm{~g} \mathrm{~mol}^{-1}}=1.48 \mathrm{~mol} \\
& \mathrm{X}_{\mathrm{O}_2}=\frac{\mathrm{n}_{\mathrm{O}_2}}{\mathrm{n}_{\mathrm{O}_2}+\mathrm{n}_{\mathrm{CO}_2}}=\frac{1.64}{3.12}=0.53 \\
&
\end{aligned}
$

$
\mathrm{X}_{\mathrm{CO}_2}=\frac{\mathrm{n}_{\mathrm{CO}_2}}{\mathrm{n}_{\mathrm{O}_2}+\mathrm{n}_{\mathrm{CO}_2}}=\frac{1.48}{3.12}=0.47
$
$
\begin{aligned}
& \mathrm{P}_{\mathrm{O}_2}=\mathrm{X}_{\mathrm{O}_2} \times \text { Total pressure } \\
& =0.53 \times 9.21 \mathrm{~atm}=4.88 \mathrm{~atm} \\
& \mathrm{P}_{\mathrm{CO}_2}=\mathrm{X}_{\mathrm{CO}_2} \times \text { Total pressure } \\
& =0.47 \times 9.21 \mathrm{~atm}=4.33 \mathrm{~atm}
\end{aligned}
$
Question 55.
A combustible gas Is stored in a metal tank at a pressure of $2.98 \mathrm{~atm}$ at $25^{\circ} \mathrm{C}$. The tank can withstand a maximum pressure of $12 \mathrm{~atm}$ after which it will explode. The building in which the tank has been stored catches fire. Now predict whether the tank will blow up first or start melting? (Melting point of the metal = $1100 \mathrm{~K})$
Answer:
$
\begin{aligned}
& \mathrm{T}_1=298 \mathrm{~K} \\
& \mathrm{P}_1=2.98 \mathrm{~atm} ; \\
& \mathrm{T}_2=1100 \mathrm{~K} \\
& \mathrm{P}_2=? \\
& \frac{P_1}{T_1}=\frac{P_2}{T_2} \\
& \mathrm{P}_2=\frac{P_1}{T_1} \times T_2 \\
& =\frac{2.98 \mathrm{~atm}}{298 \mathrm{~K}} \times 1100 \mathrm{~K}=11 \mathrm{~atm}
\end{aligned}
$
At $1100 \mathrm{~K}$, the pressure of the gas inside the tank will become $11 \mathrm{~atm}$. Given that tank can withstand a maximum pressure of $12 \mathrm{~atm}$, the tank will start melting first.
In Text Questions - Evaluate Yourself
Question 1.
Freon-I 2, the compound widely used in the refrigerator system as coolant causes depletion of ozone layer. Now It has been replaced by eco-friendly compounds. Consider $1.5 \mathrm{dm}^3$ sample of gaseous freon at a pressure of $0.3 \mathrm{~atm}$. If the pressure is changed to $1.2 \mathrm{~atm}$. at a constant temperature, what will be the volume of the gas increased or decreased?
Answer:
Volume of freon $\left(\mathrm{V}_1\right)=1.5 \mathrm{dm}^3$

Pressure $\left(\mathrm{P}_1\right)=0.3 \mathrm{~atm}$
' $T$ ' is constant
$\mathrm{P}_2=1.2 \mathrm{~atm}$
$\mathrm{V}_2=$ ?
$\mathrm{P}_1 \mathrm{~V}_1=\mathrm{P}_2 \mathrm{~V}_2$
$\mathrm{V}_2=\frac{P_1 V_1}{P_2}$
$\mathrm{V}_2=\frac{\mathrm{P}_1 \mathrm{~V}_1}{\mathrm{P}_2}$
$=\frac{0.3 \mathrm{~atm} \times 1.5 \mathrm{dm}^3}{1.2 \mathrm{~atm}}$
$=0.375 \mathrm{dm}^3$
Volume decreased from $1.5 \mathrm{dm}^3$ to $0.375 \mathrm{dm}^3$
Question 2.
Inside a certain automobile engine, the volume of air in a cylinder is $0.375 \mathrm{dm}^3$, when the pressure is 1.05 $\mathrm{atm}$. When the gas is compressed to a volume of $0.125 \mathrm{dm}^3$ at the same temperature, what is the pressure of the compressed air?
Answer:
$
\begin{aligned}
& \mathrm{V}_1=0.375 \mathrm{dm}^3 \\
& \mathrm{~V}_2=0.125 \mathrm{dm}^3 \\
& \mathrm{P}_1=1.05 \mathrm{~atm} \\
& \mathrm{P}_2=? \\
& \mathrm{~T}-\text { Constant } \\
& \mathrm{P}_1 \mathrm{~V}_1=\mathrm{P}_2 \mathrm{~V}_2 \\
& \mathrm{P}_2=\frac{P_1 V_1}{V_2}=\frac{10.5 \times 0.375}{0.125} \\
& =3.15 \mathrm{~atm}
\end{aligned}
$

Question 3.
A sample of gas has a volume of $3.8 \mathrm{dm}^3$ at an unknown temperature. When the sample is submerged in ice water at $0^{\circ} \mathrm{C}$, its volume gets reduced to $2.27 \mathrm{dm}^3$. What is its initial temperature?
Answer:
$
\begin{aligned}
& \mathrm{V}_1=3.8 \mathrm{dm}^3 \\
& \mathrm{~T}_2=0^{\circ} \mathrm{C}=273 \mathrm{~K} \\
& \mathrm{~T}_1=? \mathrm{~V}_2=2.27 \mathrm{dm}^3 \\
& \frac{V_1}{T_1}=\frac{V_2}{T_2} \\
& \mathrm{~T}_1=\left(\frac{\mathrm{T}_2}{\mathrm{~V}_2}\right) \times \mathrm{V}_1=\frac{273 \mathrm{~K}}{2.27 \mathrm{dm}^3} \times 3.8 \mathrm{dm}^3 \\
& \mathrm{~T}_1=457 \mathrm{~K}
\end{aligned}
$
Question 4.
An athlete in a kinesiology research study has his lung volume of $7.05 \mathrm{dm}^3$ during a deep inhalation. At this volume the lungs contain 0.312 mole of air. During exhalation the volume of his Jung decreases to $2.35 \mathrm{dm}^3$ How many moles of air does the athlete exhale during exhalation? (assume pressure and temperature remain constant)
Answer:
$
\begin{aligned}
& \mathrm{V}_1=7.05 \mathrm{dm}^3 \\
& \mathrm{~V}_2=2.35 \mathrm{dm}^3 \\
& \mathrm{n}_1=0.312 \mathrm{~mol} \\
& \mathrm{n}_1=?
\end{aligned}
$
' $\mathrm{P}$ ' and ' $\mathrm{T}$ ' are constant
$
\frac{V_1}{n_1}=\frac{V_2}{n_2}
$

$
\begin{aligned}
& \mathrm{n}_2=\left(\frac{\mathrm{n}_1}{\mathrm{~V}_1}\right) \times \mathrm{V}_2 \\
& \mathrm{n}_2=\frac{0.312 \mathrm{~mol}}{7.05 \mathrm{dm}^3} \times 2.35 \mathrm{dm}^5 \\
& \mathrm{n}_2=0.104 \mathrm{~mol}
\end{aligned}
$
Number of moles exhaled $=0.312-0.104=0.208$ moles
Question 5.
A small bubble rises from the bottom of a lake, where the temperature and pressure are $8^{\circ} \mathrm{C}$ and $6.4 \mathrm{~atm}$. to the water surface, where the temperature is $25^{\circ} \mathrm{C}$ and pressure is $1 \mathrm{~atm}$. Calculate the final volume in (ml) of the bubble, if its initial volume is $2.1 \mathrm{ml}$.
Answer:
$
\begin{aligned}
& \mathrm{T}_1=8^{\circ} \mathrm{C}=8+273=281 \mathrm{~K} \\
& \mathrm{P}_1=6.4 \mathrm{~atm} \mathrm{~V}_1=2.1 \mathrm{~mol} \\
& \mathrm{~T}_2=25^{\circ} \mathrm{C}=25+273=298 \mathrm{~K} \\
& \mathrm{P}_2=1 \mathrm{~atm} \\
& \mathrm{~V}_2=? \\
& \frac{P_1 V_1}{T_1}=\frac{P_2 V_2}{T_2} \\
& \mathrm{~V}_2=\left(\frac{\mathrm{P}_1 \mathrm{~V}_1}{\mathrm{~T}_1}\right) \times \frac{\mathrm{T}_2}{\mathrm{P}_2} \\
& =\frac{6.4 \mathrm{~atm} \times 2.1 \mathrm{~mol}}{281 \mathrm{~K}} \times \frac{298 \mathrm{~K}}{1 \mathrm{~atm}} \\
& \mathrm{~V}_2=14.25 \mathrm{ml}
\end{aligned}
$
Question 6.
(a) $\mathrm{A}$ mixture of $\mathrm{He}$ and $\mathrm{O}_2$ were used in the 'air' tanks of underwater divers for deep dives. For a particular dive $12 \mathrm{dm}^3$ of $\mathrm{O}_2$ at $298 \mathrm{~K}$, I atm. and $46 \mathrm{dm} 3$ of He, at $298 \mathrm{~K}, 1$ aim. were both pumped into a $5 \mathrm{dm}^3$ tank. Calculate the partial pressure of each gas and the total pressure In the tank at $298 \mathrm{~K}$
Answer:
$
\begin{array}{ll}
\mathrm{V}_{\mathrm{O}_2}=12 \mathrm{dm}^3 & \mathrm{~T}=298 \\
\mathrm{~V}_{\mathrm{He}}=40 \mathrm{dm} & \mathrm{P}=1 \mathrm{~atm} \\
\mathrm{P}=1 \mathrm{~atm} \\
\mathrm{~V}_{\text {total }}=5 \mathrm{dm}^3 & \\
\mathrm{P}_{\mathrm{O}_2}=\mathrm{X}_{\mathrm{O}_2} \times \mathrm{P}_{\text {total }} &
\end{array}
$

$
\begin{aligned}
& \mathrm{P}_{\text {total }} \times \mathrm{V}_{\text {total }}=1 \mathrm{~atm} \times 22.4 \mathrm{l} \\
& \therefore \mathrm{P}_{\text {total }}=\frac{1 \mathrm{~atm} \times 22.4 \mu}{5 /} \\
& \mathrm{P}_{\text {total }}=4.48 \mathrm{~atm} \\
& \therefore \mathrm{P}_{\mathrm{O}_2}=0.21 \times 4.48 \mathrm{~atm} \\
& =0.94 \mathrm{~atm} \\
& \therefore \mathrm{P}_{\mathrm{He}}=\mathrm{X}_{\mathrm{He}} \times \mathrm{P}_{\text {total }} \\
& \mathrm{X}_{\mathrm{He}}=\frac{\mathrm{n}_{\mathrm{He}}}{\mathrm{n}_{\mathrm{O}_2}+\mathrm{n}_{\mathrm{He}}} \\
& =\frac{2.05}{0.54+2.05} \\
& \mathrm{X}_{\mathrm{He}}=\frac{2.05}{2.59}=0.79 \\
& \mathrm{P}_{\mathrm{He}}=3.54 \mathrm{~atm}
\end{aligned}
$
Question 6.
(b) A sample of solid $\mathrm{KClO}_3$ (potassium chlorate) was heated in a test tube to obtain $\mathrm{O}_2$ according to the reaction $2 \mathrm{KClO}_3 \rightarrow 2 \mathrm{KCl}_{(\mathrm{s})}+3 \mathrm{O}_2$ The oxygen gas was collected by downward displacement of water at $295 \mathrm{~K}$. The total pressure of the mixture is $772 \mathrm{~mm}$ of $\mathrm{Hg}$. The vapour pressure of water is $26.7 \mathrm{mn}$ of $\mathrm{Hg}$ at $300 \mathrm{~K}$. What is the partial pressure of the oxygen gas?
Answer:
$
\begin{aligned}
& 2 \mathrm{KCl}_{3(\mathrm{~s})} \rightarrow 2 \mathrm{KCl}_{(\mathrm{s})} 3 \mathrm{O}_{3(\mathrm{~g})} \\
& \mathrm{P}_{\text {total }}=772 \mathrm{~mm} \mathrm{Hg} \\
& \mathrm{P}_{\mathrm{H}_2 \mathrm{O}}=26.7 \mathrm{~mm} \mathrm{Hg} \\
& \mathrm{P}_{\text {total }}=\mathrm{P}_{\mathrm{O}_2}+\mathrm{P}_{\mathrm{H}_2 \mathrm{O}} \\
& \mathrm{P}_{\mathrm{O}_2}=\mathrm{P}_{\text {total }}-\mathrm{P}_{\mathrm{H}_2 \mathrm{O}} \\
& \mathrm{P}_1=26.7 \mathrm{~mm} \mathrm{Hg} \\
& \mathrm{T}_1=300 \mathrm{~K} \\
& \mathrm{~T}_2=295 \mathrm{~K} \\
& \mathrm{P}_2=?
\end{aligned}
$

$
\begin{aligned}
& \frac{P_1}{T_1}=\frac{P_2}{T_2} \\
& \mathrm{P}_2=\left(\frac{\mathrm{P}_1}{\mathrm{~T}_1}\right) \mathrm{T}_2=\frac{26.7 \mathrm{~mm} \mathrm{Hg}}{300 \mathrm{~K}} \times 295 \mathrm{~K} \\
& \mathrm{P}_2=26.26 \mathrm{~mm} \mathrm{Hg} \\
& \therefore \mathrm{P}_{\mathrm{O}_2}=772-26.26 \\
& =745.74 \mathrm{~mm} \mathrm{Hg}
\end{aligned}
$
Question 7.
A flammable hydrocarbon gas of particular volume is found to diffuse through a small hole in 1.5 minutes. Under the same conditions of temperature and pressure an equal volume of bromine vapour takes $4.73 \mathrm{~min}$ to diffuse through the same hole. Calculate the molar mass of the unknown gas and suggest what this gas might be, (Given that molar mass of bromine $=159.8 \mathrm{~g} / \mathrm{mole}$ )
Answer:
$
\begin{aligned}
& \frac{\gamma_{\text {Hydrocarbon }}}{\gamma_{\text {Bromine }}}=\frac{t_{\text {Bromine }}}{\mathrm{t}_{\text {Hydrocarbon }}} \\
& \text { ( } \because \text { Volume constant) } \\
& =\frac{4.73 \text { minutes }}{1.5 \text { minutes }} \\
& =3.15 \\
& \frac{\gamma_{\text {Hydrocarbon }}}{\gamma_{\text {Bromine }}}=\sqrt{\frac{m_{\text {Bromine }}}{\mathrm{m}_{\text {Hydrocarbon }}}} \\
& 3.15=\sqrt{\frac{159.8 \mathrm{~g} \mathrm{~mol}^{-1}}{\mathrm{~m}_{\text {Hydrocarbon }}}} \\
& =3.1 \text { : } \\
&
\end{aligned}
$
Squaring on both sides and rearranging,
$
\begin{aligned}
& \mathrm{m}_{\text {Hydrocarbon }}=\frac{159.8 \mathrm{~g} \mathrm{~mol}^{-1}}{(3.15)^2} \\
& \mathrm{~m}_{\text {Hydrocarbon }}=16.1 \mathrm{~g} \mathrm{~mol}^{-1} \\
& \mathrm{n}(12)+(2 \mathrm{n}+2) 1=16\left(\text { general formula for hydrocarbon } \mathrm{C}_{\mathrm{n}} \mathrm{H}_{2 \mathrm{n}+2}\right) \\
& 12 \mathrm{n}+2 \mathrm{n}+2=16 \\
& 14 \mathrm{n}=16-2 \\
& 14 \mathrm{n}=14 \\
& \mathrm{n}=1
\end{aligned}
$
The hydrocarbon is $\mathrm{C}_1 \mathrm{H}_{2(1)}+2=\mathrm{CH}_4$

Question 8 .
Critical temperature of $\mathrm{H}_2 \mathrm{O}, \mathrm{NH}_3$ and $\mathrm{CO}_2$ are $647.4,405.5$ and $304.2 \mathrm{~K}$, respectively. When we start cooling from a temperature of $700 \mathrm{~K}$ which will liquefy first and which will liquefy finally?
Answer:
Critical temperature of a gas is defined as the temperature above which it cannot be liquefied even at high pressures.
$\therefore$ When cooling starts from $700 \mathrm{~K}, \mathrm{H}_2 \mathrm{O}$ vill liquefied first, then followed by ammonia and finally carbon dioxide will liquefied.
In-Text Example Problems
Question 9.
In the below figure, let us find the missing parameters [volume in (b) and pressure in (c)] $\mathrm{P}_1=1 \mathrm{~atm} \mathrm{P}_2=2 \mathrm{~atm} \mathrm{P}_3=$ ? atm
$
\begin{aligned}
& \mathrm{V}_1 \mathrm{Idm} 3 \mathrm{~V}_2=? \operatorname{dm} 3 \mathrm{~V}_3=0.25 \mathrm{dm}^3 \\
& \mathrm{~T}_1=298 \mathrm{~K} \mathrm{~T}_2=298 \mathrm{~K} \mathrm{~T}_3=298 \mathrm{~K}
\end{aligned}
$
Solution:
According to Boyle's law, at constant temperature for a given mass of gas at constant temperature,

$
\begin{aligned}
& \mathrm{P}_1 \mathrm{~V}_1=\mathrm{P}_2 \mathrm{~V}_2=\mathrm{P}_3 \mathrm{~V}_3 \\
& \mathrm{I} \mathrm{atm} \times 1 \mathrm{dm}^3=2 \mathrm{~atm} \times \mathrm{V}_2=\mathrm{P}_3 \times 0.25 \mathrm{dm}^3 \\
& \therefore 2 \mathrm{~atm} \times \mathrm{V}_2=1 \mathrm{~atm} \times 1 \mathrm{dm}^3 \\
& \mathrm{~V}_2=\frac{1 \mathrm{~atm} \times 1 \mathrm{dm}^3}{2 \mathrm{~atm}} \\
& \mathrm{~V}_2=0.5 \mathrm{dm}^3 \\
& \text { and } \mathrm{P}_3 \times 0.25 \mathrm{dm}^3=1 \mathrm{~atm} \times 1 \mathrm{dm}^3 \\
& \mathrm{P}_3=\frac{1 \mathrm{~atm} \times 1 \mathrm{dm}^3}{0.25 \mathrm{dm}^3} \\
& \mathrm{P}_3=4 \mathrm{~atm}
\end{aligned}
$
Question 10 .
In the below figure, let us find the missing parameters [volume in (b) and temperature in (c)]
$
\begin{aligned}
& \mathrm{P}_1=1 \mathrm{~atm} \mathrm{P}_2=1 \mathrm{~atm} \mathrm{P}_3=1 \mathrm{~atm} \\
& \mathrm{~V}_1=0.3 \mathrm{dm}^3 \mathrm{~V}_2=? \mathrm{dm}^3 \mathrm{~V}_3=0.15 \mathrm{dm}^3 \\
& \mathrm{~T}_1=200 \mathrm{~K} \mathrm{~T}_2=300 \mathrm{~K} \mathrm{~T}_3=? \mathrm{~K}
\end{aligned}
$
Answer:

According to Charles law,
$
\begin{aligned}
& \mathrm{T}_3=100 \mathrm{k} \\
&
\end{aligned}
$
Question 11.
Calculate the pressure exerted by 2 moles of sulphur hexafluoride in a steel vessel of volume $6 \mathrm{dm}^3$ at $70^{\circ} \mathrm{C}$ assuming It is an ideal gas.
Answer:
We will use the ideal gas equation for this calculation as below:
$
\begin{aligned}
& P=\frac{n R T}{V} \\
& =\frac{2 \mathrm{~mol} \times 0.0821 \mathrm{~L} \mathrm{~atm} \cdot \mathrm{K}^{-1} \cdot \mathrm{mol}^{-1} \times(70+273 \mathrm{~K})}{6 \mathrm{dm}^3} \\
& =9.39 \mathrm{~atm} .
\end{aligned}
$
Question 12.
A mixture of gases contains 4.76 mole of Ne, 0.74 mole of Ar and 2.5 mole of Xe. Calculate the partial pressure of gases, if the total pressure is $2 \mathrm{~atm}$. at a fixed temperature. Solve this problem using Dalton's law.
Answer:
$
\mathrm{P}_{\mathrm{Ne}}=\mathrm{X}_{\mathrm{Ne}} \mathrm{P}_{\mathrm{Total}}
$

$
\begin{aligned}
& \mathrm{X}_{\mathrm{Ne}}=\frac{\mathrm{n}_{\mathrm{Ne}}}{\mathrm{n}_{\mathrm{Ne}}+\mathrm{n}_{\mathrm{Ar}}+\mathrm{n}_{\mathrm{Xe}}} \\
& =\frac{4.76}{4.76+0.74+2.5}=0.595 \\
& \mathrm{X}_{\mathrm{Ar}}=\frac{\mathrm{n}_{\mathrm{Ar}}}{\mathrm{n}_{\mathrm{Ne}}+\mathrm{n}_{\mathrm{Ar}}+\mathrm{n}_{\mathrm{Xe}}} \\
& =\frac{0.74}{4.76+0.74+2.5}=0.093 \\
& \mathrm{X}_{\mathrm{Xe}}=\frac{\mathrm{u}_{\mathrm{Xe}}}{\mathrm{n}_{\mathrm{Ne}}+\mathrm{n}_{\mathrm{Ar}}+\mathrm{n}_{\mathrm{Xe}}} \\
& =\frac{2.5}{4.76+0.74+2.5}=0.312 \\
& \mathrm{P}_{\mathrm{Ne}}=\mathrm{X}_{\mathrm{Ne}} \mathrm{P}_{\text {Total }}=0.595 \times 2=1.19 \mathrm{~atm} \text {. } \\
& \mathrm{P}_{\mathrm{Ne}}=\mathrm{X}_{\mathrm{Ne}} \mathrm{P}_{\text {Total }}=0.093 \times 2=0.186 \mathrm{~atm} \text {. } \\
& \mathrm{P}_{\mathrm{Ne}}=\mathrm{X}_{\mathrm{Ne}} \mathrm{P}_{\mathrm{Total}}=0.312 \times 2=0.624 \mathrm{~atm} \text {. } \\
&
\end{aligned}
$
Question 13.
An unknown gas diffuses at a rate of 0.5 time that of nitrogen at the same temperature and pressure. Calculate the molar mass of the unknown gas.
Answer:
$
\begin{aligned}
& \frac{\text { rate }_{\text {unknown }}}{\text { rate } \mathrm{N}_2}=\sqrt{\frac{\mathrm{M}_{\mathrm{N}_2}}{\mathrm{M}_{\text {unknown }}}} \\
& 0.5=\sqrt{\frac{28 \mathrm{~g} \mathrm{~mol}^{-1}}{\mathrm{M}_{\text {unknown }}}}
\end{aligned}
$
Squaring on both sides
$
\begin{aligned}
& (0.5)^2=\frac{28 \mathrm{~g} \mathrm{~mol}^{-1}}{\mathrm{M}_{\text {unknown }}} \\
& \Rightarrow \mathrm{M}_{\text {unknown }}=\frac{28}{0.25}=112 \mathrm{~g} \mathrm{~mol}^{-1}
\end{aligned}
$
Question 14
If a scuba diver takes a breath at the surface filling his lungs with $5.82 \mathrm{dm} 3$ of air what volume will the air in his lungs occupy when he drives to a depth, where the pressure is $1.92 \mathrm{~atm}$. (assume temperature is constant and the pressure at the surface is exactly $1 \mathrm{~atm}$.)
Solution :
Temperature $=$ Constant

Pressure at the surface $=1 \mathrm{~atm}-\mathrm{P}_1$
Pressure at the depth $=1.92 \mathrm{~atm}-\mathrm{P}_2$
Vdlume of air breathing at the surface of the air $=5.82 \mathrm{dm}^3-\mathrm{V}_1$
Volume of air breathing at the depth $=\mathrm{V}_2=$ ?
$
\begin{aligned}
& \therefore V_2=\frac{\mathrm{P}_1 \mathrm{~V}_1}{\mathrm{P}_2}=\frac{1 \mathrm{~atm} \times 5.82 \mathrm{dm}^3}{1.92 \mathrm{~atm}} \\
& \mathrm{~V}_1=3.03 \mathrm{dm}^3
\end{aligned}
$
The volume of air scuba diver's lung occupy $=3.03 \mathrm{dm}^3$
Question 15.
Inside a certain automobile engine, the volume of air in a cylinder is $0.475 \mathrm{dm}^3$, when the pressure is 1.05 $\mathrm{atm}$. When the gas is compressed, the pressure increased to $5.65 \mathrm{~atm}$. at the same temperature. What is the volume of compressed air?
Solution:
Volume of air in the cylinder $0.475 \mathrm{dm}^3-\mathrm{V}_1$
Pressure of air $\mathrm{P}_1=1.05 \mathrm{~atm}$
Increased pressure $\mathrm{P}_2=5.65 \mathrm{~atm}$
Volume of air compressed $\mathrm{V}_2=$ ?
$
\begin{aligned}
& P_1 V_1=P_2 V_2 \\
& \therefore V_2=\frac{P_1 V_1}{P_2}=\frac{1.05 \mathrm{~atm} \times 0.475 \mathrm{dm}^3}{5.65 \mathrm{~atm}} \\
& \mathrm{~V}_2=0.08827 \mathrm{dm}^3
\end{aligned}
$
Compressed volume of air $=0.08827 \mathrm{dm}^3$