NCERT Solutions for Class 11 Physics Chapter 12 Kinetic Theory

Q12.1 Estimate the fraction of molecular volume to the actual volume occupied by oxygen gas at STP. Take the diameter of an oxygen molecule to be 3 Å.

Answer:

The diameter of an oxygen molecule, d = 3 Å.

The actual volume of a mole of oxygen molecules V _actual is

${V}_{actual}=N_A\frac{4}{3}\pi {\left(\frac{d}{2}\right)}^3$ $$
$V_{actual}=6.023\times {10}^{23}\times \frac{4}{3}\pi \times {\left(\frac{3\times {10}^{-10}}{2}\right)}^3$
$V_{actual}=8.51\times {10}^{-6}{m}^3$
$V_{actual}=8.51\times {10}^{-3}litres$

The volume occupied by a mole of oxygen gas at STP is V _molar = 22.4 litres

$\frac{V_{actual}}{V_{molar}}=\frac{8.51\times {10}^{-3}}{22.4}$

$\frac{V_{actual}}{V_{molar}}=3.8\times {10}^{-4}$

Q12.2 Molar volume is the volume occupied by 1 mol of any (ideal) gas at standard temperature and pressure (STP: 1 atmospheric pressure, $0^{0}C$ ). Show that it is $22.4Litres$ .

Answer:

As per the ideal gas equation

$$\begin{gathered} PV=nRT \\ V=\frac{nRT}{P} \end{gathered}$$

For one mole of a gas at STP we have

$V=\frac{1\times 8.314\times 273}{1.013\times {10}^5}$
$V=0.0224m^3$

$V=22.4litres$

Q12.3 Figure 13.8 shows plot of $PV/T$ versus P for $1.00\times {10}^{-3}$ kg of oxygen gas at two different temperatures.

(a) What does the dotted plot signify?
(b) Which is true: $T_1>T_{2}or{T}_1<T_2$ ?
(c) What is the value of $PV/T$ where the curves meet on the y-axis?
(d) If we obtained similar plots for $1.00\times {10}^{-3}$ kg of hydrogen, would we get the same value of $PV/T$ at the point where the curves meet on the y-axis? If not, what mass of hydrogen yields the same value of $PV/T$ (for low pressure high temperature region of the plot) ? (Molecular mass of $H_2=2.02\mu$ , of $O_2=32.0\mu$, $R=8.31Jmo{l}^{-1}{K}^{-1}$.)

Answer:

(a) The dotted plot corresponds to the ideal gas behaviour.

(b) We know the behaviour of a real gas tends close to that of ideal gas as its temperature increases and since the plot corresponding to temperature T ₁ is closer to the horizontal line that the one corresponding to T ₂ we conclude T ₁ is greater than T ₂.

(c) As per the ideal gas equation

$\frac{PV}{T}=nR$

The molar mass of oxygen = 32 g

$n=\frac{1}{32}$

R = 8.314

$nR=\frac{1}{32}\times 8.314$
$nR=0.256J{K}^{-1}$

(d) If we obtained similar plots for $1.00\times {10}^{-3}$ kg of hydrogen we would not get the same value of $PV/T$ at the point where the curves meet on the y-axis as 1 g of Hydrogen would contain more moles than 1 g of Oxygen because of having smaller molar mass.

Molar Mass of Hydrogen M = 2 g

mass of hydrogen

$m=\frac{PV}{T}\frac{M}{R}=0.256\times \frac{2}{8.314}=5.48\times {10}^{-5}Kg$

Q12.4 An oxygen cylinder of volume $30$ litres has an initial gauge pressure of $15$ atm and a temperature of ${27}^{0}C$ . After some oxygen is withdrawn from the cylinder, the gauge pressure drops to $11$ atm and its temperature drops to ${17}^{0}C$ . Estimate the mass of oxygen taken out of the cylinder ( $R=8.31Jmo{l}^{-1}{K}^{-1}$ , molecular mass of $O_2=32\mu$ ).

Answer:

Initial volume, V ₁ = Volume of Cylinder = 30 l

Initial Pressure P ₁ = 15 atm

Initial Temperature T ₁ = 27 ^o C = 300 K

The initial number of moles n ₁ inside the cylinder is

$n_1=\frac{P_1{V}_1}{R{T}_1}$
$n_1=\frac{15\times 1.013\times {10}^5\times 30\times {10}^{-3}}{8.314\times 300}$
$n_1=18.28$

Final volume, V ₂ = Volume of Cylinder = 30 l

Final Pressure P ₂ = 11 atm

Final Temperature T ₂ = 17 ^o C = 290 K

The final number of moles n ₂ inside the cylinder is

$n_2=\frac{P_2{V}_2}{R{T}_2}$
$n_2=\frac{11\times 1.013\times {10}^5\times 30\times {10}^{-3}}{8.314\times 290}$
$n_2=13.86$

Moles of oxygen taken out of the cylinder = n ₂ -n ₁ = 18.28 - 13.86 = 4.42

Mass of oxygen taken out of the cylinder m is

$m=4.42\times 32$
$m=141.44g$

Q12.5 An air bubble of volume $1.0c{m}^3$ rises from the bottom of a lake $40m$ deep at a temperature of ${12}^{0}C$ . To what volume does it grow when it reaches the surface, which is at a temperature of ${35}^{0}C$ ?

Answer:

Initial Volume of the bubble, V ₁ = 1.0 cm ³

Initial temperature, T ₁ = 12 ^o C = 273 + 12 = 285 K

The density of water is ${\rho }_w={10}^{3}kg{m}^{-3}$

Initial Pressure is P ₁

Depth of the bottom of the lake = 40 m

$\begin{array}{rl}& P_1=\text{ Atmospheric Pressure }+\text{ Pressure due to water }\\ & P_1=P_{atm}+{\rho }_{w}gh\\ & P_1=1.013\times {10}^5+{10}^3\times 9.8\times 40\\ & P_1=4.93\times {10}^5\mathrm{Pa}\end{array}$

Final Temperature, T ₂ = 35 ^o C = 35 + 273 = 308 K

Final Pressure = Atmospheric Pressure $=1.013\times {10}^{5}Pa$

Let the final volume be V ₂

As the number of moles inside the bubble remains constant, we have

$\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}$

$V_2=\frac{P_1{T}_2{V}_1}{P_2{T}_1}$

$V_2=\frac{4.93\times {10}^5\times 308\times 1}{1.013\times {10}^5\times 285}$

$V_2=5.26c{m}^3$

Q12.6 Estimate the total number of air molecules (inclusive of oxygen, nitrogen, water vapour and other constituents) in a room of capacity $25.0m^3$ at a temperature of ${27}^{0}C$ and $1atm$ pressure.

Answer:

The volume of the room, V = 25.0 m ³

Temperature of the room, T = 27 ^o C = 300 K

The pressure inside the room, P = 1 atm

Let the number of moles of air molecules inside the room be n

$\begin{array}{rl}& n=\frac{PV}{RT}\\ & n=\frac{1.013\times {10}^5\times 25}{8.314\times 300}\\ & n=1015.35\end{array}$

Avogadro's Number, $N_A=6.022\times {10}^{23}$
Number of molecules inside the room is N

$\begin{array}{rl}& N=n{N}_A\\ & N=1015.35\times 6.022\times {10}^{23}\\ & N=6.114\times {10}^{26}\end{array}$

Q12.7 Estimate the average thermal energy of a helium atom at (i) room temperature ( ${27}^{0}C$ ), (ii) the temperature on the surface of the Sun ( $6000K$ ), (iii) the temperature of $10$ million kelvin (the typical core temperature in the case of a star).

Answer:

The average energy of a Helium atom is given as $\frac{3kT}{2}$ since it is monoatomic

(i)

$E=\frac{3kT}{2}$

$E=\frac{3\times 1.38\times {10}^{-23}\times 300}{2}$

$E=6.21\times {10}^{-21}J$

(ii)

$E=\frac{3kT}{2}$

$E=\frac{3\times 1.38\times {10}^{-23}\times 6000}{2}$

$E=1.242\times {10}^{-19}J$

(iii)

$E=\frac{3kT}{2}$

$E=\frac{3\times 1.38\times {10}^{-23}\times {10}^7}{2}$

$E=2.07\times {10}^{-16}J$

Q12.8 Three vessels of equal capacity have gases at the same temperature and pressure. The first vessel contains neon (monatomic), the second contains chlorine (diatomic), and the third contains uranium hexafluoride (polyatomic). Do the vessels contain an equal number of respective molecules? Is the root mean square speed of molecules the same in the three cases? If not, in which case is v _rms the largest?

Answer:

As per Avogadro's Hypothesis, under similar conditions of temperature and pressure, equal volumes of gases contain an equal number of molecules. Since the volume of the vessels is the same and all vessels are kept at the same conditions of pressure and temperature, they would contain an equal number of molecules.

Root mean square velocity is given as

$v_{rms}=\sqrt{\frac{3kT}{m}}$

As we can see, v_rms is inversely proportional to the square root of the molar mass; the root mean square velocity will be maximum in the case of Neon as its molar mass is the least.

Q12.9 At what temperature is the root mean square speed of an atom in an argon gas cylinder equal to the rms speed of a helium gas atom at $-{20}^{0}C$ ? (atomic mass of $Ar=39.9\mu$ , of $He=4.0\mu$ ).

Answer:

As we know root mean square velocity is given as $v_{rms}=\sqrt{\frac{3RT}{M}}$

Let at temperature T the root mean square speed of an atom in an argon cylinder equal to the rms speed of a helium gas atom at $-{20}^{0}C$

$\sqrt{\frac{3R\times T}{39.9}}=\sqrt{\frac{3R\times 253}{4}}$

$T=2523.7K$

Q12.10 Estimate the mean free path and collision frequency of a nitrogen molecule in a cylinder containing nitrogen at $2.0atm$ and temperature ${17}^{0}C$ . Take the radius of a nitrogen molecule to be roughly $1.0A$ . Compare the collision time with the time the
molecule moves freely between two successive collisions (Molecular mass of $N_2=28.0\mu$ ).

Answer:

Pressure, P = 2atm

Temperature, T = 17 ^o C

The radius of the Nitrogen molecule, r=1 Å.

The molecular mass of N ₂ = 28 u

The molar mass of N ₂ = 28 g

From the ideal gas equation

$PV=nRT$

$\frac{n}{V}=\frac{P}{RT}$

The above tells us about the number of moles per unit volume; the number of molecules per unit volume would be given as

${n}^{\prime }=\frac{N_{A}n}{V}=\frac{6.022\times {10}^{23}\times 2\times 1.013\times {10}^5}{8.314\times (17+273)}$

$n^{\prime }=5.06\times {10}^{25}$

The mean free path $\lambda$ is given as

$\lambda =\frac{1}{\sqrt{2}\pi n^{\prime }{d}^2}$

$\lambda =\frac{1}{\sqrt{2}\times \pi \times 5.06\times {10}^{25}\times (2\times 1\times {10}^{-10}{)}^2}$

$\lambda =1.11\times {10}^{-7}m$

The root mean square velocity v _rms is given as

${v}_{rms}=\sqrt{\frac{3RT}{M}}$

$v_{rms}=\sqrt{\frac{3\times 8.314\times 290}{28\times {10}^{-3}}}$

$v_{rms}=508.26m{s}^{-1}$

The time between collisions T is given as

$T=\frac{1}{CollisionFrequency}$

$T=\frac{1}{\nu }$

$T=\frac{\lambda }{v_{rms}}$

$T=\frac{1.11\times {10}^{-7}}{508.26}$

$T=2.18\times {10}^{-10}s$

Collision time T' is equal to the average time taken by a molecule to travel a distance equal to its diameter

${T}^{\prime }=\frac{d}{v_{rms}}$

$T^{\prime }=\frac{2\times 1\times {10}^{-10}}{508.26}$

$T^{\prime }=3.935\times {10}^{-13}s$

The ratio of the average time between collisions to the collision time is

$\begin{array}{rl}\frac{T}{T^{\prime }}& =\frac{2.18\times {10}^{-10}}{3.935\times {10}^{-13}}\\ \frac{T}{T^{\prime }}& =554\end{array}$
Thus, we can see that the time between collisions is much larger than the collision time.

NCERT Solutions for Class 11 Physics Chapter 12 Kinetic Theory - Additional Questions

Q 1) A metre-long narrow bore held horizontally (and closed at one end) contains a $76cm$ long mercury thread, which traps a $15cm$ column of air. What happens if the tube is held vertically with the open end at the bottom?

Answer:

Initially, the pressure of the 15 cm long air column is equal to the atmospheric pressure, P ₁ = 1 atm = 76 cm of Mercury

Let the crossectional area of the tube be x cm ²

The initial volume of the air column, V ₁ = 15x cm ³

Let's assume that once the tube is held vertical, y cm of Mercury flows out of it.

The pressure of the air column after y cm of Mercury has flown out of the column P ₂ = 76 - (76 - y) cm of Mercury = y cm of mercury

Final volume of air column V ₂ = (24 + y)x cm ³

Since the temperature of the air column does not change

$$\begin{gathered} {P}_1{V}_1=P_2{V}_2 \\ 76\times 15x=y\times (24+y)x \\ 1140=y^2+24y \\ {y}^2+24y-1140=0 \end{gathered}$$

Solving the above quadratic equation, we get y = 23.8 cm or y = -47.8 cm

Since a negative amount of mercury cannot flow out of the column, y cannot be negative. Therefore, y = 23.8 cm.

Length of the air column = y + 24 = 47.8 cm.

Therefore, once the tube is held vertically, 23.8 cm of Mercury flows out of it, and the length of the air column becomes 47.8 cm

Q 2) From a certain apparatus, the diffusion rate of hydrogen has an average value of $28.7c{m}^3{s}^{-1}$. The diffusion of another gas under the same conditions is measured to have an average rate of $7.2c{m}^3{s}^{-1}$. Identify the gas.

Answer:

As per Graham's Law of diffusion, if two gases of Molar Mass M ₁ and M ₂ diffuse with rates R ₁ and R ₂ respectively, their diffusion rates are related by the following equation

$\frac{R_1}{R_2}=\sqrt{\frac{M_2}{M_1}}$

In the given question

R ₁ = 28.7 cm ³ s ^-1

R ₂ = 7.2 cm ³ s ^-1

M ₁ = 2 g

${M}_2=M_1{\left(\frac{R_1}{R_2}\right)}^2$

$M_2=2\times {\left(\frac{28.7}{7.2}\right)}^2$

$M_2=31.78g$

The above Molar Mass is close to 32, therefore, the gas is Oxygen.

Q 3) A gas in equilibrium has uniform density and pressure throughout its volume. This is strictly true only if there are no external influences. A gas column under gravity, for example, does not have uniform density (and pressure). As you might expect, its density decreases with height. The precise dependence is given by the so-called law of atmospheres $n_2=n_{1}exp[-mg(h_2-h_1)/k_{b}T]$ where n₂, n₁ refer to number density at heights h₂ and h₁ respectively. Use this relation to derive the equation for the sedimentation equilibrium of a suspension in a liquid column: $n_2=n_{1}exp[-mg{N}_A(\rho -{\rho }^{{}^{\prime }})(h_2-h_1)/(\rho RT)]$ where $\rho$ is the density of the suspended particle, and ${\rho }^{{}^{\prime }}$ , that of the surrounding medium. [ $N_A$ s Avogadro’s number, and R the universal gas constant.] [Hint: Use Archimedes' principle to find the apparent weight of the suspended particle.]

Answer:

$n_2=n_{1}exp[-mg(h_2-h_1)/k_{b}T]$ $(i)$

Let the suspended particles be spherical and have a radius r

The gravitational force acting on the suspended particles would be

$F_G=\frac{4}{3}\pi r^3\rho g$

The buoyant force acting on them would be

$F_B=\frac{4}{3}\pi r^3{\rho }^{\prime }g$

The net force acting on the particles become

${F}_{net}=F_G-F_B$

$F_{net}=\frac{4}{3}\pi r^3\rho g-\frac{4}{3}\pi r^3{\rho }^{\prime }g$

$F_{net}=\frac{4}{3}\pi r^{3}g(\rho -{\rho }^{\prime })$

Replacing mg in equation (i) with the above equation, we get

${n}_2=n_{1}exp[-\frac{4}{3}\pi r^{3}g(\rho -{\rho }^{\prime })(h_2-h_1)/k_{b}T]$

$n_2=n_{1}exp\left[\frac{-\frac{4}{3}\pi r^{3}g(\rho -{\rho }^{\prime })(h_2-h_1)}{\frac{RT}{N_A}}\right]$

$n_2=n_{1}exp\left[\frac{-mg{N}_A(\rho -{\rho }^{\prime })(h_2-h_1)}{RT{\rho }^{\prime }}\right]$

The above is the equation to be derived

Q 4) Given below are densities of some solids and liquids. Give rough estimates of the size of their atoms :

Answer:

Let one mole of a substance of atomic radius r and density $\rho$ have molar mass M

Let us assume the atoms to be spherical

Avogadro's number is $N_A=6.022\times {10}^{23}$

$$\begin{gathered} {N}_A\frac{4}{3}\pi r^3\rho =M \\ r={\left(\frac{3M}{4N_A\pi \rho }\right)}^{\frac{1}{3}} \end{gathered}$$

For Carbon

$$\begin{gathered} {N}_A\frac{4}{3}\pi r^3\rho =M \\ r={\left(\frac{3\times 12.01}{4\times 6.022\times {10}^{23}\times \pi \times 2.22\times {10}^3}\right)}^{\frac{1}{3}} \\ r=1.29 \end{gathered}$$ Å.

For gold

$$\begin{gathered} {N}_A\frac{4}{3}\pi r^3\rho =M \\ r={\left(\frac{3\times 197.00}{4\times 6.022\times {10}^{23}\times \pi \times 19.32\times {10}^3}\right)}^{\frac{1}{3}} \\ r=1.59 \end{gathered}$$ Å.

For Nitrogen

$$\begin{gathered} {N}_A\frac{4}{3}\pi r^3\rho =M \\ r={\left(\frac{3\times 14.01}{4\times 6.022\times {10}^{23}\times \pi \times 1.00\times {10}^3}\right)}^{\frac{1}{3}} \\ r=1.77 \end{gathered}$$ Å.

For Lithium

$$\begin{gathered} {N}_A\frac{4}{3}\pi r^3\rho =M \\ r={\left(\frac{3\times 6.94}{4\times 6.022\times {10}^{23}\times \pi \times 0.53\times {10}^3}\right)}^{\frac{1}{3}} \\ r=1.73 \end{gathered}$$ Å.

For Fluorine

$$\begin{gathered} {N}_A\frac{4}{3}\pi r^3\rho =M \\ r={\left(\frac{3\times 19.00}{4\times 6.022\times {10}^{23}\times \pi \times 1.14\times {10}^3}\right)}^{\frac{1}{3}} \\ r=1.88 \end{gathered}$$ Å.