# NCERT Solutions for Class 11 Physics Chapter 13 - Kinetic Theory

**13.1. 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 3Å.**

**Ans.** Diameter of an oxygen molecule

d = 3Å

$$\text{Radius}, r=\frac{d}{2}\\r =\frac{3}{2}\text{A}^\degree=1.5\text{A}^\degree\\=1.5×10^{\normalsize-8}\text{cm}\\\text{Actual volume occupied by 1 mole of oxygen gas at STP = 22400 cm}^3\\\text{Molecular volume of oxygen gas,}\\\text{V}=\frac{4}{3}\pi r^{3}\text{N}\\\text{Where, N is Avogadro’s number} = 6.023 × 10^{23} \text{molecules/mole}\\\therefore\text{V}=\frac{4}{3}×3.14×(1.5×10^{\normalsize-8})^{3}×6.023×10^{23}\\=8.51\space\text{cm}^{3}$$

Therefore, ratio of the molecular volume to the actual volume of oxygen

$$=\frac{8.51}{22400}$$

= 3.8 × 10^{–4}

**13.2. Molar volume is the volume occupied by 1 mol of any (ideal) gas at standard temperature and pressure (STP : 1 atmospheric pressure, 0°C). Show that it is 22.4 litres.**

**Ans.** The ideal gas equation relating pressure (P), volume (V), and absolute temperature (T) is :

PV = nRT

where, R is the universal constant

= 8.314 J mol^{–1}K^{–1}

n = Number of moles = 1

T = Standard temperature = 273 K

P = Standard pressure

= 1 atm = 1.013 × 10^{5} Nm^{–2}

$$\therefore\space\text{V}=\frac{\text{nRT}}{\text{P}}\\=\frac{1×8.314×273}{1.013×10^{5}}$$

= 0.0224 m^{3}

= 22.4 litres

Therefore, the molar volume of a gas at STP is 22.4 litres.

**Hence Proved**

$$\textbf{13.3. Figure 13.8 shows plot of}\space\frac{\textbf{PV}}{\textbf{T}}\space\textbf{versus P for} 1.00 × 10^{\normalsize –3}\space\textbf{kg}\\\textbf{ of oxygen gas at two different temperatures.}$$

**(a) What does the dotted plot signify?**

**(b) Which is true: T _{1} > T_{2} or T_{1} < T_{2} ?**

$$\textbf{(c) What is the value of}\space\frac{\textbf{PV}}{\textbf{T}}\space\textbf{where the curves meet on the y-axis?}$$

**(d) If we obtained similar plots for 1.00 × 10 ^{–3} kg of hydrogen, would we get the same value of $$\frac{\text{PV}}{\text{T}}$$ **

**at the point where the curves ****meet on the y-axis? If not, what mass of hydrogen yields the same value of $$\frac{\text{PV}}{\text{T}}$$**

**(for low pressure high temperature region ****of the plot) ? (Molecular mass of H _{2} = 2.02 u, **

**of O**

_{2}= 32.0 u, R = 8.31 J mo1^{–1}K^{–1}.)**Ans.** (a) The dotted straight line, which is parallel to the P-axis indicates that the value of $$\frac{\text{PV}}{\text{T}}$$ remains constant even when P is modified,

which corresponds to ‘ideal’ gas behaviour.

(b) As shown in the figure, as T_{1} is closer to the dotted line i.e. closer to ideal behaviour of gas,

Hence, T_{1} > T_{2} (higher the temperature, more ideal behaviour of the gas.)

(c) We use PV = nRT

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

Mass of the gas

= 1 × 10^{-3} kg = 1 g

Molecular mass of O_{2}

= 32 g/mol

Hence,

$$\text{Number of mole}=\frac{\text{Given weight}}{\text{Molecular weight}}\\=\frac{1}{32}\\\text{So,}\space \text{nR}=\frac{1}{32}×8.314=0.256\space\text{J/K}\\\text{Hence,}\\\text{Value of}\space\frac{\text{PV}}{\text{T}}=0.256\space\text{J/K}$$

(d) The number of moles represented by 1g of H_{2} is not the same as the number of moles represented by 1g of CO_{2}.

For example, the molecular mass of H_{2} is 2g/mol.

As a result , $$\frac{1}{32}$$

moles of H_{2} are required

(as per the question)

Therefore,

Mass of hydrogen required

= Number of moles × Molecular mass of H_{2}

$$=\frac{1}{32}×2\\=\frac{1}{16}\text{(g)}$$

= 0.0625 g

= 6.3 × 10^{-5} kg

**13.4. An oxygen cylinder of volume 30 litres has an initial gauge pressure of 15 atm and a temperature of 27° C. After some oxygen is withdrawn from the cylinder, the gauge pressure drops to 11 atm and its temperature drops to 17° C. Estimate the mass of oxygen taken out of the cylinder (R = 8.31 J mol ^{–1} K^{–1}, molecular mass of O_{2} = 32 u).**

**Ans.** Volume of gas,

V_{1} = 30 litres = 30 × 10^{–3} m^{3}

Gauge pressure,

P_{1} = 15 atm = 15 × 1.013 × 10^{5} Pa

Temperature, T_{1} = 270°C = 300 K

Universal gas constant,

R = 8.314 J mol^{-1} K^{-1}

Let the initial number of moles

of oxygen gas in the cylinder be n_{1}

The gas equation is given as follows :

P_{1}V_{1} = n_{1}RT_{1}

$$n_1=\frac{P_1V_1}{\text{RT}_1}\\=\frac{(15.195×10^{5}×30×10^{\normalsize-3})}{(8.314×300)}\\=18.276\\\text{But}\space n_1=\frac{m_1}{\text{M}}$$

Where, m_{1} = Initial mass of oxygen

M = Molecular mass of oxygen

= 32 g

Thus, m_{1} = n_{1}M

= 18.276 × 32 = 584.83 g

After some oxygen is withdrawn from the cylinder, the pressure and temperature reduce.

Volume, V_{2} = 30 litres = 30 × 10^{–3} m^{3}

Gauge pressure,

P_{2} = 11 atm

= 11 × 1.013 × 10^{5} Pa

= 11.143 × 10^{5} Pa

T emperature, T_{2} = 17° C = 290 K

Let n_{2} be the number of moles of oxygen left in the cylinder.

The gas equation is given as :

P_{2}V_{2} = n_{2}RT_{2}

$$\text{Hence,} \space n_2=\frac{P_2V_2}{\text{RT}_2}\\=\frac{(11.143×10^{5}×30×10^{\normalsize-3})}{(8.314×290)}\\=13.86\\\text{But}\space n_2=\frac{m_2}{\text{M}}$$

Where, m_{2} is the mass of oxygen remaining in the cylinder.

Therefore, m_{2} = n_{2} × M = 13.86 × 32

= 443.52 g

The mass of oxygen taken out of the cylinder is given by the relation:

Initial mass of oxygen in the cylinder – Final mass of oxygen in the cylinder

= m_{1} – m_{2}

= 584.83 g – 443.52 g

We get, = 141.31 g

= 0.141 kg

Hence, 0.141 kg of oxygen is taken out of the cylinder.

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

**Ans.** Volume of the air bubble, V_{1} = 1.0 cm^{3}

= 1.0 × 10^{–6} m^{3}

Air bubble rises to height,

d = 40 m

Temperature at a depth of 40 m,

T_{1} = 12° C = 285 K

Temperature at the surface of the lake,

T_{2} = 35° C = 308 K

Pressure on the surface of the lake :

P_{2} = 1 atm = 1 × 1.013 × 10^{5} Pa

The pressure at the depth of 40 m :

P_{1} = 1 atm + drg

Where,

r is the density of water

= 10^{3} kg/m^{3}

g is the acceleration due to gravity

= 9.8 m/s^{2}

Hence,

P_{1} = 1.013 × 10^{5} + 40 × 10^{3} × 9.8

= 493300 Pa

$$\text{By gas law}\space\frac{\text{P}_1\text{V}_1}{\text{T}_1}=\frac{\text{P}_{2}\text{V}_{2}}{\text{T}_{2}}\\\text{Where, V}_2 \text{is the volume of the air bubble when it reaches the surface.}\\\text{V}_{2}=\frac{\text{P}_1\text{V}_1\text{T}_1}{\text{T}_1\text{P}_2}\\=\frac{493300×1×10^{\normalsize-6}×308}{(285×1.013×10^{5})}$$

= 5.263 × 10^{-6} m^{3} or 5.263 cm^{3}

Hence, when the air bubble reaches the surface, its volume becomes 5.263 cm^{3}

**13.6. Estimate the total number of air molecules (inclusive of oxygen, nitrogen, water vapour and other constituents) in a room of capacity 25.0 m3 at a temperature of 27°C and 1 atm pressure.**

**Ans.** Volume of the room,

V = 25.0 m^{3}

Temperature of the room,

T = 27° C = 300 K

Pressure in the room,

P = 1 atm = 1 × 1.013 × 10^{5} Pa

The ideal gas equation relating pressure (P), Volume (V), and absolute temperature (T) can be written as :

PV = (k_{B}NT)

Where, k_{B} is Boltzmann constant = (1.38 × 10^{–23}) J/K and N is the number of air molecules in the room.

Therefore,

$$\text{N}=\frac{(\text{PV})}{k_s\text{T}}\\=\frac{(1.013×10^{5}×25)}{(1.38×10^{\normalsize-23}×300)}$$

= 6.11 × 10^{26} molecules

Hence, the total number of air molecules in the given room is 6.11 × 10^{26}.

**13.7. Estimate the average thermal energy of a helium atom at (i) room temperature (27° C), (ii) the temperature on the surface of the Sun (6000 K), (iii) the temperature of 10 million kelvin (the typical core temperature in the case of a star).**

**Ans.** (i) At room temperature,

T = 27° C = 300 K

Average thermal energy

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

Where, k is the Boltzmann constant

= 1.38 × 10^{–23} J/K

Thermal energy

$$=\frac{3}{2}\text{kT}\\=\frac{3}{2}×1.38×10^{\normalsize-23}×300$$

= 6.21 × 10–21 J

Therefore, the average thermal energy of a helium atom at room temperature of 27° C is 6.21 × 10^{–21} J

(ii) Temperature on the surface of the Sun,

T = 6000 K

Average thermal energy

$$=\frac{3}{2}\text{kT}\\=\frac{3}{2}×1.38×10^{\normalsize-23}×6000$$

= 1.242 × 10^{-19} J

Therefore, the average thermal energy of a helium atom on the surface of the Sun is 1.242 × 10^{–19} J.

(iii) At temperature,

T = 10^{7} K

Average thermal energy

$$=\frac{3}{2}\text{kT}\\=\frac{3}{2}×1.38×10^{\normalsize-23}×10^{7}$$

= 2.07 × 10^{–16} J

Therefore, the average thermal energy of a helium atom at the core of a star is 2.07 × 10^{–16} J

**13.8. 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 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 vrms the largest ?**

**Ans.** All three vessels are of the same size and have the same capacity.

As a result, each gas has the same pressure, volume and temperature as the others.

Then, the three vessels will contain an equal quantity of the corresponding molecules, according to Avogadro’s law.

This number is equal to N = 6.023 × 1023, which is Avogadro’s number.

The relationship between the root mean square speed (v_{rms}) of a gas of mass m and temperature T is :

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

Where, k and T are constants for the supplied gases.

As a result, vrms is dependent on the mass of the atoms, i.e., v_{rms} ∝ (1/m)^{1/2}.

As a result, the root mean square speed of the molecules in each of the three scenarios differs.

The mass of neon is the smallest among neon, chlorine, and uranium hexafluoride.

As a result, among the supplied gases, neon has the fastest root mean square speed.

**13.9. 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°C? (atomic mass of Ar = 39.9 u, of He = 4.0 u).**

**Ans. **Given,

Temperature of the helium atom,

T_{He} = – 20°C = 253 K

Atomic mass of argon,

M_{Ar} = 39.9 u

Atomic mass of helium,

M_{He} = 4.0 u

Let (v_{rms})Ar be the rms speed of argon and Let (v_{rms})He be the rms speed of helium.

The rms speed of argon is given by :

$$(v_{rms})_{\text{Ar}}=\sqrt{\frac{3\text{RT}_{\text{AR}}}{\text{M}_{\text{AR}}}}\space...\text{(i)}$$

Where,

R is the universal gas constant and T_{Ar} is temperature of argon gas.

The rms speed of helium is given by :

$$(v_{rms})_{\text{He}}=\sqrt{\frac{3\text{RT}_{\text{He}}}{\text{M}_{\text{He}}}}\space...(\text{ii})\\\text{Given that,}\\\text{(v}_{rms})\text{Ar} = (v_{rms})\text{He}\\\sqrt{\frac{3\text{RT}_{\text{Ar}}}{\text{M}_{\text{Ar}}}}=\sqrt{\frac{3\text{RT}_{\text{He}}}{\text{M}_{\text{He}}}}\\\frac{\text{T}_{\text{Ar}}}{\text{M}_{Ar}}=\frac{\text{T}_{\text{He}}}{\text{M}_{\text{He}}}\\\text{T}_{\text{Ar}}=\frac{\text{T}_{\text{He}×\text{M}_{\text{Ar}}}}{\text{M}_{\text{He}}}\\=\frac{253}{4}×39.9$$

= 2523.675

= 2.52 × 103 K

Hence, the temperature of the argon atom is 2.52 × 10^{3} K.

**13.10. Estimate the mean free path and collision frequency of a nitrogen molecule in a cylinder containing nitrogen at 2.0 atm and temperature 17° C. Take the radius of a nitrogen molecule to be roughly 1.0Å. Compare the collision time with the time the molecule moves freely between two successive collisions (Molecular mass of N _{2} = 28.0 u).**

**Ans.** Pressure inside the cylinder containing nitrogen P = 2.0 atm

= 2.026 × 10^{5} Pa

Temperature inside the cylinder,

T = 17°C

= (17 + 273) K = 290 K

Radius of a nitrogen molecule,

r = 1.0 Å = 1 × 10^{–10} m

Diameter

, d = 2 × 1 × 10^{–10} = 2 × 10^{–10} m

Molecular mass of nitrogen,

M = 28.0 g = 28 × 10^{–3} kg

The root mean square speed of nitrogen is given by the relation :

$$v_{rms}=\sqrt{\frac{3\text{RT}}{\text{M}}}\\\text{Where, R is the universal gas constant}\\\text{= 8.314 J mol}^{\normalsize–1} K^{\normalsize–1}\\\text{Hence},\space v_{rms}=\sqrt{\frac{3×8.314×290}{28×10^{\normalsize-3}}}\\=\text{ 508.26 m/s}\\\text{The mean free path (l) is given by relation:}\\\text{l}=\sqrt{\frac{\text{K}_{\text{T}}}{\sqrt{2}×n^{2}×\text{P}}}\\\text{Where, k is the Boltzmann constant }\\= 1.38 × 10^{\normalsize–23} \text{kg m}^{2} s^{\normalsize–2} \text{K}^{\normalsize–1}$$

$$\\\text{Hence,}\space\text{l}=\frac{(1.38×10^{\normalsize-23}×290)}{(\sqrt{2}×3.14×(2×10^{\normalsize-10})^{2}×2.026×10^{5}}\\= 1.11 × 10–7 \text{m}\\ \text{Now, Collision frequency}\\=\frac{v_{rms}}{l}\\=\frac{508.26}{1.11×10^{\normalsize-7}}\\=4.58×10^{9}\text{s}^{\normalsize-1}\\\text{Collision time is given as :}\\\text{T}=\frac{d}{v_{rms}}\\=\frac{2×10^{\normalsize-10}}{508.26}$$

= 3.93 × 10^{–13} s

Time taken between successive collisions :

$$\text{T'}=\frac{l}{v_{rms}}=\frac{1.11×10^{\normalsize-7}}{508.26}\\=2.18×10^{\normalsize-10}\text{s}\\\frac{\text{T'}}{\text{T}}=\frac{2.18×10^{\normalsize-10}}{3.93×10^{\normalsize-13}}$$

554 ≈ 500

Therefore, the time taken between successive collisions is 500 times the time taken for a collision.

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

**Ans.** Length of the narrow bore,

L = 1 m = 100 cm

Length of the mercury thread,

l = 76 cm

Length of the air column

between mercury and the closed end, l_{a} = 15 cm

Since, the bore is held vertically in air with the open end at the bottom, the mercury length that occupies the air space is :

= 100 – (76 + 15)

= 9 cm

Therefore, the total length of the air column

= 15 + 9 = 24 cm

Let h cm of mercury flow out as a result of atmospheric pressure

So, length of the air column in the bore

= (24 + h) cm

And, length of the mercury column

= (76 – h) cm

Initial pressure,

V_{1} = 15 cm^{3}

Final pressure, P_{2} = 76 – (76 – h)

= h cm of mercury

Final volume, V_{2} = (24 + h) cm^{3}

Temperature remains constant throughout the process

Therefore,

P_{1}V_{1} = P_{2}V_{2}

On substituting, w

e get,

76 × 15 = h (24 + h)

∴ h^{2} + 24h – 11410 = 0

∴ h = 23.8 cm or – 47.8 cm

But height cannot be negative. Hence, 23.8 cm of mercury will flow out from the bore.

Length of the air column = 24 + 23.8 = 47.8 cm

**13.12. From a certain apparatus, the diffusion rate of hydrogen has an average value of 28.7 cm ^{3} s^{–1}. The diffusion of another gas under the same conditions is measured to have an average rate of 7.2 cm^{3} s^{–1}. Identify the gas.**

**[Hint ****: Use Graham’s law of diffusion: R _{1}/R_{2} = (M_{2}/M_{1})^{1/2}, where R_{1}, R_{2} are diffusion rates of gases 1 and 2, and M_{1} and M_{2} their respective molecular masses. The law is a simple consequence of kinetic theory.]**

**Ans.** Given

Rate of diffusion of hydrogen,

R_{1} = 28.7 cm^{3} s^{–1}

Rate of diffusion of another gas,

R_{2} = 7.2 cm^{3} s^{–1}

According to Graham’s Law of diffusion,

We have,

$$\frac{\text{R}_1}{\text{R}_2}=\sqrt{\frac{\text{M}_{2}}{\text{M}_{1}}}$$

Where,

M_{1} is the molecular mass of hydrogen = 2.020 g

M_{2} is the molecular mass of the unknown gas

$$\therefore\space\text{M}_{2}=\text{M}_{1}\bigg(\frac{R_1}{R_2}\bigg)^{2}\\=2.02\bigg(\frac{28.7}{7.2}\bigg)^{2}$$

= 32.09 g

32 g is the molecular mass of oxygen.

Hence, the unknown gas is oxygen.

**13.13. 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})/kBT] where n_{2}, n_{1} refer to number density at heights h_{2} and h_{1} respectively. Use this relation to derive the equation for sedimentation equilibrium of a suspension in a liquid column :**

**n**

_{2}= n_{1}exp [ – mg N_{A}(ρ – ρ') (h_{2}– h_{1})/(ρRT)]**where ρ is the density of the suspended particle, and ρ′ ,that of surrounding medium. [N**

_{A}is Avogadro’s number, and R the universal gas constant.]**[Hint ****: Use Archimedes principle to find the apparent weight of the suspended particle.]**

**Ans.** According to the law of atmospheres, we have

$$n_2=n_1\text{exp}\bigg[\frac{\text{-mg}(h_2-h_1)}{k_B\text{T}}\bigg]\space...\text{(i)}$$

Where n_{1} is the number density at height h_{1} n_{2} is the number density at height h_{2}

mg is the weight of the particles suspended in the gas column.

The suspended particle experiences an apparent weight because of the liquid displaced.

According to Archimedes principle

Apparent weight = Weight of the water displaced – weight of the suspended particle

= mg – m’g

$$=\text{= mg – Vρ’g = mg –}\bigg(\frac{m}{\rho}\bigg)\rho'\text{g}$$

= mg (1 – (ρ’/ρ)) ...(ii)

ρ’= Density of the water

ρ = Density of the suspended particle

m’ = Mass of the suspended particle

m = Mass of the water displaced

V = Volume of a suspended particle

Boltzmann’ s constant (k_{B})

$$=\frac{\text{R}}{\text{N}_{\text{A}}}\space...(iii)\\\text{Substitutingequation (ii) and equation (iii) in equation (i)}\\n_2=n_1\text{exp}\bigg[\frac{\text{-mg(h}_2-\text{h}_1)}{k_B\text{T}}\bigg]\\n_2=n_1\space\text{exp}\bigg[\frac{\text{-mg}(1-\rho'/\rho)(h_2-h_1)\text{N}_{\text{A}}}{\text{N}_{\text{A}}k_{B}\text{T}}\bigg]\\\text{n}_2=\text{n}_1\text{exp}\bigg[\frac{\text{-mgN}_{\text{A}}(\rho-\rho')(h_2-h_1)}{\rho\text{RT}}\bigg]$$

**13.14. Given below are densities of some solids and liquids. Give rough estimates of the size of their atoms :**

Substance | Atomic Mass (u) |
Density (10^{3} Kg m^{–3}) |

Carbon (diamond) | 12.01 | 2.22 |

Gold | 197.00 | 19.32 |

Nitrogen (liquid) | 14.01 | 1.00 |

Lithium | 6.94 | 0.53 |

Fluorine (liquid) | 19.00 | 1.44 |

**[Hint : Assume the atoms to ****be ‘tightly packed’ in a solid or liquid phase, and use the known value of Avogadro’s number. You should, however, not take the actual numbers you obtain for various atomic sizes too literally. Because of the crudeness of the tight packing approximation, the results only indicate that atomic sizes are in the range of a few Å].**

**Ans.** If r is the radius of the atom then the volume of each atom

$$=\frac{4}{3}\pi r^{3}\\\text{Volume of all the substance}\\=\frac{4}{3}\pi r^{3}×\text{N}=\frac{\text{M}}{\rho}$$

M is the atomic mass of the substance

ρ is the density of the substance

One mole of the substance has 6.023 × 10^{23} atoms

$$\text{r}=\bigg(\frac{3\text{M}}{4\pi\rho N}\bigg)^{1/3}\\=\bigg(\frac{3\text{M}}{4\pi\rho×6.023×10^{23}}\bigg)^{1/3}\\\text{For carbon,}\\\text{M}=12.01×10^{\normalsize-3}\\\text{and}\space\rho=2.22×10^{3}\text{kg m}^{\normalsize-3}\\r=\bigg(\frac{3×12.01×10^{\normalsize-3}}{4×\frac{22}{7}×2.22×10^{3}×6.03×10^{23}}\bigg)^{1/3}\\=\bigg(\frac{36.03×10^{\normalsize-3}}{167.94×10^{26}}\bigg)^{1/3}$$

= 1.29 × 10^{–10} m

= 1.29 Å

For gold,

M = 197 × 10^{–3} kg

and ρ = 19.32 × 10^{3} kg m^{–3}

$$r=\bigg(\frac{3×197×10^{\normalsize-3}}{4×\frac{22}{7}×19.32×10^{3}×6.023×10^{23}}\bigg)^{1/3}$$

= 1.59 × 10^{–10} m = 1.59 Å

For lithium,

M = 6.94 × 10^{–3} kg

and ρ = 0.53 × 10^{3} kg/m^{3}

$$r=\bigg(\frac{3×6.94×10^{\normalsize-3}}{3.14×0.53×10^{3}×6.023×10^{23}}\bigg)^{1/3}\\\text{=1.73 × 10}^{\normalsize–10} \text{m} = 1.73 \text{Å}\\\text{For nitrogen (liquid),}\\\text{M}=14.01×10^{\normalsize-3}\text{kg}\\\text{and}\space\rho=1.00×10^{3}\text{kg/m}^{3}\\\text{r}=\bigg(\frac{3×14.01×10^{\normalsize-3}}{3.14×1.00×10^{3}×6.023×10^{23}}\bigg)^{1/3}\\=1.77×10^{\normalsize-10}\text{m}=1.77\text{Å}$$

For fluorine (liquid),

M = 19.00 × 10^{-3} kg

and ρ = 1.14 × 10^{3} kg/m^{3}

$$r=\bigg(\frac{3×19×10^{\normalsize-3}}{3.14×1.14×10^{3}×6.023×10^{23}}\bigg)^{1/3}$$

= 1.88 × 10^{–10} m = 1.88 Å

Above calculation gives the rough estimate of the size of different atoms.

## NCERT Solutions for Class 11 Physics Chapter 13 Free PDF Download

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