13.12 From a certain apparatus, the diffusion rate of hydrogen has an average value of 28.7 cm3 s–1. The diffusion of another gas under the same conditions is measured to have an average rate of 7.2 cm3 s–1. Identify the gas. [Hint : Use Graham’s law of diffusion: R1/R2 = ( M2 /M1 )1/2, where R1, R2 are diffusion rates of gases 1 and 2, and M1 and M2 their respective molecular masses. The law is a simple consequence of kinetic theory.]

13.12 Rate of diffusion of hydrogen, R1= 28.7 cm3 s–1 Rate of diffusion of another gas, R2 = 7.2 cm3 s–1 According to Graham's law of diffusion, we have: R1R2= M2M1, where M1= molecular mass of hydrogen = 2.02 g and M2 is the molecular mass of the unknown gas M2=M1* (R1R2)2= 2.02 * (28.77.2)2= 32.09 = Molecular mass of Oxygen Hence, the unknown gas is Oxygen.

What will be the effect on the root mean square velocity of oxygen molecules if the temperagture is doubled and oxygen molecule dissociates into atomic oxygen?

The velocity of atomic oxygen doubles

The velocity of atomic oxygen remains same

The velocity of atomic oxygen becomes half

The velocity of atomic oxygen becomes four times

Kinetic Theory – NCERT Solutions for Class 11 Physics Ch 12

Q: 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).

13.9 Temperature of the helium atom, THe = – 20 °C = 253 K and temperature of argon atom be = TAr Atomic mass of helium, MHe = 4.0 u Atomic mass of Argon, MAr = 39.9 u Let VrmsAr be the rms speed of Argon and VrmsHe be the rms speed of Helium From the relation of Vrms=3kTM we get rms speed of Argon, VrmsAr=3kTArMAr rms speed of Helium, VrmsHe=3kTHeMHe Since both the speeds are equal, we get 3kTArMAr = 3kTHeMHe or TArMAr = THeMHe or TAr = THe×MArMHe = 253×39.94 = 2523.675 K = 2.523 ×103 K

Q: 13.5 An air bubble of volume 1.0 cm3 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?

13.5 Volume of the air bubble , V1 = 1.0 cm3 = 1 ×10-6 m3 Height achieved by bubble, d = 40 m Temperature at a depth of 40 m, T1 = 12 ? = 285 K Temperature at the surface of the lake, T2 = 35 ? = 308 K The pressure at the surface of the lake, P2 = 1 atm = 1.013 ×105 Pa The pressure at 40m depth: P1 = 1 atm + d ?g Where ? is the density of water = 103 kg/ m3 G = acceleration due to gravity = 9.8 m/ s2 Hence P1 = 1.013 ×105+(40×103×9.8) = 493300 Pa From the relation P1V1T1 = P2V2T2 , where V2 is the volume of bubble when it reaches the surface, we get, V2 = P1V1T2T1P2 = 493300×1×10-6×308285×1.013×105 = 5.263 × 10-6m3=5.263cm3

Q: 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?

13.8 (a) According to Avogadro's law, the three vessels will contain an equal number of the respective molecules. This number is equal to Avogadro's number, N = 6.023 *1023 (b) The root mean square speed ( Vrms) of a gas of mass m and temperature T is given by the relation Vrms=3kTm . Where k is Boltzmann constant. For the given gases, k and T are constants. Hence Vrms depends only on the mass of the atoms Therefore, the root mean square speed of the molecules in the three cases is not the same. Among Neon, Chlorine and Uranium hexafluoride, the mass of the neon is the smallest, so Neon will have the highest root mean square speed among the given gases.

Q: 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?

13.11 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, lo = 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 Hence, total length of the air column = 15 + 9 = 24 cm Let h cm of mercury flow out as a result of atmospheric pressure. Length of the air column in the bore = 24 + h cm Length of the mercury column = 76 – h cm Initial pressure, P1 = 76 cm of mercury Initial volume, V1= 15 cm3 Final pressure, P2 = 76 – (76 – h) = h cm of mercury Final volume, V2 = (24 + h) cm3 Since temperature remained constant throughout the process, P1V1= P2V2 or 76 *15 = h * (24+h) Or, h2 + 24h - 76 *15 =0 h = -24±242+4*76*152= -24±71.672 So h = 23.84 cm or -47.84 cm Since height cannot be negative, so h = 23.84 cm 23.84 cm of mercury will flow out from the bore and (76-23.84) 52.16 cm of mercury will remain in the bore. The length of the air column will be 24 + 23.84 = 47.84 cm

Updated on Jul 14, 2025 17:39 IST

By Pallavi Pathak, Assistant Manager Content

NCERT Solutions for Class 11 Physics Chapter 12 – Kinetic Theory provides a comprehensive study material for this chapter. Kinetic Theory Class 11 offers a detailed explanation and step-by-step solutions to all NCERT textbook questions. The chapter covers the basic postulates of the kinetic theory, the molecular structure of gases, and how temperature relates to molecular motion.
These solutions are ideal for the CBSE Board exam preparation and entrance exams like NEET and JEE Main. It simplifies the complex topics such as mean free path, pressure derivation, and degrees of freedom. If you want to access the comprehensive material for Class 11 Physics, do check out the Class 11 Physics Notes. Here, you will get the important topics, solved examples, and chapter-wise NCERT notes. Also, it has the PDFs for all the chapters.

Table of content

NCERT Class 11 Physics Chapter 12 Kinetic Theory: Key Topics, Weightage, and Important Formulas
Class 11 Physics kinetic Theory NCERT Solution PDF: Download free PDF
Chapter 12 Kinetic Theory Important Formulas & Concepts
NCERT Questions and Answers for Physics Class 11 Kinetic Theory

NCERT Class 11 Physics Chapter 12 Kinetic Theory: Key Topics, Weightage, and Important Formulas

Kinetic Theory is an important chapter in Class 11 Physics. The students can check here the topics covered in this chapter:

Exercise	Topics Covered
12.1	Introduction
12.2	Molecular Nature of Matter
12.3	Behaviour of Gases
12.4	Kinetic Theory of an Ideal Gas
12.5	Law of Equipartition of Energy
12.6	Specific Heat Capacity
12.7	Mean Free Path

Class 11 Physics Kinetic Theory Weightage in NEET, JEE Main

Exam	Number of Questions	Weightage
NEET	1 question	2%
JEE Main	2 questions	2-3%

Class 11 Physics kinetic Theory NCERT Solution PDF: Download free PDF

Shiksha provides the NCERT Solution for Class 11 Physics Chapter Kinetic Theory in PDF format. Students can download it from the link given below and access it from anywhere without any requirement for internet connectivity.

Download Here: NCERT Solution for Class XI Physics Chapter Kinetic Theory PDF

Chapter 12 Kinetic Theory Important Formulas & Concepts

The following are the important formulas related to the kinetic theory class 11:

Ideal Gas Equation:
$PV = nRT$
Kinetic Interpretation of Pressure:
$P = \frac{1}{3} \rho \bar{v}^2$
where $\rho$ is density, $\bar{v}$ is the root mean square speed
Average Kinetic Energy of a Gas Molecule:
$KE = \frac{3}{2} k_B T$
RMS Speed:
$v_{\text{rms}} = \sqrt{\frac{3k_B T}{m}} = \sqrt{\frac{3RT}{M}}$
Degrees of Freedom (f):
- Monoatomic gas: $f = 3$
- Diatomic gas: $f = 5$
- Polyatomic gas: $f = 6$
Law of Equipartition of Energy:
$\text{Total energy} = \frac{f}{2}k_BT$ per molecule
Specific Heat Relation (Mayer’s Formula):
$C_P - C_V = R$
Mean Free Path (λ):
$\lambda = \frac{1}{\sqrt{2} \pi d^2 n}$
where $d$ is the molecular diameter and $n$ is number density

NCERT Questions and Answers for Physics Class 11 Kinetic Theory

See here all the NCERT solutions for kinetic theory of gases class 11:

Q.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 3 Å.

Ans.13.1

Diameter of the oxygen molecule, d = 3Å

Radius, r = $\frac{d}{2}$ = 1.5 Å = 1.5 $\times 10^{- 8}$ cm

Actual volume occupied by 1 mole of oxygen gas at STP = 22.4 lit = 22400 ${c m}^{3}$

Molecular volume of oxygen gas, V = $\frac{4}{3} π r^{3}$ N,

where N = Avogadro’s number = 6.023 $\times 10^{23}$ molecules/mole

Therefore, V = $\frac{4}{3}$ $\times 3.1416 \times ({1.5 \times 10^{- 8})}^{3}$ $\times$ 6.023 $\times 10^{23}$ = 8.51 ${c m}^{3}$

Ratio of the molecular volume to the actual volume of oxygen = $\frac{8.51}{22400}$ = 3.8 $\times 10^{- 4}$

Q.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.13.2

The ideal gas equation relating to pressure (P), volume (V) and absolute temperature (T) is given by the relation

PV = nRT, where

R is the universal gas constant = 8.314 J /mol/K

N = number of moles = 1 (given)

T = standard temperature = 273 K

P = standard pressure = 1 atm = 1.013 $\times 10^{5}$ N/ $m^{2}$

Therefore V = $\frac{n R T}{P}$ = $\frac{1 \times 8.314 \times 273}{1.013 \times 10^{5}}$ = 0.02240 $m^{3}$ = 22.4 litres

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

Q.13.3 Figure 13.8 shows plot of PV/T versus P for 1.00×10–3 kg of oxygen gas at two different temperatures.

(a) What does the dotted plot signify?

(b) Which is true: T₁ > T₂ or T₁ < T₂?

(c) What is the value of PV/T where the curves meet on the y-axis?

(d) If we obtained similar plots for 1.00×10^–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 H₂ = 2.02 u, of O₂ = 32.0 u, R = 8.31 J mo1^–1 K^–1.)

Ans.13.3

(a) The dotted plot in the graph signifies the ideal behaviour of the gas, i.e. $\frac{P V}{T}$ = $μ$ R, (where $μ$ is the number of moles and R is the universal gas constant) is a constant quantity. It is not dependent on the pressure of the gas. The curve of the gas at temperature $T_{1}$ is closer to the dotted plot than the curve of the gas at temperature $T_{2 .}$

(b) A real gas approaches the behaviour of an ideal gas when its temperature increases. Therefore $T_{1}$ > $T_{2}$ is true for the given plot.

(c) The value of the ratio $\frac{P V}{T}$ , where the two curves meet, is $μ$ R. This is because the ideal gas equation is given as $\frac{P V}{T}$ = $μ$ R. From the given data, we have

Molecular mass of oxygen = 32 g

Mass of oxygen = 1 $\times 10^{- 3}$ = 1 g

R = 8.314 J/mole/K

Then $\frac{P V}{T}$ = $\frac{1}{32} \times 8.314$ = 0.26 J/k

Hence, the value of the ratio $\frac{P V}{T}$ , where the curves meet on the y-axis, is 0.26 J/K.

(d) If we obtain similar plots for 1.00 $\times 10^{- 3}$ kg of hydrogen, then we will not get the same values of $\frac{P V}{T}$ at the point where the curves meet on the y-axis. This is because the molecular mass of hydrogen (2.02 u) is different from the oxygen (32.0 u). We have:

$\frac{P V}{T} = 0.26 J / K$ .

Given, molecular mass (M) of $H_{2}$ = 2.02 u

$\frac{P V}{T}$ = $μ$ R at constant temperature, where $μ$ = $\frac{m}{M}$

M = mass of $H_{2}$

Then, m = $\frac{P V}{T} \times \frac{M}{R}$ = $\frac{0.26 \times 2.02}{8.31}$ = 6.3 $\times 10^{- 2}$ g = 6.3 $\times 10^{- 5}$ kg

Hence, 6.3 $\times 10^{- 5}$ kg of $H_{2}$ will yield the same value of $\frac{P V}{T}$

Q.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 O₂ = 32 u).

Ans.13.4

Volume of oxygen, $V_{1}$ = 30 litres = 30 $\times 10^{- 3} m^{3}$

Gauge pressure, $P_{1}$ = 15 atm = 15 $\times 1.013 \times 10^{5}$ Pa

Temperature, $T_{1}$ = 27 $℃$ = 300 K

Universal gas constant, R = 8.314 J/mole/K

Let the initial number of moles of oxygen gas cylinder be $n_{1}$

From the gas equation, we get $P_{1} V_{1} = n_{1} R T_{1}$

$n_{1} = \frac{P_{1} V_{1}}{R T_{1}}$ = $\frac{15 \times 1.013 \times 10^{5} \times 30 \times 10^{- 3}}{8.314 \times 300}$ = 18.276

But $n_{1}$ = $\frac{m_{1}}{M}$ , where $m_{1}$ = initial mass of oxygen

M = Molecular mass of oxygen = 32 g

$m_{1}$ = $n_{1}$ M = 18.276 $\times 32 = 584.84$ g

After some oxygen is withdrawn from the cylinder, the pressure and temperature reduces, but volume remained unchanged.

Hence, Volume, $V_{2}$ = 30 litres = 30 $\times 10^{- 3} m^{3}$

Gauge pressure, $P_{2}$ = 11 atm = 11 $\times 1.013 \times 10^{5}$ Pa

Temperature, $T_{2}$ = 17 $℃$ = 290 K

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

From the gas equation, we get $P_{2} V_{2} = n_{2} R T_{2}$

$n_{2} = \frac{P_{2} V_{2}}{R 2}$ = $\frac{11 \times 1.013 \times 10^{5} \times 30 \times 10^{- 3}}{8.314 \times 290}$ = 13.864

But $n_{2}$ = $\frac{m_{2}}{M}$ , where $m_{2}$ = remaining mass of oxygen

M = Molecular mass of oxygen = 32 g

$m_{2}$ = $n_{2}$ M = 13.864 $\times 32 = 443.65$ g

So the mass of oxygen taken out = $m_{1} - m_{2}$ = 141.19 gm = 0.141 kg

Commonly asked questions

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).

13.9 Temperature of the helium atom, $T_{H e}$ = – 20 °C = 253 K and temperature of argon atom be = $T_{A r}$

Atomic mass of helium, $M_{H e}$ = 4.0 u

Atomic mass of Argon, $M_{A r}$ = 39.9 u

Let ${V_{r m s}}_{A r}$ be the rms speed of Argon and ${V_{r m s}}_{H e}$ be the rms speed of Helium

From the relation of $V_{r m s} = \sqrt{\frac{3 k T}{M}}$ we get

rms speed of Argon, ${V_{r m s}}_{A r} = \sqrt{\frac{3 k T_{A r}}{M_{A r}}}$

rms speed of Helium, ${V_{r m s}}_{H e} = \sqrt{\frac{3 k T_{H e}}{M_{H e}}}$

Since both the speeds are equal, we get

$\sqrt{\frac{3 k T_{A r}}{M_{A r}}}$ = $\sqrt{\frac{3 k T_{H e}}{M_{H e}}}$ or $\frac{T_{A r}}{M_{A r}}$ = $\frac{T_{H e}}{M_{H e}}$ or $T_{A r}$ = $\frac{T_{H e} \times M_{A r}}{M_{H e}}$ = $\frac{253 \times 39.9}{4}$ = 2523.675 K = 2.523 $\times 10^{3}$ K

13.5 An air bubble of volume 1.0 cm3 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?

13.5 Volume of the air bubble $,$ $V_{1}$ = 1.0 ${c m}^{3}$ = 1 $\times 10^{- 6}$ $m^{3}$

Height achieved by bubble, d = 40 m

Temperature at a depth of 40 m, $T_{1}$ = 12 $?$ = 285 K

Temperature at the surface of the lake, $T_{2}$ = 35 $?$ = 308 K

The pressure at the surface of the lake, $P_{2}$ = 1 atm = 1.013 $\times 10^{5}$ Pa

The pressure at 40m depth: $P_{1}$ = 1 atm + d $? g$

Where $?$ is the density of water = $10^{3}$ kg/ $m^{3}$

G = acceleration due to gravity = 9.8 m/ $s^{2}$

Hence $P_{1}$ = 1.013 $\times 10^{5} + (40 \times 10^{3} \times 9.8)$ = 493300 Pa

From the relation $\frac{P_{1} V_{1}}{T_{1}}$ = $\frac{P_{2} V_{2}}{T_{2}}$ , where $V_{2}$ is the volume of bubble when it reaches the surface, we get, $V_{2}$ = $\frac{P_{1} V_{1} T_{2}}{T_{1} P_{2}}$ = $\frac{493300 \times 1 \times 10^{- 6} \times 308}{285 \times 1.013 \times 10^{5}}$ $=$ 5.263 $\times$ $10^{- 6} m^{3} = 5.263 {c m}^{3}$

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?

13.8 (a) According to Avogadro's law, the three vessels will contain an equal number of the respective molecules. This number is equal to Avogadro's number, N = 6.023 $* 10^{23}$

(b) The root mean square speed ( $V_{r m s})$ of a gas of mass m and temperature T is given by the relation $V_{r m s} = \sqrt{\frac{3 k T}{m}}$ . Where k is Boltzmann constant. For the given gases, k and T are constants. Hence $V_{r m s}$ depends only on the mass of the atoms

Therefore, the root mean square speed of the molecules in the three cases is not the same. Among Neon, Chlorine and Uranium hexafluoride, the mass of the neon is the smallest, so Neon will have the highest root mean square speed among the given gases.

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?

13.11 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, $l_{o}$ = 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

Hence, total length of the air column = 15 + 9 = 24 cm

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

Length of the air column in the bore = 24 + h cm

Length of the mercury column = 76 – h cm

Initial pressure, $P_{1}$ = 76 cm of mercury

Initial volume, $V_{1}$ = 15 ${c m}^{3}$

Final pressure, $P_{2}$ = 76 – (76 – h) = h cm of mercury

Final volume, $V_{2}$ = (24 + h) ${c m}^{3}$

Since temperature remained constant throughout the process,

$P_{1} V_{1}$ = $P_{2} V_{2}$ or 76 $* 15$ = h $* (24 + h)$

Or, $h^{2}$ + 24h - 76 $* 15$ =0

h = $\frac{- 24 \pm \sqrt{24^{2} + 4 * 76 * 15}}{2}$ = $\frac{- 24 \pm 71.67}{2}$

So h = 23.84 cm or -47.84 cm

Since height cannot be negative, so h = 23.84 cm

23.84 cm of mercury will flow out from the bore and (76-23.84) 52.16 cm of mercury will remain in the bore. The length of the air column will be 24 + 23.84 = 47.84 cm

13.12 From a certain apparatus, the diffusion rate of hydrogen has an average value of 28.7 cm³ s^–1. The diffusion of another gas under the same conditions is measured to have an average rate of 7.2 cm³ s^–1. Identify the gas.

[Hint : Use Graham’s law of diffusion: R1/R2 = ( M₂ /M₁ )^1/2, where R₁, R₂ are diffusion rates of gases 1 and 2, and M₁ and M₂ their respective molecular masses. The law is a simple consequence of kinetic theory.]

13.12 Rate of diffusion of hydrogen, $R_{1}$ = 28.7 cm³ s^–1

Rate of diffusion of another gas, $R_{2}$ = 7.2 cm³ s^–1

According to Graham's law of diffusion, we have:

$\frac{R_{1}}{R_{2}}$ = $\sqrt{\frac{M_{2}}{M_{1}}}$ , where $M_{1}$ = molecular mass of hydrogen = 2.02 g and $M_{2}$ is the molecular mass of the unknown gas

$M_{2} = M_{1} * ({\frac{R_{1}}{R_{2}})}^{2}$ = 2.02 $* ({\frac{28.7}{7.2})}^{2}$ = 32.09 = 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

n₂ = n₁ exp [ -mg (h₂ – h₁)/ k_BT]

where n₂, n₁ refer to number density at heights h₂ and h1 respectively. Use this relation to derive the equation for sedimentation equilibrium of a suspension in a liquid column:

n₂ = n₁ exp [ -mg N_A ( $ρ$ - $ρ^{'}$ ) (h₂ –h₁₎/ ( $ρ$ RT)]

where $ρ$ is the density of the suspended particle, and 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.]

13.13 According to law of atmospheres, we have

n₂ = n₁ exp [ -mg (h₂ – h₁)/ k_BT] …..(i)

where $n_{1}$ is the number of density at height $h_{1}$ and $n_{2}$ is the number of density at height $h_{2}$

mg is the weight of the particle suspended in the gas column

Density of the medium = $?$

Density of the suspended particle = $?^{'}$

Mass of one suspended particle = m’

Mass of medium displaced = m

Volume of the suspended particle = V

According to Archimedes’s principle for a particle suspended in a liquid column, the effective weight of the suspended particle is given as:

Weight of the medium displaced – weight of the suspended particle = mg – m’g

= mg- V $?^{'} g$ = mg – ( $\frac{m}{?}$ ) $?^{'} g$

= mg(1- $\frac{?^{'}}{?}$ ) ……(ii)

Gas constant R = $k_{B}$ N

$k_{B} = \frac{R}{N}$ …(iii)

Substituting from (ii) and (iii) in (i), we get

n₂ = n₁ exp [ -mg (h₂ – h₁)/ k_BT]

= n1 exp [ $[- m g (- \frac{?^{'}}{?}) (h 2 - h 1) \frac{N}{R T}]$ ]

= n1 exp[ $[- m g (? - ?^{'}) (h 2 - h 1) \frac{N}{R T ?}]$ ]

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 ( $10^{3} K g m^{- 3})$
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	1.14

[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].

13.14 Let the atomic mass of a substance be = M and the density of the substance be = $?$

Avogadro's number, N = 6.023 $* 10^{23}$

Volume of N number of molecules = $\frac{4}{3} ? r^{3} N$ ……. (i)

Volume of one mole of a substance = $\frac{M}{?}$ …… (ii)

Equating (i) and (ii), we get

$\frac{4}{3} ? r^{3} N = \frac{M}{?}$

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 3 Å.

13.1 Diameter of the oxygen molecule, d = 3Å

Radius, r = $\frac{d}{2}$ = 1.5 Å = 1.5 $* 10^{- 8}$ cm

Actual volume occupied by 1 mole of oxygen gas at STP = 22.4 lit = 22400 ${c m}^{3}$

Molecular volume of oxygen gas, V = $\frac{4}{3} ? r^{3}$ N,

where N = Avogadro's number = 6.023 $* 10^{23}$ molecules/mole

Therefore, V = $\frac{4}{3}$ $* 3.1416 * ({1.5 * 10^{- 8})}^{3}$ $*$ 6.023 $* 10^{23}$ = 8.51 ${c m}^{3}$

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.

13.2 The ideal gas equation relating to pressure (P), volume (V) and absolute temperature (T) is given by the relation

PV = nRT, where

R is the universal gas constant = 8.314 J /mol/K

N = number of moles = 1 (given)

T = standard temperature = 273 K

P = standard pressure = 1 atm = 1.013 $* 10^{5}$ N/ $m^{2}$

Therefore V = $\frac{n R T}{P}$ = $\frac{1 * 8.314 * 273}{1.013 * 10^{5}}$ = 0.02240 $m^{3}$ = 22.4 litres

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

13.3 Figure 13.8 shows plot of PV/T versus P for 1.00×10–3 kg of oxygen gas at two different temperatures.

(a) What does the dotted plot signify?

(b) Which is true: T₁ > T₂ or T₁ < T₂?

(c) What is the value of PV/T where the curves meet on the y-axis?

(d) If we obtained similar plots for 1.00×10^–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 H₂ = 2.02 u, of O₂ = 32.0 u, R = 8.31 J mo1^–1 K^–1.)

13.3 (a) The dotted plot in the graph signifies the ideal behaviour of the gas, i.e. $\frac{P V}{T}$ = $?$ R, (where $?$ is the number of moles and R is the universal gas constant) is a constant quantity. It is not dependent on the pressure of the gas. The curve of the gas at temperature $T_{1}$ is closer to the dotted plot than the curve of the gas at temperature $T_{2 .}$

(b) A real gas approaches the behaviour of an ideal gas when its temperature increases. Therefore $T_{1}$ > $T_{2}$ is true for the given plot.

(c) The value of the ratio $\frac{P V}{T}$ , where the two curves meet, is $?$ R. This is because the ideal gas equation is given as $\frac{P V}{T}$ = $?$ R. From the given data, we have

Molecular mass of oxygen = 32 g

Mass of oxygen = 1 $\times 10^{- 3}$ = 1 g

R = 8.314 J/mole/K

Then $\frac{P V}{T}$ = $\frac{1}{32} \times 8.314$ = 0.26 J/k

Hence, the value of the ratio $\frac{P V}{T}$ , where the curves meet on the y-axis, is 0.26 J/K.

$\frac{P V}{T} = 0.26 J / K$ .

Given, molecular mass (M) of $H_{2}$ = 2.02 u

$\frac{P V}{T}$ = $?$ R at constant temperature, where $?$ = $\frac{m}{M}$

M = mass of $H_{2}$

Then, m = $\frac{P V}{T} \times \frac{M}{R}$ = $\frac{0.26 \times 2.02}{8.31}$ = 6.3 $\times 10^{- 2}$ g = 6.3 $\times 10^{- 5}$ kg

Hence, 6.3 $\times 10^{- 5}$ kg of $H_{2}$ will yield the same value of $\frac{P V}{T}$

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 O₂ = 32 u).

13.4 Volume of oxygen, $V_{1}$ = 30 litres = 30 $\times 10^{- 3} m^{3}$

Gauge pressure, $P_{1}$ = 15 atm = 15 $\times 1.013 \times 10^{5}$ Pa

Temperature, $T_{1}$ = 27 $?$ = 300 K

Universal gas constant, R = 8.314 J/mole/K

Let the initial number of moles of oxygen gas cylinder be $n_{1}$

From the gas equation, we get $P_{1} V_{1} = n_{1} R T_{1}$

$n_{1} = \frac{P_{1} V_{1}}{R T_{1}}$ = $\frac{15 \times 1.013 \times 10^{5} \times 30 \times 10^{- 3}}{8.314 \times 300}$ = 18.276

But $n_{1}$ = $\frac{m_{1}}{M}$ , where $m_{1}$ = initial mass of oxygen

M = Molecular mass of oxygen = 32 g

$m_{1}$ = $n_{1}$ M = 18.276 $\times 32 = 584.84$ g

After some oxygen is withdrawn from the cylinder, the pressure and temperature reduces, but volume remained unchanged.

Hence, Volume, $V_{2}$ = 30 litres = 30 $\times 10^{- 3} m^{3}$

Gauge pressure, $P_{2}$ = 11 atm = 11 $\times 1.013 \times 10^{5}$ Pa

Temperature, $T_{2}$ = 17 $?$ = 290 K

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

From the gas equation, we get $P_{2} V_{2} = n_{2} R T_{2}$

$n_{2} = \frac{P_{2} V_{2}}{R 2}$ = $\frac{11 \times 1.013 \times 10^{5} \times 30 \times 10^{- 3}}{8.314 \times 290}$ = 13.864

But $n_{2}$ = $\frac{m_{2}}{M}$ , where $m_{2}$ = remaining mass of oxygen

M = Molecular mass of oxygen = 32 g

$m_{2}$ = $n_{2}$ M = 13.864 $\times 32 = 443.65$ g

So the mass of oxygen taken out = $m_{1} - m_{2}$ = 141.19 gm = 0.141 kg

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 $m^{3}$ at a temperature of 27 °C and 1 atm pressure.

13.6 Volume of the room, V = 25.0 $m^{3}$

Temperature of the room, T = 27 $?$ = 300 K

Pressure of the room, P = 1 atm = 1 $\times$ 1.013 $\times 10^{5}$ Pa

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

PV = $k_{B}$ NT, where $k_{B}$ is Boltzmann constant = 1.38 $\times 10^{- 23}$ $m^{2} k g s^{- 2} K^{- 1}$

N is the number of air molecules in the room.

N = $\frac{P V}{k_{B} T}$ = $\frac{1 \times 1.013 \times 10^{5} \times 25}{1.38 \times 10^{- 23} \times 300}$ = 6.117 $\times 10^{26}$

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).

13.7 (i) At room temperature, T = 27 $?$ = 300 K

$k_{B}$ is Boltzmann constant = 1.38 $\times 10^{- 23}$ $m^{2} k g s^{- 2} K^{- 1}$

Average thermal energy = $\frac{3}{2} k_{B} T$ = $\frac{3 \times 1.38 \times 10^{- 23} \times 300}{2}$ = 6.21 $\times 10^{- 21}$ J

Hence, the average thermal energy of a helium atom at room temperature is 6.21 $\times 10^{- 21}$ J

(ii) On the surface of the Sun, T = 6000 K

Hence average thermal energy = $\frac{3}{2} k_{B} T$ = $\frac{3 \times 1.38 \times 10^{- 23} \times 6000}{2}$ = 1.242 $\times 10^{- 19}$ J

Hence, the average thermal energy of a helium atom on the surface of the Sun is 1.242 $\times 10^{- 19}$ J

(iii) Inside the core of a star, T = $10^{7}$ K

Hence average thermal energy = $\frac{3}{2} k_{B} T$ = $\frac{3 \times 1.38 \times 10^{- 23} \times 10^{7}}{2}$ = 2.07 $\times 10^{- 16}$ J

Hence, the average thermal energy of a helium atom inside the core of a star is 2.07 $\times 10^{- 16}$ J

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 $?$ . Take the radius of a nitrogen molecule to be roughly 1.0 Å. Compare the collision time with the time the molecule moves freely between two successive collisions (Molecular mass of $N_{2}$ = 28.0 u).

13.10 Pressure inside the cylinder containing nitrogen, P = 2.0 atm = 2 $\times 1.013 \times 10^{5}$ Pa

Temperature inside the cylinder, T = 17 $? = 290 K$

Radius of nitrogen molecule, r = 1.0 Å = 1 $\times 10^{- 10}$ m

Diameter of nitrogen molecule, d = 2 $\times 10^{- 10}$ m

Molecular mass of nitrogen molecule, M = 28 u = 28 g (assume) = 28 $\times 10^{- 3}$ kg

The root means square speed of nitrogen is given by the relation

R is the universal gas constant = 8.314 J/mole/K

Hence

The mean free path $l$ is given by

$l ? = \frac{k T}{? 2 ? d^{2} P}$ where k = Boltzmann constant = 1.38 $\times 10^{- 23}$ kg- $m^{2} s^{- 2} K^{- 1}$

$\overset{?}{l} = \frac{1.38 \times 10^{- 23} \times 290}{? 2 ? {(2 \times 10^{- 10})}^{2} \times 2 \times 1.013 \times 10^{5}}$ = 1.11 $\times 10^{- 7}$ m

Collision frequency = $\frac{V_{r m s}}{\overset{?}{l}}$ = $\frac{508.26}{1.11 \times 10^{- 7}}$ = 4.57 $10^{9}$ /s

Collision time, T = $\frac{d}{V_{r m s}}$ = $\frac{2 \times 10^{- 10}}{508.26}$ S= 3.93 $\times 10^{- 13}$ s

Time taken between successive collision T’ = $\frac{\overset{?}{l}}{V_{r m s}}$ = $\frac{1.11 \times 10^{- 7}}{508.26}$ s = 2.18 $\times 10^{- 10}$
s $\frac{T ’}{T}$ = $\frac{2.18 \times 10^{- 10}}{3.93 \times 10^{- 13}}$ = 555.71

Hence, the time taken between successive collisions is 556 times the time taken for a collision.

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