Calculate the r.m.s. speed of smoke particles of mass 5 × 10 - 17 kg in Brownian motion in air at NTP . Given : Boltzmann constant k = 1 . 38 × 10 - 23 JK - 1 .

According to Kinetic theory of gas, Average kinetic energy of gas molecule = 1 2 m v 2 r m s = 3 2 k B T Therefore, v r m s = 3 k B T m = 3 × 1 . 38 × 10 - 23 × 273 5 × 10 - 17 v r m s = 1 . 5 × 10 - 2 m / s

There are 5 . 5 × 10 24 hydrogen molecules in a 100 cm 3 flask. If the mass of each molecule be 3 . 32 × 10 - 27 kg and rms speed 300 m s - 1 , find the pressure of the flask in atmosphere.

Pressure of flask is given by P = 1 3 m N V v 2 r m s P = 1 3 3 . 32 × 10 - 27 × 5 . 5 × 10 24 10 - 4 × ( 300 ) 2 P = 1643400 3 × 10 1 P = 5478 × 10 3 P = 5 . 478 × 10 6 N / m 2

A container of capacity 1 1 litre contains 2 . 5 × 10 23 nitrogen molecules, each of mass 4 . 65 × 10 - 26 kg and rms speed 500 ms - 1 . Find the pressure and the total kinetic energy of gas.

Pressure is given by P = 1 3 m n v 2 r m s Mass of nitrogen molecule, m = 4 . 65 × 10 - 26 k g RMS speed, v r m s = 500 m / s n = number of molecules /m 3 n = 2 . 5 × 10 23 × 10 3 = 2 . 5 × 10 26 Therefore, P = 1 3 × 4 . 65 × 10 - 26 × 2 . 5 × 10 26 × ( 500 ) 2 P = 9 . 69 N / m 2 Total kinetic energy, K . E = 1 2 M N c 2 Where, N = no. of molecules of the gas ⇒ K . E = 1 2 × 4 . 65 × 10 - 26 × 2 . 5 × 10 23 × ( 500 ) 2 = 1 . 45 × 10 3 J

The average kinetic energy of a molecule of hydrogen at 0 ° C is 5 . 64 × 10 - 21 J and molar gas constant ( R ) = 8 . 32 J K - 1 . Calculate Avogadro number.

Here, T = 0 ∘ C = 273 K Kinetic energy per molecules = 1 2 m υ 2 = 5 . 64 × 10 - 21 J R = 8 . 31 J m o l - 1 K - 1 N = ? As Kinetic Energy per gram mole = 3 2 R T ∴ 5 . 64 × 10 - 21 × N = 32 × 8 . 31 × 273 N = 3 × 8 . 31 × 273 2 × 5 . 64 × 10 - 21 N = 6 . 033 × 10 23

HARD

11th West Bengal Board

IMPORTANT

Earn 100

Estimate the fraction of molecular volume to the actual volume occupied by oxygen gas at $N . T . P .$ . Take the diameter of an oxygen molecule to be $3 \dot{A}$ .

Important Questions on Kinetic Theory of Gases

MEDIUM

11th West Bengal Board

IMPORTANT

A Text Book of PHYSICS PART I : CLASS 11>Chapter 20 - Kinetic Theory of Gases>EXERCISE>Q 14

Calculate the r.m.s. speed of smoke particles of mass

5 \times 10^{- 17} kg

in Brownian motion in air at

NTP

. Given : Boltzmann constant

k = 1.38 \times 10^{- 23} {JK}^{- 1}

HARD

11th West Bengal Board

IMPORTANT

A Text Book of PHYSICS PART I : CLASS 11>Chapter 20 - Kinetic Theory of Gases>EXERCISE>Q 15

Assuming the air to be composed of nitrogen molecules. Find the mean free path of air molecules at

S . T . P .

Given : radius of nitrogen molecule

r = 1.7 \times 10^{- 10} m

MEDIUM

11th West Bengal Board

IMPORTANT

A Text Book of PHYSICS PART I : CLASS 11>Chapter 20 - Kinetic Theory of Gases>EXERCISE>Q 1

If water molecules in

1 g

of water were distributed uniformly over the surface of the earth how many such molecules would there be per unit area of the earth's surface? Radius of the earth

= 6.4 \times 10^{8} cm

and Avogadro number

= 6.03 \times 10^{23}

HARD

11th West Bengal Board

IMPORTANT

A Text Book of PHYSICS PART I : CLASS 11>Chapter 20 - Kinetic Theory of Gases>EXERCISE>Q 2

There are

5.5 \times 10^{24}

hydrogen molecules in a

100 {cm}^{3}

flask. If the mass of each molecule be

3.32 \times 10^{- 27} kg

and

rms

speed

300 m s^{- 1}

, find the pressure of the flask in atmosphere.

HARD

11th West Bengal Board

IMPORTANT

A Text Book of PHYSICS PART I : CLASS 11>Chapter 20 - Kinetic Theory of Gases>EXERCISE>Q 3

A container of capacity

1

1 litre contains

2.5 \times 10^{23}

nitrogen molecules, each of mass

4.65 \times 10^{- 26} kg

and

rms

speed

500 {ms}^{- 1}

. Find the pressure and the total kinetic energy of gas.

HARD

11th West Bengal Board

IMPORTANT

A Text Book of PHYSICS PART I : CLASS 11>Chapter 20 - Kinetic Theory of Gases>EXERCISE>Q 4

Calculate the number of molecules per

{cm}^{3}

of an ideal gas at

27 ° C

and

1 cm

of mercury pressure. Mean kinetic energy per molecule at

27 ° C = 4 \times 10^{- 4} erg

and density of mercury

= 13.6 g {cm}^{- 3}

HARD

11th West Bengal Board

IMPORTANT

A Text Book of PHYSICS PART I : CLASS 11>Chapter 20 - Kinetic Theory of Gases>EXERCISE>Q 5

Find out the number of molecules in a gas contained in a volume of

1 {cm}^{3}

at a pressure of

10^{- 3}

atmosphere and at temperature of

- 27 ° C

R_{0} = 8.31

Joule

{mole}^{- 1} K^{- 1}

; Avogadro number

= 6.03 \times 10^{23}

HARD

11th West Bengal Board

IMPORTANT

A Text Book of PHYSICS PART I : CLASS 11>Chapter 20 - Kinetic Theory of Gases>EXERCISE>Q 6

The average kinetic energy of a molecule of hydrogen at

0 ° C

5.64 \times 10^{- 21} J

and molar gas constant

(R) = 8.32 J K^{- 1}

. Calculate Avogadro number.

Estimate the fraction of molecular volume to the actual volume occupied by oxygen gas at N.T.P.. Take the diameter of an oxygen molecule to be 3A˙.

Important Questions on Kinetic Theory of Gases

Learn and Prepare for any exam you want !

Estimate the fraction of molecular volume to the actual volume occupied by oxygen gas at $N . T . P .$ . Take the diameter of an oxygen molecule to be $3 \dot{A}$ .