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

The average translational energy and the rms speed of molecules in a gas at 300 K are 6 . 21 × 10 - 21 J and 484 m s - 1 respectively. What are the corresponding values at 600 K (assuming ideal gas behaviour)?

The average translational K E = 3 2 k T which is directly proportional to T , while r.m.s speed of molecules is given by v r m s = 3 R T M i . e . , v r m s = T When temperature of gas is increased from 300 K to 600 K (i.e. 2 times), the average translational KE will increase to 2 times and r.m.s speed to 2 o r 1 . 414 times. &there4; Average translational K . E = 2 × 6 . 21 × 10 - 21 J = = 12 . 42 × 10 - 21 J And RMS speed v r m s = ( 1 . 414 ) ( 484 ) m / s = 684 m / s

HARD

11th West Bengal Board

IMPORTANT

Earn 100

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 $= 6.03 \times 10^{23}$ .

Important Questions on Kinetic Theory of Gases

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.

HARD

11th West Bengal Board

IMPORTANT

A Text Book of PHYSICS PART I : CLASS 11>Chapter 20 - Kinetic Theory of Gases>EXERCISE>Q 7

Calculate

the $K . E .$ of translation of an oxygen molecule at $27 ° C$ and

the total $K . E .$ of a mole of oxygen at $27 ° C$ . Avogadro number $= 6.023 \times 10^{23}$ and Boltzmann's constant $= 1.37 \times 10^{- 23} {JK}^{- 1}$ ; Oxygen molecule is diatomic.

HARD

11th West Bengal Board

IMPORTANT

A Text Book of PHYSICS PART I : CLASS 11>Chapter 20 - Kinetic Theory of Gases>EXERCISE>Q 8

With what speed would one mole of oxygen at

300 K

be moving in order that the translational

K . E

. of its centre of mass be equal to the total

K . E

of all its molecules? Molecular weight of oxygen

= 32

HARD

11th West Bengal Board

IMPORTANT

A Text Book of PHYSICS PART I : CLASS 11>Chapter 20 - Kinetic Theory of Gases>EXERCISE>Q 9

The average translational energy and the

rms

speed of molecules in a gas at

300 K

are

6.21 \times 10^{- 21} J

and

484 m s^{- 1}

respectively. What are the corresponding values at

600 K

(assuming ideal gas behaviour)?

HARD

11th West Bengal Board

IMPORTANT

A Text Book of PHYSICS PART I : CLASS 11>Chapter 20 - Kinetic Theory of Gases>EXERCISE>Q 10

Two closed vessels of equal volume contain air at a pressure of

300 kpa

and temperature

27 ° C

. The two are connected by a narrow tube of negligible volume. If one of the vessels is now maintained at

27 ° C

and the other at

127 ° C

, what will be the pressure in the vessels?

HARD

11th West Bengal Board

IMPORTANT

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

A vessel of volume

2 L

contains

0.1

mole of oxygen and

0.2

moles of carbon dioxide. If the temperature of the mixture is

300 K

, find the pressure.

HARD

11th West Bengal Board

IMPORTANT

A Text Book of PHYSICS PART I : CLASS 11>Chapter 20 - Kinetic Theory of Gases>EXERCISE>Q 12

Two vessels of equal volume contain ideal gas at pressures and temperatures

p_{1}, T_{1}

and

p_{2}, T_{2}

respectively. They are now connected by a narrow tube of negligible volume. Find the relation between the equilibrium pressure

p

and equilibrium temperature

T

HARD

11th West Bengal Board

IMPORTANT

A Text Book of PHYSICS PART I : CLASS 11>Chapter 20 - Kinetic Theory of Gases>EXERCISE>Q 13

Show that the molecular collision frequency for a given type of gas in a container of constant volume varies as the square root of the absolute temperature.

Find out the number of molecules in a gas contained in a volume of 1 cm3 at a pressure of 10-3 atmosphere and at temperature of -27°C. R0=8.31 Joule mole-1 K-1; Avogadro number =6.03×1023.

Important Questions on Kinetic Theory of Gases

Learn and Prepare for any exam you want !

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 $= 6.03 \times 10^{23}$ .