The kinetic energy of 1 kg of oxygen at 300 K is 1 . 356 × 10 6 J . Find the kinetic energy of 4 kg of oxygen at 400 K .

Given, m = 1 kg , T = 300 K , KE = 1 · 356 × 10 6 J Now, M new = 4 kg T new = 400 K KE of 1 mole = 3 2 RT KE of n mole = 3 2 nRT = 3 2 m M RT KE new = 3 2 4000 M , × R × 400 ...................(i) KE = 3 2 1000 M × R × 300 .....................(ii) Dividing Eq. ( i ) from Eq. ( i i ) , we get KE new KE = 4000 × 400 1000 × 300 KE New = 16 3 KE = 16 3 × 1 . 356 × 10 6 KE new = 7 . 232 × 10 6 J

The kinetic energy of 1 kg of oxygen at 300 K is 1 . 356 × 10 6 J . Find the kinetic energy of 4 kg of oxygen at 400 K .

Given, m = 1 kg , T = 300 K , KE = 1 · 356 × 10 6 J Now, M new = 4 kg T new = 400 K KE of 1 mole = 3 2 RT KE of n mole = 3 2 nRT = 3 2 m M RT KE new = 3 2 4000 M , × R × 400 ...................(i) KE = 3 2 1000 M × R × 300 .....................(ii) Dividing Eq. ( i ) from Eq. ( i i ) , we get KE new KE = 4000 × 400 1000 × 300 KE New = 16 3 KE = 16 3 × 1 . 356 × 10 6 KE new = 7 . 232 × 10 6 J

HARD

Earn 100

The kinetic energy of $1 kg$ of oxygen at $300 K$ is $1.356 \times 10^{6} J$ . Find the kinetic energy of $4 kg$ of oxygen at $400 K$ .

Important Questions on Kinetic Theory

EASY

Physics>Heat and Thermodynamics>Kinetic Theory>Law of Equipartition of Energy

Which of the following shows the correct relationship between the pressure

' P'

and density

ρ

of an ideal gas at constant temperature ?

EASY

Physics>Heat and Thermodynamics>Kinetic Theory>Law of Equipartition of Energy

A container is divided into two equal part I and II by a partition with small hole of diameter d. The two partitions are filled with same ideal gas, but held at temperature

T_{I} = 150 K

and

T_{I I} = 300 K

by connecting to heat reservoirs. Let

λ_{I}

and

λ_{I I}

be the mean free paths of the gas particles in the two parts such that

d > > λ_{I}

and

d > > λ_{I I} .

Then

\frac{λ_{I}}{λ_{I I}}

is close to.

MEDIUM

Physics>Heat and Thermodynamics>Kinetic Theory>Law of Equipartition of Energy

A gas molecule of mass

M

at the surface of the earth has kinetic energy equivalent to

0 ° C

. If it were to go up straight without colliding with any other molecules, how high would it rise? Assume that the height attained is much less than the radius of the earth. (

k_{B}

is Boltzmann constant)

MEDIUM

Physics>Heat and Thermodynamics>Kinetic Theory>Law of Equipartition of Energy

The temperature, at which the root mean square velocity of hydrogen molecules equals their escape their escape velocity from the earth, is closest to:

[

Boltzmann Constant

k_{B} = 1.38 \times 10^{- 23} J / K

Avogadro number

N_{A} = 6.02 \times 10^{26} / k g

Radius of Earth:

6.4 \times 10^{6} m

Gravitational acceleration on Earth

= 10 m s^{- 2}]

EASY

Physics>Heat and Thermodynamics>Kinetic Theory>Law of Equipartition of Energy

The molecules of a given mass of gas have RMS velocity of

200

m s^{- 1}

27^{o} C

and

1.0 \times 10^{5} N m^{- 2}

pressure. When the temperature and pressure of the gas are respectively,

127^{o} C

and

0.05 \times 10^{5} N m^{- 2}

, the r.m.s. velocity of its molecules in

m s^{- 1}

is:

MEDIUM

Physics>Heat and Thermodynamics>Kinetic Theory>Law of Equipartition of Energy

10^{22}

gas molecules each of mass

10^{- 26}

kg collides with a surface (perpendicular to it) elastically per second over an area

1 m^{2}

with a speed

10^{4} m / s,

the pressure exerted by the gas molecules will be of the order of:

EASY

Physics>Heat and Thermodynamics>Kinetic Theory>Law of Equipartition of Energy

The temperature of an ideal gas is increased from

100 K

400 K

. If the

r m s

speed of the gas molecule is

V

100 K

, then at

400 K

it becomes

EASY

Physics>Heat and Thermodynamics>Kinetic Theory>Law of Equipartition of Energy

The mean free path of molecules of a gas (radius ‘

r

’) is inversely proportional to:

MEDIUM

Physics>Heat and Thermodynamics>Kinetic Theory>Law of Equipartition of Energy

An ideal gas is enclosed in a cylinder at pressure of

2

atm and temperature,

300 K .

The mean time between two successive collisions is

6 \times 10^{- 8} s .

If the pressure is doubled and temperature is increased to

500 K,

the mean time between two successive collisions will be close to:

MEDIUM

Physics>Heat and Thermodynamics>Kinetic Theory>Law of Equipartition of Energy

15 g

mass of nitrogen gas is enclosed in a vessel at a temperature,

27^{o} C

. The amount of heat transferred to the gas, so that

R . M . S .

velocity of molecules is doubled, is about.

[R = 8 .3 J {(K mole)}^{- 1}]

MEDIUM

Physics>Heat and Thermodynamics>Kinetic Theory>Law of Equipartition of Energy

25 \times 10^{- 3} m^{3}

volume cylinder is filled with

1

mol of

O_{2}

gas at room temperature

(300 K)

. The molecular diameter of

O_{2}

, and its root mean square speed, are found to be

0.3 n m

and

200 m s^{- 1}

, respectively. What is the average collision rate (per second) for an

O_{2}

molecule?

EASY

Physics>Heat and Thermodynamics>Kinetic Theory>Law of Equipartition of Energy

Given below are two statements:

Statement I: In a diatomic molecule, the rotational energy at a given temperature obeys Maxwell's distribution.

Statement II : In a diatomic molecule, the rotational energy at a given temperature equals the translational kinetic energy for each molecule.

In the light of the above statements, choose the correct answer from the options given below:

EASY

Physics>Heat and Thermodynamics>Kinetic Theory>Law of Equipartition of Energy

The number density of molecules of a gas depends on their distance r from the origin as,

n (r) = n_{0} e^{- α r^{4}} .

Then the numer of molecules is proportional to:

EASY

Physics>Heat and Thermodynamics>Kinetic Theory>Law of Equipartition of Energy

Increase in temperature of a gas filled in a container will lead to

EASY

Physics>Heat and Thermodynamics>Kinetic Theory>Law of Equipartition of Energy

For given gas at

1 a t m

pressure,

r m s

speed of the molecules is

200 m / s

127 ° C .

2 a t m

pressure and at

227 ° C,

the

r m s

speed of the molecules will be:

EASY

Physics>Heat and Thermodynamics>Kinetic Theory>Law of Equipartition of Energy

The root mean square speed of smoke particles each of mass

5 \times 10^{- 17} k g

in their Brownian motion in air at

N . T . P

EASY

Physics>Heat and Thermodynamics>Kinetic Theory>Law of Equipartition of Energy

H C l

molecule has rotational, translational and vibrational motions. If the rms velocity of

H C l

molecules in its gaseous phase is

\bar{ν}

, m is its mass and

k_{B}

is Boltzmann's constant, then its temperature will be:

HARD

Physics>Heat and Thermodynamics>Kinetic Theory>Law of Equipartition of Energy

The number of gas molecules striking per second per square meter of the top surface of a table placed in a room at

20 ° C

and

1

atmospheric pressure is of the order of (

k_{b} = 1.4 \times 10^{- 23} J K^{- 1}

and the average mass of an air molecule is

5 \times 10^{- 27} kg

)

HARD

Physics>Heat and Thermodynamics>Kinetic Theory>Law of Equipartition of Energy

Consider an ideal gas confined in an isolated closed chamber. As the gas undergoes an adiabatic expansion, the average time of collision between molecules increases as $V^{q}$ , where $V$ is the volume of the gas. The value of $q$ is:

$(γ = \frac{C_{P}}{C_{v}})$

EASY

Physics>Heat and Thermodynamics>Kinetic Theory>Law of Equipartition of Energy

A man is climbing up a spiral staircase. His degrees of freedom are

The kinetic energy of 1kg of oxygen at 300K is 1.356×106J. Find the kinetic energy of 4kg of oxygen at 400K.

Important Questions on Kinetic Theory

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The kinetic energy of $1 kg$ of oxygen at $300 K$ is $1.356 \times 10^{6} J$ . Find the kinetic energy of $4 kg$ of oxygen at $400 K$ .