QBManager

Kinetic Theory of Gases

139292 If the degrees of freedom of a gas are $f$ then the ratio of its specific heat $\frac{C_{p}}{C_{v}}$ is given by

1

1 + \frac{2}{f}

2

1 - \frac{2}{f}

3

1 + \frac{1}{f}

4

1 - \frac{1}{f}

Explanation:

A We know that,
Ratio of two specific heat of gas is:-
$\frac{C_{p}}{C_{v}} = γ$
We know that,
$γ = 1 + \frac{2}{f}$
Where, $f =$ degree of Freedom of gas

AMU-2002

Kinetic Theory of Gases

139293 At which of the following temperature would the molecules of a gas have twice, the average kinetic energy they have at $20^{\circ} C$ ?

1

313^{\circ} C

2

373^{\circ} C

3

393^{\circ} C

4

586^{\circ} C

Explanation:

A Given that,
$T_{1} = 20^{\circ} C = 273 + 20$
$T_{2} = 273 + T$
$E_{1} = E$
$E_{2} = 2 E$
We know that,
Kinetic energy $(E) = \frac{3}{2} RT$
$E \propto T$
$\frac{E_{1}}{E_{2}} = \frac{T_{1}}{T_{2}}$
$\frac{E}{2 E} = \frac{273 + 20}{273 + T}$
$586 = 273 + T$
$T = 313^{\circ} C$

AMU-2002

Kinetic Theory of Gases

139295 The gases carbon-monoxide (CO) and nitrogen at the same temperature have kinetic energies $E_{1}$ and $E_{2}$ respectively. Then

1

E_{1} = E_{2}

2

E_{1} > E_{2}

3

E_{1} < E_{2}

4

E_{1}

and

E_{2}

cannot be compared

Explanation:

A We know that,
$KE \propto \frac{3}{2} kT$
Here, Hence,
$CO$ and $N_{2}$ both are diatomic gas
${({KE}_{1})}_{Co} = \frac{5}{2} kT$
${({KE}_{2})}_{N 2} = \frac{5}{2} kT$
Hence,
For same Temperature, $E_{1} = E_{2}$ .

MHT-CET 2005

Kinetic Theory of Gases

139296 The degrees of freedom of a molecule of a triatomic gas are

1 2

2 4

3 6

4 8

Explanation:

C Degree of Freedom (f) $= 3 N - n$
Where, $N =$ total no of particle
$n =$ holonomic constraints
For triatomic molecule-
no of particle $(N) = 3$
and Separation between 3 atom is Fixed i.e.
no of holonomic constraints $(n) = 3$ Hence,
$f = 3 N - n$
$= 3 \times 3 - 3$
$= 9 - 3$
$f = 6$
Degree of Freedom of diatomic molecule is 6.

AIPMT-1999

Kinetic Theory of Gases

139292 If the degrees of freedom of a gas are $f$ then the ratio of its specific heat $\frac{C_{p}}{C_{v}}$ is given by

1

1 + \frac{2}{f}

2

1 - \frac{2}{f}

3

1 + \frac{1}{f}

4

1 - \frac{1}{f}

Explanation:

A We know that,
Ratio of two specific heat of gas is:-
$\frac{C_{p}}{C_{v}} = γ$
We know that,
$γ = 1 + \frac{2}{f}$
Where, $f =$ degree of Freedom of gas

AMU-2002

Kinetic Theory of Gases

139293 At which of the following temperature would the molecules of a gas have twice, the average kinetic energy they have at $20^{\circ} C$ ?

1

313^{\circ} C

2

373^{\circ} C

3

393^{\circ} C

4

586^{\circ} C

Explanation:

A Given that,
$T_{1} = 20^{\circ} C = 273 + 20$
$T_{2} = 273 + T$
$E_{1} = E$
$E_{2} = 2 E$
We know that,
Kinetic energy $(E) = \frac{3}{2} RT$
$E \propto T$
$\frac{E_{1}}{E_{2}} = \frac{T_{1}}{T_{2}}$
$\frac{E}{2 E} = \frac{273 + 20}{273 + T}$
$586 = 273 + T$
$T = 313^{\circ} C$

AMU-2002

Kinetic Theory of Gases

139295 The gases carbon-monoxide (CO) and nitrogen at the same temperature have kinetic energies $E_{1}$ and $E_{2}$ respectively. Then

1

E_{1} = E_{2}

2

E_{1} > E_{2}

3

E_{1} < E_{2}

4

E_{1}

and

E_{2}

cannot be compared

Explanation:

A We know that,
$KE \propto \frac{3}{2} kT$
Here, Hence,
$CO$ and $N_{2}$ both are diatomic gas
${({KE}_{1})}_{Co} = \frac{5}{2} kT$
${({KE}_{2})}_{N 2} = \frac{5}{2} kT$
Hence,
For same Temperature, $E_{1} = E_{2}$ .

MHT-CET 2005

Kinetic Theory of Gases

139296 The degrees of freedom of a molecule of a triatomic gas are

1 2

2 4

3 6

4 8

Explanation:

C Degree of Freedom (f) $= 3 N - n$
Where, $N =$ total no of particle
$n =$ holonomic constraints
For triatomic molecule-
no of particle $(N) = 3$
and Separation between 3 atom is Fixed i.e.
no of holonomic constraints $(n) = 3$ Hence,
$f = 3 N - n$
$= 3 \times 3 - 3$
$= 9 - 3$
$f = 6$
Degree of Freedom of diatomic molecule is 6.

AIPMT-1999

Kinetic Theory of Gases

139292 If the degrees of freedom of a gas are $f$ then the ratio of its specific heat $\frac{C_{p}}{C_{v}}$ is given by

1

1 + \frac{2}{f}

2

1 - \frac{2}{f}

3

1 + \frac{1}{f}

4

1 - \frac{1}{f}

Explanation:

A We know that,
Ratio of two specific heat of gas is:-
$\frac{C_{p}}{C_{v}} = γ$
We know that,
$γ = 1 + \frac{2}{f}$
Where, $f =$ degree of Freedom of gas

AMU-2002

Kinetic Theory of Gases

139293 At which of the following temperature would the molecules of a gas have twice, the average kinetic energy they have at $20^{\circ} C$ ?

1

313^{\circ} C

2

373^{\circ} C

3

393^{\circ} C

4

586^{\circ} C

Explanation:

A Given that,
$T_{1} = 20^{\circ} C = 273 + 20$
$T_{2} = 273 + T$
$E_{1} = E$
$E_{2} = 2 E$
We know that,
Kinetic energy $(E) = \frac{3}{2} RT$
$E \propto T$
$\frac{E_{1}}{E_{2}} = \frac{T_{1}}{T_{2}}$
$\frac{E}{2 E} = \frac{273 + 20}{273 + T}$
$586 = 273 + T$
$T = 313^{\circ} C$

AMU-2002

Kinetic Theory of Gases

139295 The gases carbon-monoxide (CO) and nitrogen at the same temperature have kinetic energies $E_{1}$ and $E_{2}$ respectively. Then

1

E_{1} = E_{2}

2

E_{1} > E_{2}

3

E_{1} < E_{2}

4

E_{1}

and

E_{2}

cannot be compared

Explanation:

A We know that,
$KE \propto \frac{3}{2} kT$
Here, Hence,
$CO$ and $N_{2}$ both are diatomic gas
${({KE}_{1})}_{Co} = \frac{5}{2} kT$
${({KE}_{2})}_{N 2} = \frac{5}{2} kT$
Hence,
For same Temperature, $E_{1} = E_{2}$ .

MHT-CET 2005

Kinetic Theory of Gases

139296 The degrees of freedom of a molecule of a triatomic gas are

1 2

2 4

3 6

4 8

Explanation:

C Degree of Freedom (f) $= 3 N - n$
Where, $N =$ total no of particle
$n =$ holonomic constraints
For triatomic molecule-
no of particle $(N) = 3$
and Separation between 3 atom is Fixed i.e.
no of holonomic constraints $(n) = 3$ Hence,
$f = 3 N - n$
$= 3 \times 3 - 3$
$= 9 - 3$
$f = 6$
Degree of Freedom of diatomic molecule is 6.

AIPMT-1999

Kinetic Theory of Gases

139292 If the degrees of freedom of a gas are $f$ then the ratio of its specific heat $\frac{C_{p}}{C_{v}}$ is given by

1

1 + \frac{2}{f}

2

1 - \frac{2}{f}

3

1 + \frac{1}{f}

4

1 - \frac{1}{f}

Explanation:

A We know that,
Ratio of two specific heat of gas is:-
$\frac{C_{p}}{C_{v}} = γ$
We know that,
$γ = 1 + \frac{2}{f}$
Where, $f =$ degree of Freedom of gas

AMU-2002

Kinetic Theory of Gases

139293 At which of the following temperature would the molecules of a gas have twice, the average kinetic energy they have at $20^{\circ} C$ ?

1

313^{\circ} C

2

373^{\circ} C

3

393^{\circ} C

4

586^{\circ} C

Explanation:

A Given that,
$T_{1} = 20^{\circ} C = 273 + 20$
$T_{2} = 273 + T$
$E_{1} = E$
$E_{2} = 2 E$
We know that,
Kinetic energy $(E) = \frac{3}{2} RT$
$E \propto T$
$\frac{E_{1}}{E_{2}} = \frac{T_{1}}{T_{2}}$
$\frac{E}{2 E} = \frac{273 + 20}{273 + T}$
$586 = 273 + T$
$T = 313^{\circ} C$

AMU-2002

Kinetic Theory of Gases

139295 The gases carbon-monoxide (CO) and nitrogen at the same temperature have kinetic energies $E_{1}$ and $E_{2}$ respectively. Then

1

E_{1} = E_{2}

2

E_{1} > E_{2}

3

E_{1} < E_{2}

4

E_{1}

and

E_{2}

cannot be compared

Explanation:

A We know that,
$KE \propto \frac{3}{2} kT$
Here, Hence,
$CO$ and $N_{2}$ both are diatomic gas
${({KE}_{1})}_{Co} = \frac{5}{2} kT$
${({KE}_{2})}_{N 2} = \frac{5}{2} kT$
Hence,
For same Temperature, $E_{1} = E_{2}$ .

MHT-CET 2005

Kinetic Theory of Gases

139296 The degrees of freedom of a molecule of a triatomic gas are

1 2

2 4

3 6

4 8

Explanation:

C Degree of Freedom (f) $= 3 N - n$
Where, $N =$ total no of particle
$n =$ holonomic constraints
For triatomic molecule-
no of particle $(N) = 3$
and Separation between 3 atom is Fixed i.e.
no of holonomic constraints $(n) = 3$ Hence,
$f = 3 N - n$
$= 3 \times 3 - 3$
$= 9 - 3$
$f = 6$
Degree of Freedom of diatomic molecule is 6.

AIPMT-1999