QBManager

Kinetic Theory of Gases

139142 Which statements are correct about degrees of freedom?

1 B and C only

2 B and D only

3 A and B only

4 C and D only

Explanation:

B A.A molecule with $n$ degrees of freedom can exchange energy in n different ways.
B.From Maxmell's law of equipartition of energy, each degree of freedom is associated with an energy $\frac{1}{2} RT / mole$ . C. A monatomic molecule has no rotational degrees.
D.Also Neglecting vibrational degrees, ${CH}_{4}$ , polyatomic molecule has 3 translational and 3 rotational degrees, making it 6 as total degrees.
Hence, option (b) is correct.

Shift-II

Kinetic Theory of Gases

139146 Sound travels in a mixture of two moles of helium and $n$ moles of hydrogen. If $r m s$ speed of gas molecules in the mixture is $\sqrt{2}$ times the speed of sound, then the value of $n$ will be

1 1

2 2

3 3

4 4

Explanation:

B Given, $n_{1} = 2, n_{2} = n$ and
$v_{rms} = \sqrt{2} v_{sound}$
$v_{rms} = \sqrt{\frac{3 RT}{M}}, v_{sound} = \sqrt{\frac{γ RT}{M}}$
$∴ \frac{3 RT}{M} = 2 \times \frac{γ RT}{M} from (i)$
$γ = \frac{3}{2} \Rightarrow \frac{C_{p}}{C_{v}} = γ = \frac{3}{2}$
$C_{P} = \frac{n_{1} C_{p_{1}} + n_{2} C_{p_{2}}}{n_{1} + n_{2}}, C_{V} = \frac{n_{1} C_{v_{1}} + n_{2} C_{v_{2}}}{n_{1} + n_{2}}$
$\frac{C_{P}}{C_{V}} = \frac{n_{1} C_{p_{1}} + n_{2} C_{p_{2}}}{n_{1} C_{v_{1}} + n_{2} C_{v_{2}}} = \frac{2 (\frac{5}{2} R) + n (\frac{7}{2} R)}{2 (\frac{3}{2} R) + n (\frac{5}{2} R)}$
$\frac{3}{2} = \frac{10 + 7 n}{6 + 5 n}$
$18 + 15 n = 20 + 14 n$
$n = 2$

Shift-II

Kinetic Theory of Gases

139148 If the pressure is constant the temperature at which root mean square velocity of a gas will be half of its value at $0^{\circ} C$ is

1

- 273^{\circ} C

or

0 K

2

- 273 K

or

- 546^{\circ} C

3

- {204.75}^{\circ} C

or

68.25 K

4

100 K

or

- 173^{\circ} C

Explanation:

C Given,
$T_{1} = 0^{\circ} C = 273 K$
${(v_{1})}_{rms} = v_{o}$
${(v_{2})}_{rms} = \frac{v_{o}}{2}$
We know that,
$∴ v_{rms} = \sqrt{\frac{3 RT}{M}} \Rightarrow v_{rms} \propto \sqrt{T}$
$∴ \frac{{(v_{1})}_{rms}}{{(v_{2})}_{rms}} = \frac{\sqrt{T_{1}}}{\sqrt{T_{2}}}$
$\frac{v_{o}}{v_{o}} = \sqrt{\frac{273}{T_{2}}} \Rightarrow 2 = \sqrt{\frac{273}{T_{2}}}$
$4 = \frac{273}{T_{2}} \Rightarrow T_{2} = \frac{273}{4} = 68.25 K$
$T_{2} = 68.25 K or (- {204.75}^{\circ} C)$

Assam CEE-31.07.2022

Kinetic Theory of Gases

139149 At what temperature, an oxygen molecule has the same r.m.s velocity as the hydrogen molecule has at $20 K$ ?

1

160 K

2

320 K

3

293 K

4

347 K

Explanation:

B Same r.m.s velocity,
${(v_{O_{2}})}_{rms} = {(v_{H_{2}})}_{rms} = v$
Molecular mass of $O_{2} = (M_{O_{2}}) = 32$
Molecular mass of $H_{2} = (M_{H_{2}}) = 2$
$T_{H_{2}} = 20 K$
$∵ v_{rms} \propto \sqrt{\frac{T}{M}}$
$\frac{{(v_{O_{2}})}_{rms}}{(v_{H_{2}})} = \frac{\sqrt{\frac{T_{O_{2}}}{M_{O_{2}}}}}{\sqrt{\frac{T_{H_{2}}}{M_{H_{2}}}}}$
$\frac{v}{v} = \sqrt{\frac{T_{O_{2}} / 32}{20 / 2}}$
$1 = \sqrt{\frac{T_{O_{2}} / 32}{20 / 2}}$
$T_{O_{2}} = 320 K$

TS EAMCET 19.07.2022

Kinetic Theory of Gases

139142 Which statements are correct about degrees of freedom?

1 B and C only

2 B and D only

3 A and B only

4 C and D only

Explanation:

B A.A molecule with $n$ degrees of freedom can exchange energy in n different ways.
B.From Maxmell's law of equipartition of energy, each degree of freedom is associated with an energy $\frac{1}{2} RT / mole$ . C. A monatomic molecule has no rotational degrees.
D.Also Neglecting vibrational degrees, ${CH}_{4}$ , polyatomic molecule has 3 translational and 3 rotational degrees, making it 6 as total degrees.
Hence, option (b) is correct.

Shift-II

Kinetic Theory of Gases

139146 Sound travels in a mixture of two moles of helium and $n$ moles of hydrogen. If $r m s$ speed of gas molecules in the mixture is $\sqrt{2}$ times the speed of sound, then the value of $n$ will be

1 1

2 2

3 3

4 4

Explanation:

B Given, $n_{1} = 2, n_{2} = n$ and
$v_{rms} = \sqrt{2} v_{sound}$
$v_{rms} = \sqrt{\frac{3 RT}{M}}, v_{sound} = \sqrt{\frac{γ RT}{M}}$
$∴ \frac{3 RT}{M} = 2 \times \frac{γ RT}{M} from (i)$
$γ = \frac{3}{2} \Rightarrow \frac{C_{p}}{C_{v}} = γ = \frac{3}{2}$
$C_{P} = \frac{n_{1} C_{p_{1}} + n_{2} C_{p_{2}}}{n_{1} + n_{2}}, C_{V} = \frac{n_{1} C_{v_{1}} + n_{2} C_{v_{2}}}{n_{1} + n_{2}}$
$\frac{C_{P}}{C_{V}} = \frac{n_{1} C_{p_{1}} + n_{2} C_{p_{2}}}{n_{1} C_{v_{1}} + n_{2} C_{v_{2}}} = \frac{2 (\frac{5}{2} R) + n (\frac{7}{2} R)}{2 (\frac{3}{2} R) + n (\frac{5}{2} R)}$
$\frac{3}{2} = \frac{10 + 7 n}{6 + 5 n}$
$18 + 15 n = 20 + 14 n$
$n = 2$

Shift-II

Kinetic Theory of Gases

139148 If the pressure is constant the temperature at which root mean square velocity of a gas will be half of its value at $0^{\circ} C$ is

1

- 273^{\circ} C

or

0 K

2

- 273 K

or

- 546^{\circ} C

3

- {204.75}^{\circ} C

or

68.25 K

4

100 K

or

- 173^{\circ} C

Explanation:

C Given,
$T_{1} = 0^{\circ} C = 273 K$
${(v_{1})}_{rms} = v_{o}$
${(v_{2})}_{rms} = \frac{v_{o}}{2}$
We know that,
$∴ v_{rms} = \sqrt{\frac{3 RT}{M}} \Rightarrow v_{rms} \propto \sqrt{T}$
$∴ \frac{{(v_{1})}_{rms}}{{(v_{2})}_{rms}} = \frac{\sqrt{T_{1}}}{\sqrt{T_{2}}}$
$\frac{v_{o}}{v_{o}} = \sqrt{\frac{273}{T_{2}}} \Rightarrow 2 = \sqrt{\frac{273}{T_{2}}}$
$4 = \frac{273}{T_{2}} \Rightarrow T_{2} = \frac{273}{4} = 68.25 K$
$T_{2} = 68.25 K or (- {204.75}^{\circ} C)$

Assam CEE-31.07.2022

Kinetic Theory of Gases

139149 At what temperature, an oxygen molecule has the same r.m.s velocity as the hydrogen molecule has at $20 K$ ?

1

160 K

2

320 K

3

293 K

4

347 K

Explanation:

B Same r.m.s velocity,
${(v_{O_{2}})}_{rms} = {(v_{H_{2}})}_{rms} = v$
Molecular mass of $O_{2} = (M_{O_{2}}) = 32$
Molecular mass of $H_{2} = (M_{H_{2}}) = 2$
$T_{H_{2}} = 20 K$
$∵ v_{rms} \propto \sqrt{\frac{T}{M}}$
$\frac{{(v_{O_{2}})}_{rms}}{(v_{H_{2}})} = \frac{\sqrt{\frac{T_{O_{2}}}{M_{O_{2}}}}}{\sqrt{\frac{T_{H_{2}}}{M_{H_{2}}}}}$
$\frac{v}{v} = \sqrt{\frac{T_{O_{2}} / 32}{20 / 2}}$
$1 = \sqrt{\frac{T_{O_{2}} / 32}{20 / 2}}$
$T_{O_{2}} = 320 K$

TS EAMCET 19.07.2022

Kinetic Theory of Gases

139142 Which statements are correct about degrees of freedom?

1 B and C only

2 B and D only

3 A and B only

4 C and D only

Explanation:

B A.A molecule with $n$ degrees of freedom can exchange energy in n different ways.
B.From Maxmell's law of equipartition of energy, each degree of freedom is associated with an energy $\frac{1}{2} RT / mole$ . C. A monatomic molecule has no rotational degrees.
D.Also Neglecting vibrational degrees, ${CH}_{4}$ , polyatomic molecule has 3 translational and 3 rotational degrees, making it 6 as total degrees.
Hence, option (b) is correct.

Shift-II

Kinetic Theory of Gases

139146 Sound travels in a mixture of two moles of helium and $n$ moles of hydrogen. If $r m s$ speed of gas molecules in the mixture is $\sqrt{2}$ times the speed of sound, then the value of $n$ will be

1 1

2 2

3 3

4 4

Explanation:

B Given, $n_{1} = 2, n_{2} = n$ and
$v_{rms} = \sqrt{2} v_{sound}$
$v_{rms} = \sqrt{\frac{3 RT}{M}}, v_{sound} = \sqrt{\frac{γ RT}{M}}$
$∴ \frac{3 RT}{M} = 2 \times \frac{γ RT}{M} from (i)$
$γ = \frac{3}{2} \Rightarrow \frac{C_{p}}{C_{v}} = γ = \frac{3}{2}$
$C_{P} = \frac{n_{1} C_{p_{1}} + n_{2} C_{p_{2}}}{n_{1} + n_{2}}, C_{V} = \frac{n_{1} C_{v_{1}} + n_{2} C_{v_{2}}}{n_{1} + n_{2}}$
$\frac{C_{P}}{C_{V}} = \frac{n_{1} C_{p_{1}} + n_{2} C_{p_{2}}}{n_{1} C_{v_{1}} + n_{2} C_{v_{2}}} = \frac{2 (\frac{5}{2} R) + n (\frac{7}{2} R)}{2 (\frac{3}{2} R) + n (\frac{5}{2} R)}$
$\frac{3}{2} = \frac{10 + 7 n}{6 + 5 n}$
$18 + 15 n = 20 + 14 n$
$n = 2$

Shift-II

Kinetic Theory of Gases

139148 If the pressure is constant the temperature at which root mean square velocity of a gas will be half of its value at $0^{\circ} C$ is

1

- 273^{\circ} C

or

0 K

2

- 273 K

or

- 546^{\circ} C

3

- {204.75}^{\circ} C

or

68.25 K

4

100 K

or

- 173^{\circ} C

Explanation:

C Given,
$T_{1} = 0^{\circ} C = 273 K$
${(v_{1})}_{rms} = v_{o}$
${(v_{2})}_{rms} = \frac{v_{o}}{2}$
We know that,
$∴ v_{rms} = \sqrt{\frac{3 RT}{M}} \Rightarrow v_{rms} \propto \sqrt{T}$
$∴ \frac{{(v_{1})}_{rms}}{{(v_{2})}_{rms}} = \frac{\sqrt{T_{1}}}{\sqrt{T_{2}}}$
$\frac{v_{o}}{v_{o}} = \sqrt{\frac{273}{T_{2}}} \Rightarrow 2 = \sqrt{\frac{273}{T_{2}}}$
$4 = \frac{273}{T_{2}} \Rightarrow T_{2} = \frac{273}{4} = 68.25 K$
$T_{2} = 68.25 K or (- {204.75}^{\circ} C)$

Assam CEE-31.07.2022

Kinetic Theory of Gases

139149 At what temperature, an oxygen molecule has the same r.m.s velocity as the hydrogen molecule has at $20 K$ ?

1

160 K

2

320 K

3

293 K

4

347 K

Explanation:

B Same r.m.s velocity,
${(v_{O_{2}})}_{rms} = {(v_{H_{2}})}_{rms} = v$
Molecular mass of $O_{2} = (M_{O_{2}}) = 32$
Molecular mass of $H_{2} = (M_{H_{2}}) = 2$
$T_{H_{2}} = 20 K$
$∵ v_{rms} \propto \sqrt{\frac{T}{M}}$
$\frac{{(v_{O_{2}})}_{rms}}{(v_{H_{2}})} = \frac{\sqrt{\frac{T_{O_{2}}}{M_{O_{2}}}}}{\sqrt{\frac{T_{H_{2}}}{M_{H_{2}}}}}$
$\frac{v}{v} = \sqrt{\frac{T_{O_{2}} / 32}{20 / 2}}$
$1 = \sqrt{\frac{T_{O_{2}} / 32}{20 / 2}}$
$T_{O_{2}} = 320 K$

TS EAMCET 19.07.2022

Kinetic Theory of Gases

139142 Which statements are correct about degrees of freedom?

1 B and C only

2 B and D only

3 A and B only

4 C and D only

Explanation:

B A.A molecule with $n$ degrees of freedom can exchange energy in n different ways.
B.From Maxmell's law of equipartition of energy, each degree of freedom is associated with an energy $\frac{1}{2} RT / mole$ . C. A monatomic molecule has no rotational degrees.
D.Also Neglecting vibrational degrees, ${CH}_{4}$ , polyatomic molecule has 3 translational and 3 rotational degrees, making it 6 as total degrees.
Hence, option (b) is correct.

Shift-II

Kinetic Theory of Gases

139146 Sound travels in a mixture of two moles of helium and $n$ moles of hydrogen. If $r m s$ speed of gas molecules in the mixture is $\sqrt{2}$ times the speed of sound, then the value of $n$ will be

1 1

2 2

3 3

4 4

Explanation:

B Given, $n_{1} = 2, n_{2} = n$ and
$v_{rms} = \sqrt{2} v_{sound}$
$v_{rms} = \sqrt{\frac{3 RT}{M}}, v_{sound} = \sqrt{\frac{γ RT}{M}}$
$∴ \frac{3 RT}{M} = 2 \times \frac{γ RT}{M} from (i)$
$γ = \frac{3}{2} \Rightarrow \frac{C_{p}}{C_{v}} = γ = \frac{3}{2}$
$C_{P} = \frac{n_{1} C_{p_{1}} + n_{2} C_{p_{2}}}{n_{1} + n_{2}}, C_{V} = \frac{n_{1} C_{v_{1}} + n_{2} C_{v_{2}}}{n_{1} + n_{2}}$
$\frac{C_{P}}{C_{V}} = \frac{n_{1} C_{p_{1}} + n_{2} C_{p_{2}}}{n_{1} C_{v_{1}} + n_{2} C_{v_{2}}} = \frac{2 (\frac{5}{2} R) + n (\frac{7}{2} R)}{2 (\frac{3}{2} R) + n (\frac{5}{2} R)}$
$\frac{3}{2} = \frac{10 + 7 n}{6 + 5 n}$
$18 + 15 n = 20 + 14 n$
$n = 2$

Shift-II

Kinetic Theory of Gases

139148 If the pressure is constant the temperature at which root mean square velocity of a gas will be half of its value at $0^{\circ} C$ is

1

- 273^{\circ} C

or

0 K

2

- 273 K

or

- 546^{\circ} C

3

- {204.75}^{\circ} C

or

68.25 K

4

100 K

or

- 173^{\circ} C

Explanation:

C Given,
$T_{1} = 0^{\circ} C = 273 K$
${(v_{1})}_{rms} = v_{o}$
${(v_{2})}_{rms} = \frac{v_{o}}{2}$
We know that,
$∴ v_{rms} = \sqrt{\frac{3 RT}{M}} \Rightarrow v_{rms} \propto \sqrt{T}$
$∴ \frac{{(v_{1})}_{rms}}{{(v_{2})}_{rms}} = \frac{\sqrt{T_{1}}}{\sqrt{T_{2}}}$
$\frac{v_{o}}{v_{o}} = \sqrt{\frac{273}{T_{2}}} \Rightarrow 2 = \sqrt{\frac{273}{T_{2}}}$
$4 = \frac{273}{T_{2}} \Rightarrow T_{2} = \frac{273}{4} = 68.25 K$
$T_{2} = 68.25 K or (- {204.75}^{\circ} C)$

Assam CEE-31.07.2022

Kinetic Theory of Gases

139149 At what temperature, an oxygen molecule has the same r.m.s velocity as the hydrogen molecule has at $20 K$ ?

1

160 K

2

320 K

3

293 K

4

347 K

Explanation:

B Same r.m.s velocity,
${(v_{O_{2}})}_{rms} = {(v_{H_{2}})}_{rms} = v$
Molecular mass of $O_{2} = (M_{O_{2}}) = 32$
Molecular mass of $H_{2} = (M_{H_{2}}) = 2$
$T_{H_{2}} = 20 K$
$∵ v_{rms} \propto \sqrt{\frac{T}{M}}$
$\frac{{(v_{O_{2}})}_{rms}}{(v_{H_{2}})} = \frac{\sqrt{\frac{T_{O_{2}}}{M_{O_{2}}}}}{\sqrt{\frac{T_{H_{2}}}{M_{H_{2}}}}}$
$\frac{v}{v} = \sqrt{\frac{T_{O_{2}} / 32}{20 / 2}}$
$1 = \sqrt{\frac{T_{O_{2}} / 32}{20 / 2}}$
$T_{O_{2}} = 320 K$

TS EAMCET 19.07.2022

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