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Kinetic Theory of Gases

139248 The temperature at which the velocity of oxygen will be half that of hydrogen at NTP is

1

1092^{\circ} C

2

1492^{\circ} C

3

273 K

4

819^{\circ} C

Explanation:

D Let temperature of oxygen $= T_{1} K$ at which velocity of oxygen equal to velocity of hydrogen.
At NTP, T $= 273 K$
$∴ \sqrt{\frac{3 {RT}_{1}}{M_{O_{2}}}} = \frac{1}{2} \sqrt{\frac{3 RT}{M_{H_{2}}}}$
Taking square both side, we get -
$\frac{T_{1}}{M_{O_{2}}} = \frac{1}{4} \times \frac{T}{M_{H_{2}}}$
$\frac{T_{1}}{32} = \frac{1}{4} \times \frac{273}{2}$
$\frac{T_{1}}{32} = \frac{273}{8}$
$T_{1} = 4 \times 273$
$T_{1} = 1092 K$
$T_{1} = 1092 - 273 = 819^{\circ} C$
$T_{1} = 819^{\circ} C$

J and K CET- 2002

Kinetic Theory of Gases

139250 The speed of sound through a gaseous medium bears a constant ratio with the rms speed of its molecules. This constant ratio is :

1

γ

2

\sqrt{\frac{2 γ}{3}}

3

γ - 1

4

\sqrt{\frac{γ}{3}}

Explanation:

D Given, The root mean square velocity of molecule is,
$v_{rms} = \sqrt{\frac{3 RT}{M}}$
$∵ PV = RT and ρ = \frac{M}{V}$
$v_{rms} = \sqrt{\frac{3 PM}{ρ M}}$
$v_{rms} = \sqrt{\frac{3 P}{ρ}}$
$Speed of sound (v) = \sqrt{\frac{γ P}{ρ}}$
So,
$\frac{v_{s}}{v_{rms}} = \frac{\sqrt{\frac{γ P}{ρ}}}{\sqrt{\frac{3 P}{ρ}}} = \sqrt{\frac{γ}{3}}$

Karnataka CET-2001

Kinetic Theory of Gases

139252 For a gas molecule with 6 degrees of freedom the law of equipartition of energy gives the following relation between the molar specific heat $(C_{V})$ and constant $(R)$

1

C_{V} = \frac{R}{2}

2

C_{V} = R

3

C_{V} = 2 R

4

C_{V} = 3 R

Explanation:

D Given, degrees of freedom (f) $= 6$ We know that,
$C_{P} - C_{V} = R$
And, $\frac{C_{P}}{C_{V}} = γ = 1 + \frac{2}{f}$
$\frac{C_{P}}{C_{V}} = 1 + \frac{2}{6} = \frac{4}{3}$
$C_{P} = \frac{4}{3} C_{V}$
Putting value of $C_{P}$ in equation (i), we have-
$\frac{4}{3} C_{V} - C_{V} = R$
$\frac{1}{3} C_{V} = R$
$C_{V} = 3 R$

J and K CET- 2008

Kinetic Theory of Gases

139253 If the density of hydrogen gas at NTP is $0.0000893 gm / {cm}^{3}$ , then the root mean square velocity of molecules of hydrogen molecules at NTP is

1

1840 cm / \sec

2

184.0 cm / \sec

3

1840 m / \sec

4

184.0 m / \sec

Explanation:

C Given, density $(ρ) = 0.0000893 gm / {cm}^{3}$
$ρ = 8.93 \times 10^{- 5} \times (\frac{10^{- 3}}{10^{- 6}}) kg / m^{3}$
$ρ = 8.93 \times 10^{- 2} kg / m^{3}$
Pressure $(P) = 1.013 \times 10^{5} N / m^{2}$
$V_{rms} = \sqrt{\frac{3 P}{ρ}} = \sqrt{\frac{3 \times 1.013 \times 10^{5}}{8.93 \times 10^{- 2}}} m / s$
$v_{rms} = 1.844 \times 10^{3} m / s$
$v_{rms} ≅ 1840 m / s$

J and K CET- 2001

Kinetic Theory of Gases

139248 The temperature at which the velocity of oxygen will be half that of hydrogen at NTP is

1

1092^{\circ} C

2

1492^{\circ} C

3

273 K

4

819^{\circ} C

Explanation:

D Let temperature of oxygen $= T_{1} K$ at which velocity of oxygen equal to velocity of hydrogen.
At NTP, T $= 273 K$
$∴ \sqrt{\frac{3 {RT}_{1}}{M_{O_{2}}}} = \frac{1}{2} \sqrt{\frac{3 RT}{M_{H_{2}}}}$
Taking square both side, we get -
$\frac{T_{1}}{M_{O_{2}}} = \frac{1}{4} \times \frac{T}{M_{H_{2}}}$
$\frac{T_{1}}{32} = \frac{1}{4} \times \frac{273}{2}$
$\frac{T_{1}}{32} = \frac{273}{8}$
$T_{1} = 4 \times 273$
$T_{1} = 1092 K$
$T_{1} = 1092 - 273 = 819^{\circ} C$
$T_{1} = 819^{\circ} C$

J and K CET- 2002

Kinetic Theory of Gases

139250 The speed of sound through a gaseous medium bears a constant ratio with the rms speed of its molecules. This constant ratio is :

1

γ

2

\sqrt{\frac{2 γ}{3}}

3

γ - 1

4

\sqrt{\frac{γ}{3}}

Explanation:

D Given, The root mean square velocity of molecule is,
$v_{rms} = \sqrt{\frac{3 RT}{M}}$
$∵ PV = RT and ρ = \frac{M}{V}$
$v_{rms} = \sqrt{\frac{3 PM}{ρ M}}$
$v_{rms} = \sqrt{\frac{3 P}{ρ}}$
$Speed of sound (v) = \sqrt{\frac{γ P}{ρ}}$
So,
$\frac{v_{s}}{v_{rms}} = \frac{\sqrt{\frac{γ P}{ρ}}}{\sqrt{\frac{3 P}{ρ}}} = \sqrt{\frac{γ}{3}}$

Karnataka CET-2001

Kinetic Theory of Gases

139252 For a gas molecule with 6 degrees of freedom the law of equipartition of energy gives the following relation between the molar specific heat $(C_{V})$ and constant $(R)$

1

C_{V} = \frac{R}{2}

2

C_{V} = R

3

C_{V} = 2 R

4

C_{V} = 3 R

Explanation:

D Given, degrees of freedom (f) $= 6$ We know that,
$C_{P} - C_{V} = R$
And, $\frac{C_{P}}{C_{V}} = γ = 1 + \frac{2}{f}$
$\frac{C_{P}}{C_{V}} = 1 + \frac{2}{6} = \frac{4}{3}$
$C_{P} = \frac{4}{3} C_{V}$
Putting value of $C_{P}$ in equation (i), we have-
$\frac{4}{3} C_{V} - C_{V} = R$
$\frac{1}{3} C_{V} = R$
$C_{V} = 3 R$

J and K CET- 2008

Kinetic Theory of Gases

139253 If the density of hydrogen gas at NTP is $0.0000893 gm / {cm}^{3}$ , then the root mean square velocity of molecules of hydrogen molecules at NTP is

1

1840 cm / \sec

2

184.0 cm / \sec

3

1840 m / \sec

4

184.0 m / \sec

Explanation:

C Given, density $(ρ) = 0.0000893 gm / {cm}^{3}$
$ρ = 8.93 \times 10^{- 5} \times (\frac{10^{- 3}}{10^{- 6}}) kg / m^{3}$
$ρ = 8.93 \times 10^{- 2} kg / m^{3}$
Pressure $(P) = 1.013 \times 10^{5} N / m^{2}$
$V_{rms} = \sqrt{\frac{3 P}{ρ}} = \sqrt{\frac{3 \times 1.013 \times 10^{5}}{8.93 \times 10^{- 2}}} m / s$
$v_{rms} = 1.844 \times 10^{3} m / s$
$v_{rms} ≅ 1840 m / s$

J and K CET- 2001

Kinetic Theory of Gases

139248 The temperature at which the velocity of oxygen will be half that of hydrogen at NTP is

1

1092^{\circ} C

2

1492^{\circ} C

3

273 K

4

819^{\circ} C

Explanation:

D Let temperature of oxygen $= T_{1} K$ at which velocity of oxygen equal to velocity of hydrogen.
At NTP, T $= 273 K$
$∴ \sqrt{\frac{3 {RT}_{1}}{M_{O_{2}}}} = \frac{1}{2} \sqrt{\frac{3 RT}{M_{H_{2}}}}$
Taking square both side, we get -
$\frac{T_{1}}{M_{O_{2}}} = \frac{1}{4} \times \frac{T}{M_{H_{2}}}$
$\frac{T_{1}}{32} = \frac{1}{4} \times \frac{273}{2}$
$\frac{T_{1}}{32} = \frac{273}{8}$
$T_{1} = 4 \times 273$
$T_{1} = 1092 K$
$T_{1} = 1092 - 273 = 819^{\circ} C$
$T_{1} = 819^{\circ} C$

J and K CET- 2002

Kinetic Theory of Gases

139250 The speed of sound through a gaseous medium bears a constant ratio with the rms speed of its molecules. This constant ratio is :

1

γ

2

\sqrt{\frac{2 γ}{3}}

3

γ - 1

4

\sqrt{\frac{γ}{3}}

Explanation:

D Given, The root mean square velocity of molecule is,
$v_{rms} = \sqrt{\frac{3 RT}{M}}$
$∵ PV = RT and ρ = \frac{M}{V}$
$v_{rms} = \sqrt{\frac{3 PM}{ρ M}}$
$v_{rms} = \sqrt{\frac{3 P}{ρ}}$
$Speed of sound (v) = \sqrt{\frac{γ P}{ρ}}$
So,
$\frac{v_{s}}{v_{rms}} = \frac{\sqrt{\frac{γ P}{ρ}}}{\sqrt{\frac{3 P}{ρ}}} = \sqrt{\frac{γ}{3}}$

Karnataka CET-2001

Kinetic Theory of Gases

139252 For a gas molecule with 6 degrees of freedom the law of equipartition of energy gives the following relation between the molar specific heat $(C_{V})$ and constant $(R)$

1

C_{V} = \frac{R}{2}

2

C_{V} = R

3

C_{V} = 2 R

4

C_{V} = 3 R

Explanation:

D Given, degrees of freedom (f) $= 6$ We know that,
$C_{P} - C_{V} = R$
And, $\frac{C_{P}}{C_{V}} = γ = 1 + \frac{2}{f}$
$\frac{C_{P}}{C_{V}} = 1 + \frac{2}{6} = \frac{4}{3}$
$C_{P} = \frac{4}{3} C_{V}$
Putting value of $C_{P}$ in equation (i), we have-
$\frac{4}{3} C_{V} - C_{V} = R$
$\frac{1}{3} C_{V} = R$
$C_{V} = 3 R$

J and K CET- 2008

Kinetic Theory of Gases

139253 If the density of hydrogen gas at NTP is $0.0000893 gm / {cm}^{3}$ , then the root mean square velocity of molecules of hydrogen molecules at NTP is

1

1840 cm / \sec

2

184.0 cm / \sec

3

1840 m / \sec

4

184.0 m / \sec

Explanation:

C Given, density $(ρ) = 0.0000893 gm / {cm}^{3}$
$ρ = 8.93 \times 10^{- 5} \times (\frac{10^{- 3}}{10^{- 6}}) kg / m^{3}$
$ρ = 8.93 \times 10^{- 2} kg / m^{3}$
Pressure $(P) = 1.013 \times 10^{5} N / m^{2}$
$V_{rms} = \sqrt{\frac{3 P}{ρ}} = \sqrt{\frac{3 \times 1.013 \times 10^{5}}{8.93 \times 10^{- 2}}} m / s$
$v_{rms} = 1.844 \times 10^{3} m / s$
$v_{rms} ≅ 1840 m / s$

J and K CET- 2001

Kinetic Theory of Gases

139248 The temperature at which the velocity of oxygen will be half that of hydrogen at NTP is

1

1092^{\circ} C

2

1492^{\circ} C

3

273 K

4

819^{\circ} C

Explanation:

D Let temperature of oxygen $= T_{1} K$ at which velocity of oxygen equal to velocity of hydrogen.
At NTP, T $= 273 K$
$∴ \sqrt{\frac{3 {RT}_{1}}{M_{O_{2}}}} = \frac{1}{2} \sqrt{\frac{3 RT}{M_{H_{2}}}}$
Taking square both side, we get -
$\frac{T_{1}}{M_{O_{2}}} = \frac{1}{4} \times \frac{T}{M_{H_{2}}}$
$\frac{T_{1}}{32} = \frac{1}{4} \times \frac{273}{2}$
$\frac{T_{1}}{32} = \frac{273}{8}$
$T_{1} = 4 \times 273$
$T_{1} = 1092 K$
$T_{1} = 1092 - 273 = 819^{\circ} C$
$T_{1} = 819^{\circ} C$

J and K CET- 2002

Kinetic Theory of Gases

139250 The speed of sound through a gaseous medium bears a constant ratio with the rms speed of its molecules. This constant ratio is :

1

γ

2

\sqrt{\frac{2 γ}{3}}

3

γ - 1

4

\sqrt{\frac{γ}{3}}

Explanation:

D Given, The root mean square velocity of molecule is,
$v_{rms} = \sqrt{\frac{3 RT}{M}}$
$∵ PV = RT and ρ = \frac{M}{V}$
$v_{rms} = \sqrt{\frac{3 PM}{ρ M}}$
$v_{rms} = \sqrt{\frac{3 P}{ρ}}$
$Speed of sound (v) = \sqrt{\frac{γ P}{ρ}}$
So,
$\frac{v_{s}}{v_{rms}} = \frac{\sqrt{\frac{γ P}{ρ}}}{\sqrt{\frac{3 P}{ρ}}} = \sqrt{\frac{γ}{3}}$

Karnataka CET-2001

Kinetic Theory of Gases

139252 For a gas molecule with 6 degrees of freedom the law of equipartition of energy gives the following relation between the molar specific heat $(C_{V})$ and constant $(R)$

1

C_{V} = \frac{R}{2}

2

C_{V} = R

3

C_{V} = 2 R

4

C_{V} = 3 R

Explanation:

D Given, degrees of freedom (f) $= 6$ We know that,
$C_{P} - C_{V} = R$
And, $\frac{C_{P}}{C_{V}} = γ = 1 + \frac{2}{f}$
$\frac{C_{P}}{C_{V}} = 1 + \frac{2}{6} = \frac{4}{3}$
$C_{P} = \frac{4}{3} C_{V}$
Putting value of $C_{P}$ in equation (i), we have-
$\frac{4}{3} C_{V} - C_{V} = R$
$\frac{1}{3} C_{V} = R$
$C_{V} = 3 R$

J and K CET- 2008

Kinetic Theory of Gases

139253 If the density of hydrogen gas at NTP is $0.0000893 gm / {cm}^{3}$ , then the root mean square velocity of molecules of hydrogen molecules at NTP is

1

1840 cm / \sec

2

184.0 cm / \sec

3

1840 m / \sec

4

184.0 m / \sec

Explanation:

C Given, density $(ρ) = 0.0000893 gm / {cm}^{3}$
$ρ = 8.93 \times 10^{- 5} \times (\frac{10^{- 3}}{10^{- 6}}) kg / m^{3}$
$ρ = 8.93 \times 10^{- 2} kg / m^{3}$
Pressure $(P) = 1.013 \times 10^{5} N / m^{2}$
$V_{rms} = \sqrt{\frac{3 P}{ρ}} = \sqrt{\frac{3 \times 1.013 \times 10^{5}}{8.93 \times 10^{- 2}}} m / s$
$v_{rms} = 1.844 \times 10^{3} m / s$
$v_{rms} ≅ 1840 m / s$

J and K CET- 2001