QBManager

Kinetic Theory of Gases

139126 The respective speeds of the five molecules are $1, 2, 3, 4$ and $5 km \cdot s^{- 1}$ . Then the ratio of their RMS velocity and the average velocity will be

1

\sqrt{11} : \sqrt{3}

2

3 : \sqrt{11}

3

1 : 2

4

3 : 4

Explanation:

A $U_{rms} = \sqrt{\frac{{(V_{1})}^{2} + {(V_{2})}^{2} + {(V_{3})}^{2} + \dots \dots . . {(V_{n})}^{2}}{n}}$
$U_{rms} = \sqrt{\frac{(1)^{2} + (2)^{3} + (3)^{2} + (4)^{2} + (5)^{2}}{5}}$
$U_{rms} = \sqrt{\frac{55}{5}} = \sqrt{11}$
$U_{av} = \sqrt{\frac{v_{1} + v_{2} + v_{3} + v_{4} + \dots v_{n}}{n}}$
$= \sqrt{\frac{1 + 2 + 3 + 4 + 5}{5}} = \sqrt{\frac{15}{5}} = \sqrt{3}$
$V_{av} = \sqrt{3}$
$\frac{v_{rms}}{v_{av}} = \sqrt{\frac{11}{3}}$

Shift-I]

Kinetic Theory of Gases

139127 The ratio of the specific heats $\frac{C_{P}}{C_{V}} = γ$ in terms of degrees of freedom (n) is given by

1

(1 + \frac{n}{3})

2

(1 + \frac{2}{n})

3

(1 + \frac{n}{2})

4

(1 + \frac{1}{n})

Explanation:

B Degree of freedom $= n$
We know that,
$C_{V} = \frac{n}{2} R$
and
$C_{P} - C_{V} = R$
$C_{P} = R + C_{V}$
From equation (i) \& (ii)
$C_{P} = R + \frac{n}{2} R$
$C_{P} = R (1 + \frac{n}{2})$
Equation (iii) divided by eq $^{n}$ (i),
$\frac{C_{P}}{C_{V}} = \frac{R (1 + \frac{n}{2})}{\frac{n}{2} R}$
$\frac{C_{P}}{C_{V}} = \frac{2}{n} + 1$

CG PET- 2005

Kinetic Theory of Gases

139128 For a gas if ratio of specific heats at constant pressure and volume is $γ$ then value of degrees of freedom is

1

\frac{3 γ - 1}{2 γ - 1}

2

\frac{2}{γ - 1}

3

\frac{9}{2} (γ - 1)

4

\frac{25}{2} (γ - 1)

Explanation:

B We know that,
$C_{p} = (\frac{f}{2} + 1) R and C_{v} = \frac{f}{2} R$
$γ = \frac{C_{p}}{C_{v}} = \frac{(\frac{f}{2} + 1) R}{(f / 2) R}$
$= \frac{(f / 2 + 1)}{f / 2}$
$\frac{f}{2} γ = \frac{f}{2} + 1$
$\frac{f}{2} (γ - 1) = 1$
$\frac{f}{2} = \frac{1}{(γ - 1)}$
$f = \frac{2}{(γ - 1)}$

AP EAMCET (17.09.2020) Shift-II

Kinetic Theory of Gases

139129 The root mean square velocity of hydrogen molecules at $300 K$ is $1930 m / s$ . The rms velocity of oxygen molecules at $1200 K$ will be-

1

765 m / s

2

1065 m / s

3

965 m / s

4

865 m / s

Explanation:

C Given that,
${(v_{rms})}_{H} = 1930 m / s$
Temperature of Hydrogen $(T_{H}) = 300 K$
Molecular mass of Hydrogen $(M_{H}) = 2$
Molecular mass of oxygen $(M_{O_{2}}) = 32$
Temperature of oxygen $(T_{O_{2}}) = 1200 K$
We know that,
$v_{rms} = \sqrt{\frac{3 R T}{M}}$
$v_{rms} \propto \sqrt{\frac{T}{M}}$
Hence,
$\frac{{(v_{rms})}_{H}}{{(v_{rms})}_{O_{2}}} = \sqrt{\frac{T_{H}}{T_{O_{2}}} \times \frac{M_{O_{2}}}{M_{H}}}$
$\frac{1930}{{(v_{rms})}_{O_{2}}} = \sqrt{\frac{300}{1200} \times \frac{32}{2}}$
$\frac{1930}{{(v_{rms})}_{O_{2}}} = \frac{2}{1}$
${(v_{rms})}_{O_{2}} = 965 m / s .$

CG PET-22.05.2022

Kinetic Theory of Gases

139126 The respective speeds of the five molecules are $1, 2, 3, 4$ and $5 km \cdot s^{- 1}$ . Then the ratio of their RMS velocity and the average velocity will be

1

\sqrt{11} : \sqrt{3}

2

3 : \sqrt{11}

3

1 : 2

4

3 : 4

Explanation:

A $U_{rms} = \sqrt{\frac{{(V_{1})}^{2} + {(V_{2})}^{2} + {(V_{3})}^{2} + \dots \dots . . {(V_{n})}^{2}}{n}}$
$U_{rms} = \sqrt{\frac{(1)^{2} + (2)^{3} + (3)^{2} + (4)^{2} + (5)^{2}}{5}}$
$U_{rms} = \sqrt{\frac{55}{5}} = \sqrt{11}$
$U_{av} = \sqrt{\frac{v_{1} + v_{2} + v_{3} + v_{4} + \dots v_{n}}{n}}$
$= \sqrt{\frac{1 + 2 + 3 + 4 + 5}{5}} = \sqrt{\frac{15}{5}} = \sqrt{3}$
$V_{av} = \sqrt{3}$
$\frac{v_{rms}}{v_{av}} = \sqrt{\frac{11}{3}}$

Shift-I]

Kinetic Theory of Gases

139127 The ratio of the specific heats $\frac{C_{P}}{C_{V}} = γ$ in terms of degrees of freedom (n) is given by

1

(1 + \frac{n}{3})

2

(1 + \frac{2}{n})

3

(1 + \frac{n}{2})

4

(1 + \frac{1}{n})

Explanation:

B Degree of freedom $= n$
We know that,
$C_{V} = \frac{n}{2} R$
and
$C_{P} - C_{V} = R$
$C_{P} = R + C_{V}$
From equation (i) \& (ii)
$C_{P} = R + \frac{n}{2} R$
$C_{P} = R (1 + \frac{n}{2})$
Equation (iii) divided by eq $^{n}$ (i),
$\frac{C_{P}}{C_{V}} = \frac{R (1 + \frac{n}{2})}{\frac{n}{2} R}$
$\frac{C_{P}}{C_{V}} = \frac{2}{n} + 1$

CG PET- 2005

Kinetic Theory of Gases

139128 For a gas if ratio of specific heats at constant pressure and volume is $γ$ then value of degrees of freedom is

1

\frac{3 γ - 1}{2 γ - 1}

2

\frac{2}{γ - 1}

3

\frac{9}{2} (γ - 1)

4

\frac{25}{2} (γ - 1)

Explanation:

B We know that,
$C_{p} = (\frac{f}{2} + 1) R and C_{v} = \frac{f}{2} R$
$γ = \frac{C_{p}}{C_{v}} = \frac{(\frac{f}{2} + 1) R}{(f / 2) R}$
$= \frac{(f / 2 + 1)}{f / 2}$
$\frac{f}{2} γ = \frac{f}{2} + 1$
$\frac{f}{2} (γ - 1) = 1$
$\frac{f}{2} = \frac{1}{(γ - 1)}$
$f = \frac{2}{(γ - 1)}$

AP EAMCET (17.09.2020) Shift-II

Kinetic Theory of Gases

139129 The root mean square velocity of hydrogen molecules at $300 K$ is $1930 m / s$ . The rms velocity of oxygen molecules at $1200 K$ will be-

1

765 m / s

2

1065 m / s

3

965 m / s

4

865 m / s

Explanation:

C Given that,
${(v_{rms})}_{H} = 1930 m / s$
Temperature of Hydrogen $(T_{H}) = 300 K$
Molecular mass of Hydrogen $(M_{H}) = 2$
Molecular mass of oxygen $(M_{O_{2}}) = 32$
Temperature of oxygen $(T_{O_{2}}) = 1200 K$
We know that,
$v_{rms} = \sqrt{\frac{3 R T}{M}}$
$v_{rms} \propto \sqrt{\frac{T}{M}}$
Hence,
$\frac{{(v_{rms})}_{H}}{{(v_{rms})}_{O_{2}}} = \sqrt{\frac{T_{H}}{T_{O_{2}}} \times \frac{M_{O_{2}}}{M_{H}}}$
$\frac{1930}{{(v_{rms})}_{O_{2}}} = \sqrt{\frac{300}{1200} \times \frac{32}{2}}$
$\frac{1930}{{(v_{rms})}_{O_{2}}} = \frac{2}{1}$
${(v_{rms})}_{O_{2}} = 965 m / s .$

CG PET-22.05.2022

Kinetic Theory of Gases

139126 The respective speeds of the five molecules are $1, 2, 3, 4$ and $5 km \cdot s^{- 1}$ . Then the ratio of their RMS velocity and the average velocity will be

1

\sqrt{11} : \sqrt{3}

2

3 : \sqrt{11}

3

1 : 2

4

3 : 4

Explanation:

A $U_{rms} = \sqrt{\frac{{(V_{1})}^{2} + {(V_{2})}^{2} + {(V_{3})}^{2} + \dots \dots . . {(V_{n})}^{2}}{n}}$
$U_{rms} = \sqrt{\frac{(1)^{2} + (2)^{3} + (3)^{2} + (4)^{2} + (5)^{2}}{5}}$
$U_{rms} = \sqrt{\frac{55}{5}} = \sqrt{11}$
$U_{av} = \sqrt{\frac{v_{1} + v_{2} + v_{3} + v_{4} + \dots v_{n}}{n}}$
$= \sqrt{\frac{1 + 2 + 3 + 4 + 5}{5}} = \sqrt{\frac{15}{5}} = \sqrt{3}$
$V_{av} = \sqrt{3}$
$\frac{v_{rms}}{v_{av}} = \sqrt{\frac{11}{3}}$

Shift-I]

Kinetic Theory of Gases

139127 The ratio of the specific heats $\frac{C_{P}}{C_{V}} = γ$ in terms of degrees of freedom (n) is given by

1

(1 + \frac{n}{3})

2

(1 + \frac{2}{n})

3

(1 + \frac{n}{2})

4

(1 + \frac{1}{n})

Explanation:

B Degree of freedom $= n$
We know that,
$C_{V} = \frac{n}{2} R$
and
$C_{P} - C_{V} = R$
$C_{P} = R + C_{V}$
From equation (i) \& (ii)
$C_{P} = R + \frac{n}{2} R$
$C_{P} = R (1 + \frac{n}{2})$
Equation (iii) divided by eq $^{n}$ (i),
$\frac{C_{P}}{C_{V}} = \frac{R (1 + \frac{n}{2})}{\frac{n}{2} R}$
$\frac{C_{P}}{C_{V}} = \frac{2}{n} + 1$

CG PET- 2005

Kinetic Theory of Gases

139128 For a gas if ratio of specific heats at constant pressure and volume is $γ$ then value of degrees of freedom is

1

\frac{3 γ - 1}{2 γ - 1}

2

\frac{2}{γ - 1}

3

\frac{9}{2} (γ - 1)

4

\frac{25}{2} (γ - 1)

Explanation:

B We know that,
$C_{p} = (\frac{f}{2} + 1) R and C_{v} = \frac{f}{2} R$
$γ = \frac{C_{p}}{C_{v}} = \frac{(\frac{f}{2} + 1) R}{(f / 2) R}$
$= \frac{(f / 2 + 1)}{f / 2}$
$\frac{f}{2} γ = \frac{f}{2} + 1$
$\frac{f}{2} (γ - 1) = 1$
$\frac{f}{2} = \frac{1}{(γ - 1)}$
$f = \frac{2}{(γ - 1)}$

AP EAMCET (17.09.2020) Shift-II

Kinetic Theory of Gases

139129 The root mean square velocity of hydrogen molecules at $300 K$ is $1930 m / s$ . The rms velocity of oxygen molecules at $1200 K$ will be-

1

765 m / s

2

1065 m / s

3

965 m / s

4

865 m / s

Explanation:

C Given that,
${(v_{rms})}_{H} = 1930 m / s$
Temperature of Hydrogen $(T_{H}) = 300 K$
Molecular mass of Hydrogen $(M_{H}) = 2$
Molecular mass of oxygen $(M_{O_{2}}) = 32$
Temperature of oxygen $(T_{O_{2}}) = 1200 K$
We know that,
$v_{rms} = \sqrt{\frac{3 R T}{M}}$
$v_{rms} \propto \sqrt{\frac{T}{M}}$
Hence,
$\frac{{(v_{rms})}_{H}}{{(v_{rms})}_{O_{2}}} = \sqrt{\frac{T_{H}}{T_{O_{2}}} \times \frac{M_{O_{2}}}{M_{H}}}$
$\frac{1930}{{(v_{rms})}_{O_{2}}} = \sqrt{\frac{300}{1200} \times \frac{32}{2}}$
$\frac{1930}{{(v_{rms})}_{O_{2}}} = \frac{2}{1}$
${(v_{rms})}_{O_{2}} = 965 m / s .$

CG PET-22.05.2022

Kinetic Theory of Gases

139126 The respective speeds of the five molecules are $1, 2, 3, 4$ and $5 km \cdot s^{- 1}$ . Then the ratio of their RMS velocity and the average velocity will be

1

\sqrt{11} : \sqrt{3}

2

3 : \sqrt{11}

3

1 : 2

4

3 : 4

Explanation:

A $U_{rms} = \sqrt{\frac{{(V_{1})}^{2} + {(V_{2})}^{2} + {(V_{3})}^{2} + \dots \dots . . {(V_{n})}^{2}}{n}}$
$U_{rms} = \sqrt{\frac{(1)^{2} + (2)^{3} + (3)^{2} + (4)^{2} + (5)^{2}}{5}}$
$U_{rms} = \sqrt{\frac{55}{5}} = \sqrt{11}$
$U_{av} = \sqrt{\frac{v_{1} + v_{2} + v_{3} + v_{4} + \dots v_{n}}{n}}$
$= \sqrt{\frac{1 + 2 + 3 + 4 + 5}{5}} = \sqrt{\frac{15}{5}} = \sqrt{3}$
$V_{av} = \sqrt{3}$
$\frac{v_{rms}}{v_{av}} = \sqrt{\frac{11}{3}}$

Shift-I]

Kinetic Theory of Gases

139127 The ratio of the specific heats $\frac{C_{P}}{C_{V}} = γ$ in terms of degrees of freedom (n) is given by

1

(1 + \frac{n}{3})

2

(1 + \frac{2}{n})

3

(1 + \frac{n}{2})

4

(1 + \frac{1}{n})

Explanation:

B Degree of freedom $= n$
We know that,
$C_{V} = \frac{n}{2} R$
and
$C_{P} - C_{V} = R$
$C_{P} = R + C_{V}$
From equation (i) \& (ii)
$C_{P} = R + \frac{n}{2} R$
$C_{P} = R (1 + \frac{n}{2})$
Equation (iii) divided by eq $^{n}$ (i),
$\frac{C_{P}}{C_{V}} = \frac{R (1 + \frac{n}{2})}{\frac{n}{2} R}$
$\frac{C_{P}}{C_{V}} = \frac{2}{n} + 1$

CG PET- 2005

Kinetic Theory of Gases

139128 For a gas if ratio of specific heats at constant pressure and volume is $γ$ then value of degrees of freedom is

1

\frac{3 γ - 1}{2 γ - 1}

2

\frac{2}{γ - 1}

3

\frac{9}{2} (γ - 1)

4

\frac{25}{2} (γ - 1)

Explanation:

B We know that,
$C_{p} = (\frac{f}{2} + 1) R and C_{v} = \frac{f}{2} R$
$γ = \frac{C_{p}}{C_{v}} = \frac{(\frac{f}{2} + 1) R}{(f / 2) R}$
$= \frac{(f / 2 + 1)}{f / 2}$
$\frac{f}{2} γ = \frac{f}{2} + 1$
$\frac{f}{2} (γ - 1) = 1$
$\frac{f}{2} = \frac{1}{(γ - 1)}$
$f = \frac{2}{(γ - 1)}$

AP EAMCET (17.09.2020) Shift-II

Kinetic Theory of Gases

139129 The root mean square velocity of hydrogen molecules at $300 K$ is $1930 m / s$ . The rms velocity of oxygen molecules at $1200 K$ will be-

1

765 m / s

2

1065 m / s

3

965 m / s

4

865 m / s

Explanation:

C Given that,
${(v_{rms})}_{H} = 1930 m / s$
Temperature of Hydrogen $(T_{H}) = 300 K$
Molecular mass of Hydrogen $(M_{H}) = 2$
Molecular mass of oxygen $(M_{O_{2}}) = 32$
Temperature of oxygen $(T_{O_{2}}) = 1200 K$
We know that,
$v_{rms} = \sqrt{\frac{3 R T}{M}}$
$v_{rms} \propto \sqrt{\frac{T}{M}}$
Hence,
$\frac{{(v_{rms})}_{H}}{{(v_{rms})}_{O_{2}}} = \sqrt{\frac{T_{H}}{T_{O_{2}}} \times \frac{M_{O_{2}}}{M_{H}}}$
$\frac{1930}{{(v_{rms})}_{O_{2}}} = \sqrt{\frac{300}{1200} \times \frac{32}{2}}$
$\frac{1930}{{(v_{rms})}_{O_{2}}} = \frac{2}{1}$
${(v_{rms})}_{O_{2}} = 965 m / s .$

CG PET-22.05.2022