139126
The respective speeds of the five molecules are $1,2,3,4$ and $5 \mathrm{~km} \cdot \mathrm{s}^{-1}$. Then the ratio of their RMS velocity and the average velocity will be
139127
The ratio of the specific heats $\frac{C_{P}}{C_{V}}=\gamma$ in terms of degrees of freedom (n) is given by
1 $\left(1+\frac{n}{3}\right)$
2 $\left(1+\frac{2}{\mathrm{n}}\right)$
3 $\left(1+\frac{n}{2}\right)$
4 $\left(1+\frac{1}{\mathrm{n}}\right)$
Explanation:
B Degree of freedom $=\mathrm{n}$ We know that, $\mathrm{C}_{\mathrm{V}}=\frac{\mathrm{n}}{2} \mathrm{R}$ and $\mathrm{C}_{\mathrm{P}}-\mathrm{C}_{\mathrm{V}}=\mathrm{R}$ $\mathrm{C}_{\mathrm{P}}=\mathrm{R}+\mathrm{C}_{\mathrm{V}}$ From equation (i) \& (ii) $C_{P}=R+\frac{n}{2} R$ $C_{P}=R\left(1+\frac{n}{2}\right)$ Equation (iii) divided by eq ${ }^{\mathrm{n}}$ (i), $\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}=\frac{\mathrm{R}\left(1+\frac{\mathrm{n}}{2}\right)}{\frac{\mathrm{n}}{2} \mathrm{R}}$ $\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}=\frac{2}{\mathrm{n}}+1$
CG PET- 2005
Kinetic Theory of Gases
139128
For a gas if ratio of specific heats at constant pressure and volume is $\gamma$ then value of degrees of freedom is
1 $\frac{3 \gamma-1}{2 \gamma-1}$
2 $\frac{2}{\gamma-1}$
3 $\frac{9}{2}(\gamma-1)$
4 $\frac{25}{2}(\gamma-1)$
Explanation:
B We know that, $\mathrm{C}_{\mathrm{p}}=\left(\frac{\mathrm{f}}{2}+1\right) \mathrm{R} \text { and } \mathrm{C}_{\mathrm{v}}=\frac{\mathrm{f}}{2} \mathrm{R}$ $\gamma=\frac{\mathrm{C}_{\mathrm{p}}}{\mathrm{C}_{\mathrm{v}}}=\frac{\left(\frac{\mathrm{f}}{2}+1\right) \mathrm{R}}{(\mathrm{f} / 2) \mathrm{R}}$ $=\frac{(\mathrm{f} / 2+1)}{\mathrm{f} / 2}$ $\frac{\mathrm{f}}{2} \gamma=\frac{\mathrm{f}}{2}+1$ $\frac{\mathrm{f}}{2}(\gamma-1)=1$ $\frac{\mathrm{f}}{2}=\frac{1}{(\gamma-1)}$ $\mathrm{f}=\frac{2}{(\gamma-1)}$
AP EAMCET (17.09.2020) Shift-II
Kinetic Theory of Gases
139129
The root mean square velocity of hydrogen molecules at $300 \mathrm{~K}$ is $1930 \mathrm{~m} / \mathrm{s}$. The rms velocity of oxygen molecules at $1200 \mathrm{~K}$ will be-
1 $765 \mathrm{~m} / \mathrm{s}$
2 $1065 \mathrm{~m} / \mathrm{s}$
3 $965 \mathrm{~m} / \mathrm{s}$
4 $865 \mathrm{~m} / \mathrm{s}$
Explanation:
C Given that, $\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{H}}=1930 \mathrm{~m} / \mathrm{s}$ Temperature of Hydrogen $\left(\mathrm{T}_{\mathrm{H}}\right)=300 \mathrm{~K}$ Molecular mass of Hydrogen $\left(\mathrm{M}_{\mathrm{H}}\right)=2$ Molecular mass of oxygen $\left(\mathrm{M}_{\mathrm{O}_{2}}\right)=32$ Temperature of oxygen $\left(\mathrm{T}_{\mathrm{O}_{2}}\right)=1200 \mathrm{~K}$ We know that, $\mathrm{v}_{\mathrm{rms}}=\sqrt{\frac{3 R T}{M}}$ $\mathrm{v}_{\mathrm{rms}} \propto \sqrt{\frac{\mathrm{T}}{\mathrm{M}}}$ Hence, $\frac{\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{H}}}{\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{O}_{2}}}=\sqrt{\frac{\mathrm{T}_{\mathrm{H}}}{\mathrm{T}_{\mathrm{O}_{2}}} \times \frac{\mathrm{M}_{\mathrm{O}_{2}}}{\mathrm{M}_{\mathrm{H}}}}$ $\frac{1930}{\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{O}_{2}}}=\sqrt{\frac{300}{1200} \times \frac{32}{2}}$ $\frac{1930}{\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{O}_{2}}}=\frac{2}{1}$ $\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{O}_{2}}=965 \mathrm{~m} / \mathrm{s} .$
139126
The respective speeds of the five molecules are $1,2,3,4$ and $5 \mathrm{~km} \cdot \mathrm{s}^{-1}$. Then the ratio of their RMS velocity and the average velocity will be
139127
The ratio of the specific heats $\frac{C_{P}}{C_{V}}=\gamma$ in terms of degrees of freedom (n) is given by
1 $\left(1+\frac{n}{3}\right)$
2 $\left(1+\frac{2}{\mathrm{n}}\right)$
3 $\left(1+\frac{n}{2}\right)$
4 $\left(1+\frac{1}{\mathrm{n}}\right)$
Explanation:
B Degree of freedom $=\mathrm{n}$ We know that, $\mathrm{C}_{\mathrm{V}}=\frac{\mathrm{n}}{2} \mathrm{R}$ and $\mathrm{C}_{\mathrm{P}}-\mathrm{C}_{\mathrm{V}}=\mathrm{R}$ $\mathrm{C}_{\mathrm{P}}=\mathrm{R}+\mathrm{C}_{\mathrm{V}}$ From equation (i) \& (ii) $C_{P}=R+\frac{n}{2} R$ $C_{P}=R\left(1+\frac{n}{2}\right)$ Equation (iii) divided by eq ${ }^{\mathrm{n}}$ (i), $\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}=\frac{\mathrm{R}\left(1+\frac{\mathrm{n}}{2}\right)}{\frac{\mathrm{n}}{2} \mathrm{R}}$ $\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}=\frac{2}{\mathrm{n}}+1$
CG PET- 2005
Kinetic Theory of Gases
139128
For a gas if ratio of specific heats at constant pressure and volume is $\gamma$ then value of degrees of freedom is
1 $\frac{3 \gamma-1}{2 \gamma-1}$
2 $\frac{2}{\gamma-1}$
3 $\frac{9}{2}(\gamma-1)$
4 $\frac{25}{2}(\gamma-1)$
Explanation:
B We know that, $\mathrm{C}_{\mathrm{p}}=\left(\frac{\mathrm{f}}{2}+1\right) \mathrm{R} \text { and } \mathrm{C}_{\mathrm{v}}=\frac{\mathrm{f}}{2} \mathrm{R}$ $\gamma=\frac{\mathrm{C}_{\mathrm{p}}}{\mathrm{C}_{\mathrm{v}}}=\frac{\left(\frac{\mathrm{f}}{2}+1\right) \mathrm{R}}{(\mathrm{f} / 2) \mathrm{R}}$ $=\frac{(\mathrm{f} / 2+1)}{\mathrm{f} / 2}$ $\frac{\mathrm{f}}{2} \gamma=\frac{\mathrm{f}}{2}+1$ $\frac{\mathrm{f}}{2}(\gamma-1)=1$ $\frac{\mathrm{f}}{2}=\frac{1}{(\gamma-1)}$ $\mathrm{f}=\frac{2}{(\gamma-1)}$
AP EAMCET (17.09.2020) Shift-II
Kinetic Theory of Gases
139129
The root mean square velocity of hydrogen molecules at $300 \mathrm{~K}$ is $1930 \mathrm{~m} / \mathrm{s}$. The rms velocity of oxygen molecules at $1200 \mathrm{~K}$ will be-
1 $765 \mathrm{~m} / \mathrm{s}$
2 $1065 \mathrm{~m} / \mathrm{s}$
3 $965 \mathrm{~m} / \mathrm{s}$
4 $865 \mathrm{~m} / \mathrm{s}$
Explanation:
C Given that, $\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{H}}=1930 \mathrm{~m} / \mathrm{s}$ Temperature of Hydrogen $\left(\mathrm{T}_{\mathrm{H}}\right)=300 \mathrm{~K}$ Molecular mass of Hydrogen $\left(\mathrm{M}_{\mathrm{H}}\right)=2$ Molecular mass of oxygen $\left(\mathrm{M}_{\mathrm{O}_{2}}\right)=32$ Temperature of oxygen $\left(\mathrm{T}_{\mathrm{O}_{2}}\right)=1200 \mathrm{~K}$ We know that, $\mathrm{v}_{\mathrm{rms}}=\sqrt{\frac{3 R T}{M}}$ $\mathrm{v}_{\mathrm{rms}} \propto \sqrt{\frac{\mathrm{T}}{\mathrm{M}}}$ Hence, $\frac{\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{H}}}{\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{O}_{2}}}=\sqrt{\frac{\mathrm{T}_{\mathrm{H}}}{\mathrm{T}_{\mathrm{O}_{2}}} \times \frac{\mathrm{M}_{\mathrm{O}_{2}}}{\mathrm{M}_{\mathrm{H}}}}$ $\frac{1930}{\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{O}_{2}}}=\sqrt{\frac{300}{1200} \times \frac{32}{2}}$ $\frac{1930}{\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{O}_{2}}}=\frac{2}{1}$ $\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{O}_{2}}=965 \mathrm{~m} / \mathrm{s} .$
139126
The respective speeds of the five molecules are $1,2,3,4$ and $5 \mathrm{~km} \cdot \mathrm{s}^{-1}$. Then the ratio of their RMS velocity and the average velocity will be
139127
The ratio of the specific heats $\frac{C_{P}}{C_{V}}=\gamma$ in terms of degrees of freedom (n) is given by
1 $\left(1+\frac{n}{3}\right)$
2 $\left(1+\frac{2}{\mathrm{n}}\right)$
3 $\left(1+\frac{n}{2}\right)$
4 $\left(1+\frac{1}{\mathrm{n}}\right)$
Explanation:
B Degree of freedom $=\mathrm{n}$ We know that, $\mathrm{C}_{\mathrm{V}}=\frac{\mathrm{n}}{2} \mathrm{R}$ and $\mathrm{C}_{\mathrm{P}}-\mathrm{C}_{\mathrm{V}}=\mathrm{R}$ $\mathrm{C}_{\mathrm{P}}=\mathrm{R}+\mathrm{C}_{\mathrm{V}}$ From equation (i) \& (ii) $C_{P}=R+\frac{n}{2} R$ $C_{P}=R\left(1+\frac{n}{2}\right)$ Equation (iii) divided by eq ${ }^{\mathrm{n}}$ (i), $\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}=\frac{\mathrm{R}\left(1+\frac{\mathrm{n}}{2}\right)}{\frac{\mathrm{n}}{2} \mathrm{R}}$ $\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}=\frac{2}{\mathrm{n}}+1$
CG PET- 2005
Kinetic Theory of Gases
139128
For a gas if ratio of specific heats at constant pressure and volume is $\gamma$ then value of degrees of freedom is
1 $\frac{3 \gamma-1}{2 \gamma-1}$
2 $\frac{2}{\gamma-1}$
3 $\frac{9}{2}(\gamma-1)$
4 $\frac{25}{2}(\gamma-1)$
Explanation:
B We know that, $\mathrm{C}_{\mathrm{p}}=\left(\frac{\mathrm{f}}{2}+1\right) \mathrm{R} \text { and } \mathrm{C}_{\mathrm{v}}=\frac{\mathrm{f}}{2} \mathrm{R}$ $\gamma=\frac{\mathrm{C}_{\mathrm{p}}}{\mathrm{C}_{\mathrm{v}}}=\frac{\left(\frac{\mathrm{f}}{2}+1\right) \mathrm{R}}{(\mathrm{f} / 2) \mathrm{R}}$ $=\frac{(\mathrm{f} / 2+1)}{\mathrm{f} / 2}$ $\frac{\mathrm{f}}{2} \gamma=\frac{\mathrm{f}}{2}+1$ $\frac{\mathrm{f}}{2}(\gamma-1)=1$ $\frac{\mathrm{f}}{2}=\frac{1}{(\gamma-1)}$ $\mathrm{f}=\frac{2}{(\gamma-1)}$
AP EAMCET (17.09.2020) Shift-II
Kinetic Theory of Gases
139129
The root mean square velocity of hydrogen molecules at $300 \mathrm{~K}$ is $1930 \mathrm{~m} / \mathrm{s}$. The rms velocity of oxygen molecules at $1200 \mathrm{~K}$ will be-
1 $765 \mathrm{~m} / \mathrm{s}$
2 $1065 \mathrm{~m} / \mathrm{s}$
3 $965 \mathrm{~m} / \mathrm{s}$
4 $865 \mathrm{~m} / \mathrm{s}$
Explanation:
C Given that, $\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{H}}=1930 \mathrm{~m} / \mathrm{s}$ Temperature of Hydrogen $\left(\mathrm{T}_{\mathrm{H}}\right)=300 \mathrm{~K}$ Molecular mass of Hydrogen $\left(\mathrm{M}_{\mathrm{H}}\right)=2$ Molecular mass of oxygen $\left(\mathrm{M}_{\mathrm{O}_{2}}\right)=32$ Temperature of oxygen $\left(\mathrm{T}_{\mathrm{O}_{2}}\right)=1200 \mathrm{~K}$ We know that, $\mathrm{v}_{\mathrm{rms}}=\sqrt{\frac{3 R T}{M}}$ $\mathrm{v}_{\mathrm{rms}} \propto \sqrt{\frac{\mathrm{T}}{\mathrm{M}}}$ Hence, $\frac{\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{H}}}{\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{O}_{2}}}=\sqrt{\frac{\mathrm{T}_{\mathrm{H}}}{\mathrm{T}_{\mathrm{O}_{2}}} \times \frac{\mathrm{M}_{\mathrm{O}_{2}}}{\mathrm{M}_{\mathrm{H}}}}$ $\frac{1930}{\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{O}_{2}}}=\sqrt{\frac{300}{1200} \times \frac{32}{2}}$ $\frac{1930}{\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{O}_{2}}}=\frac{2}{1}$ $\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{O}_{2}}=965 \mathrm{~m} / \mathrm{s} .$
139126
The respective speeds of the five molecules are $1,2,3,4$ and $5 \mathrm{~km} \cdot \mathrm{s}^{-1}$. Then the ratio of their RMS velocity and the average velocity will be
139127
The ratio of the specific heats $\frac{C_{P}}{C_{V}}=\gamma$ in terms of degrees of freedom (n) is given by
1 $\left(1+\frac{n}{3}\right)$
2 $\left(1+\frac{2}{\mathrm{n}}\right)$
3 $\left(1+\frac{n}{2}\right)$
4 $\left(1+\frac{1}{\mathrm{n}}\right)$
Explanation:
B Degree of freedom $=\mathrm{n}$ We know that, $\mathrm{C}_{\mathrm{V}}=\frac{\mathrm{n}}{2} \mathrm{R}$ and $\mathrm{C}_{\mathrm{P}}-\mathrm{C}_{\mathrm{V}}=\mathrm{R}$ $\mathrm{C}_{\mathrm{P}}=\mathrm{R}+\mathrm{C}_{\mathrm{V}}$ From equation (i) \& (ii) $C_{P}=R+\frac{n}{2} R$ $C_{P}=R\left(1+\frac{n}{2}\right)$ Equation (iii) divided by eq ${ }^{\mathrm{n}}$ (i), $\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}=\frac{\mathrm{R}\left(1+\frac{\mathrm{n}}{2}\right)}{\frac{\mathrm{n}}{2} \mathrm{R}}$ $\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}=\frac{2}{\mathrm{n}}+1$
CG PET- 2005
Kinetic Theory of Gases
139128
For a gas if ratio of specific heats at constant pressure and volume is $\gamma$ then value of degrees of freedom is
1 $\frac{3 \gamma-1}{2 \gamma-1}$
2 $\frac{2}{\gamma-1}$
3 $\frac{9}{2}(\gamma-1)$
4 $\frac{25}{2}(\gamma-1)$
Explanation:
B We know that, $\mathrm{C}_{\mathrm{p}}=\left(\frac{\mathrm{f}}{2}+1\right) \mathrm{R} \text { and } \mathrm{C}_{\mathrm{v}}=\frac{\mathrm{f}}{2} \mathrm{R}$ $\gamma=\frac{\mathrm{C}_{\mathrm{p}}}{\mathrm{C}_{\mathrm{v}}}=\frac{\left(\frac{\mathrm{f}}{2}+1\right) \mathrm{R}}{(\mathrm{f} / 2) \mathrm{R}}$ $=\frac{(\mathrm{f} / 2+1)}{\mathrm{f} / 2}$ $\frac{\mathrm{f}}{2} \gamma=\frac{\mathrm{f}}{2}+1$ $\frac{\mathrm{f}}{2}(\gamma-1)=1$ $\frac{\mathrm{f}}{2}=\frac{1}{(\gamma-1)}$ $\mathrm{f}=\frac{2}{(\gamma-1)}$
AP EAMCET (17.09.2020) Shift-II
Kinetic Theory of Gases
139129
The root mean square velocity of hydrogen molecules at $300 \mathrm{~K}$ is $1930 \mathrm{~m} / \mathrm{s}$. The rms velocity of oxygen molecules at $1200 \mathrm{~K}$ will be-
1 $765 \mathrm{~m} / \mathrm{s}$
2 $1065 \mathrm{~m} / \mathrm{s}$
3 $965 \mathrm{~m} / \mathrm{s}$
4 $865 \mathrm{~m} / \mathrm{s}$
Explanation:
C Given that, $\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{H}}=1930 \mathrm{~m} / \mathrm{s}$ Temperature of Hydrogen $\left(\mathrm{T}_{\mathrm{H}}\right)=300 \mathrm{~K}$ Molecular mass of Hydrogen $\left(\mathrm{M}_{\mathrm{H}}\right)=2$ Molecular mass of oxygen $\left(\mathrm{M}_{\mathrm{O}_{2}}\right)=32$ Temperature of oxygen $\left(\mathrm{T}_{\mathrm{O}_{2}}\right)=1200 \mathrm{~K}$ We know that, $\mathrm{v}_{\mathrm{rms}}=\sqrt{\frac{3 R T}{M}}$ $\mathrm{v}_{\mathrm{rms}} \propto \sqrt{\frac{\mathrm{T}}{\mathrm{M}}}$ Hence, $\frac{\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{H}}}{\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{O}_{2}}}=\sqrt{\frac{\mathrm{T}_{\mathrm{H}}}{\mathrm{T}_{\mathrm{O}_{2}}} \times \frac{\mathrm{M}_{\mathrm{O}_{2}}}{\mathrm{M}_{\mathrm{H}}}}$ $\frac{1930}{\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{O}_{2}}}=\sqrt{\frac{300}{1200} \times \frac{32}{2}}$ $\frac{1930}{\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{O}_{2}}}=\frac{2}{1}$ $\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{O}_{2}}=965 \mathrm{~m} / \mathrm{s} .$
