139248
The temperature at which the velocity of oxygen will be half that of hydrogen at NTP is
1 $1092^{\circ} \mathrm{C}$
2 $1492^{\circ} \mathrm{C}$
3 $273 \mathrm{~K}$
4 $819^{\circ} \mathrm{C}$
Explanation:
D Let temperature of oxygen $=\mathrm{T}_{1} \mathrm{~K}$ at which velocity of oxygen equal to velocity of hydrogen. At NTP, T $=273 \mathrm{~K}$ $\therefore \sqrt{\frac{3 \mathrm{RT}_{1}}{\mathrm{M}_{\mathrm{O}_{2}}}}=\frac{1}{2} \sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}_{\mathrm{H}_{2}}}}$ Taking square both side, we get - $\frac{\mathrm{T}_{1}}{\mathrm{M}_{\mathrm{O}_{2}}}=\frac{1}{4} \times \frac{\mathrm{T}}{\mathrm{M}_{\mathrm{H}_{2}}}$ $\frac{\mathrm{T}_{1}}{32}=\frac{1}{4} \times \frac{273}{2}$ $\frac{\mathrm{T}_{1}}{32}=\frac{273}{8}$ $\mathrm{~T}_{1}=4 \times 273$ $\mathrm{~T}_{1}=1092 \mathrm{~K}$ $\mathrm{~T}_{1}=1092-273=819^{\circ} \mathrm{C}$ $\mathrm{T}_{1}=819^{\circ} \mathrm{C}$
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 $\gamma$
2 $\sqrt{\frac{2 \gamma}{3}}$
3 $\gamma-1$
4 $\sqrt{\frac{\gamma}{3}}$
Explanation:
D Given, The root mean square velocity of molecule is, $\quad \mathrm{v}_{\mathrm{rms}}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}}}$ $\because \mathrm{PV}=\mathrm{RT} \text { and } \rho=\frac{\mathrm{M}}{\mathrm{V}}$ $\mathrm{v}_{\mathrm{rms}}=\sqrt{\frac{3 \mathrm{PM}}{\rho \mathrm{M}}}$ $\mathrm{v}_{\text {rms }}=\sqrt{\frac{3 \mathrm{P}}{\rho}}$ $\text { Speed of sound }(\mathrm{v})=\sqrt{\frac{\gamma \mathrm{P}}{\rho}}$ So, $\frac{\mathrm{v}_{\mathrm{s}}}{\mathrm{v}_{\mathrm{rms}}}=\frac{\sqrt{\frac{\gamma \mathrm{P}}{\rho}}}{\sqrt{\frac{3 \mathrm{P}}{\rho}}}=\sqrt{\frac{\gamma}{3}}$
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 $\left(C_{V}\right)$ and constant $(R)$
1 $\mathrm{C}_{\mathrm{V}}=\frac{\mathrm{R}}{2}$
2 $\mathrm{C}_{\mathrm{V}}=\mathrm{R}$
3 $\mathrm{C}_{\mathrm{V}}=2 \mathrm{R}$
4 $\mathrm{C}_{\mathrm{V}}=3 \mathrm{R}$
Explanation:
D Given, degrees of freedom (f) $=6$ We know that, $\mathrm{C}_{\mathrm{P}}-\mathrm{C}_{\mathrm{V}}=\mathrm{R}$ And, $\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}=\gamma=1+\frac{2}{\mathrm{f}}$ $\frac{C_{P}}{C_{V}}=1+\frac{2}{6}=\frac{4}{3}$ $C_{P}=\frac{4}{3} C_{V}$ 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 $0.0000893 \mathrm{gm} / \mathrm{cm}^{3}$, then the root mean square velocity of molecules of hydrogen molecules at NTP is
NEET Test Series from KOTA - 10 Papers In MS WORD
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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 $1092^{\circ} \mathrm{C}$
2 $1492^{\circ} \mathrm{C}$
3 $273 \mathrm{~K}$
4 $819^{\circ} \mathrm{C}$
Explanation:
D Let temperature of oxygen $=\mathrm{T}_{1} \mathrm{~K}$ at which velocity of oxygen equal to velocity of hydrogen. At NTP, T $=273 \mathrm{~K}$ $\therefore \sqrt{\frac{3 \mathrm{RT}_{1}}{\mathrm{M}_{\mathrm{O}_{2}}}}=\frac{1}{2} \sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}_{\mathrm{H}_{2}}}}$ Taking square both side, we get - $\frac{\mathrm{T}_{1}}{\mathrm{M}_{\mathrm{O}_{2}}}=\frac{1}{4} \times \frac{\mathrm{T}}{\mathrm{M}_{\mathrm{H}_{2}}}$ $\frac{\mathrm{T}_{1}}{32}=\frac{1}{4} \times \frac{273}{2}$ $\frac{\mathrm{T}_{1}}{32}=\frac{273}{8}$ $\mathrm{~T}_{1}=4 \times 273$ $\mathrm{~T}_{1}=1092 \mathrm{~K}$ $\mathrm{~T}_{1}=1092-273=819^{\circ} \mathrm{C}$ $\mathrm{T}_{1}=819^{\circ} \mathrm{C}$
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 $\gamma$
2 $\sqrt{\frac{2 \gamma}{3}}$
3 $\gamma-1$
4 $\sqrt{\frac{\gamma}{3}}$
Explanation:
D Given, The root mean square velocity of molecule is, $\quad \mathrm{v}_{\mathrm{rms}}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}}}$ $\because \mathrm{PV}=\mathrm{RT} \text { and } \rho=\frac{\mathrm{M}}{\mathrm{V}}$ $\mathrm{v}_{\mathrm{rms}}=\sqrt{\frac{3 \mathrm{PM}}{\rho \mathrm{M}}}$ $\mathrm{v}_{\text {rms }}=\sqrt{\frac{3 \mathrm{P}}{\rho}}$ $\text { Speed of sound }(\mathrm{v})=\sqrt{\frac{\gamma \mathrm{P}}{\rho}}$ So, $\frac{\mathrm{v}_{\mathrm{s}}}{\mathrm{v}_{\mathrm{rms}}}=\frac{\sqrt{\frac{\gamma \mathrm{P}}{\rho}}}{\sqrt{\frac{3 \mathrm{P}}{\rho}}}=\sqrt{\frac{\gamma}{3}}$
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 $\left(C_{V}\right)$ and constant $(R)$
1 $\mathrm{C}_{\mathrm{V}}=\frac{\mathrm{R}}{2}$
2 $\mathrm{C}_{\mathrm{V}}=\mathrm{R}$
3 $\mathrm{C}_{\mathrm{V}}=2 \mathrm{R}$
4 $\mathrm{C}_{\mathrm{V}}=3 \mathrm{R}$
Explanation:
D Given, degrees of freedom (f) $=6$ We know that, $\mathrm{C}_{\mathrm{P}}-\mathrm{C}_{\mathrm{V}}=\mathrm{R}$ And, $\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}=\gamma=1+\frac{2}{\mathrm{f}}$ $\frac{C_{P}}{C_{V}}=1+\frac{2}{6}=\frac{4}{3}$ $C_{P}=\frac{4}{3} C_{V}$ 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 $0.0000893 \mathrm{gm} / \mathrm{cm}^{3}$, then the root mean square velocity of molecules of hydrogen molecules at NTP is
139248
The temperature at which the velocity of oxygen will be half that of hydrogen at NTP is
1 $1092^{\circ} \mathrm{C}$
2 $1492^{\circ} \mathrm{C}$
3 $273 \mathrm{~K}$
4 $819^{\circ} \mathrm{C}$
Explanation:
D Let temperature of oxygen $=\mathrm{T}_{1} \mathrm{~K}$ at which velocity of oxygen equal to velocity of hydrogen. At NTP, T $=273 \mathrm{~K}$ $\therefore \sqrt{\frac{3 \mathrm{RT}_{1}}{\mathrm{M}_{\mathrm{O}_{2}}}}=\frac{1}{2} \sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}_{\mathrm{H}_{2}}}}$ Taking square both side, we get - $\frac{\mathrm{T}_{1}}{\mathrm{M}_{\mathrm{O}_{2}}}=\frac{1}{4} \times \frac{\mathrm{T}}{\mathrm{M}_{\mathrm{H}_{2}}}$ $\frac{\mathrm{T}_{1}}{32}=\frac{1}{4} \times \frac{273}{2}$ $\frac{\mathrm{T}_{1}}{32}=\frac{273}{8}$ $\mathrm{~T}_{1}=4 \times 273$ $\mathrm{~T}_{1}=1092 \mathrm{~K}$ $\mathrm{~T}_{1}=1092-273=819^{\circ} \mathrm{C}$ $\mathrm{T}_{1}=819^{\circ} \mathrm{C}$
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 $\gamma$
2 $\sqrt{\frac{2 \gamma}{3}}$
3 $\gamma-1$
4 $\sqrt{\frac{\gamma}{3}}$
Explanation:
D Given, The root mean square velocity of molecule is, $\quad \mathrm{v}_{\mathrm{rms}}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}}}$ $\because \mathrm{PV}=\mathrm{RT} \text { and } \rho=\frac{\mathrm{M}}{\mathrm{V}}$ $\mathrm{v}_{\mathrm{rms}}=\sqrt{\frac{3 \mathrm{PM}}{\rho \mathrm{M}}}$ $\mathrm{v}_{\text {rms }}=\sqrt{\frac{3 \mathrm{P}}{\rho}}$ $\text { Speed of sound }(\mathrm{v})=\sqrt{\frac{\gamma \mathrm{P}}{\rho}}$ So, $\frac{\mathrm{v}_{\mathrm{s}}}{\mathrm{v}_{\mathrm{rms}}}=\frac{\sqrt{\frac{\gamma \mathrm{P}}{\rho}}}{\sqrt{\frac{3 \mathrm{P}}{\rho}}}=\sqrt{\frac{\gamma}{3}}$
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 $\left(C_{V}\right)$ and constant $(R)$
1 $\mathrm{C}_{\mathrm{V}}=\frac{\mathrm{R}}{2}$
2 $\mathrm{C}_{\mathrm{V}}=\mathrm{R}$
3 $\mathrm{C}_{\mathrm{V}}=2 \mathrm{R}$
4 $\mathrm{C}_{\mathrm{V}}=3 \mathrm{R}$
Explanation:
D Given, degrees of freedom (f) $=6$ We know that, $\mathrm{C}_{\mathrm{P}}-\mathrm{C}_{\mathrm{V}}=\mathrm{R}$ And, $\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}=\gamma=1+\frac{2}{\mathrm{f}}$ $\frac{C_{P}}{C_{V}}=1+\frac{2}{6}=\frac{4}{3}$ $C_{P}=\frac{4}{3} C_{V}$ 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 $0.0000893 \mathrm{gm} / \mathrm{cm}^{3}$, then the root mean square velocity of molecules of hydrogen molecules at NTP is
139248
The temperature at which the velocity of oxygen will be half that of hydrogen at NTP is
1 $1092^{\circ} \mathrm{C}$
2 $1492^{\circ} \mathrm{C}$
3 $273 \mathrm{~K}$
4 $819^{\circ} \mathrm{C}$
Explanation:
D Let temperature of oxygen $=\mathrm{T}_{1} \mathrm{~K}$ at which velocity of oxygen equal to velocity of hydrogen. At NTP, T $=273 \mathrm{~K}$ $\therefore \sqrt{\frac{3 \mathrm{RT}_{1}}{\mathrm{M}_{\mathrm{O}_{2}}}}=\frac{1}{2} \sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}_{\mathrm{H}_{2}}}}$ Taking square both side, we get - $\frac{\mathrm{T}_{1}}{\mathrm{M}_{\mathrm{O}_{2}}}=\frac{1}{4} \times \frac{\mathrm{T}}{\mathrm{M}_{\mathrm{H}_{2}}}$ $\frac{\mathrm{T}_{1}}{32}=\frac{1}{4} \times \frac{273}{2}$ $\frac{\mathrm{T}_{1}}{32}=\frac{273}{8}$ $\mathrm{~T}_{1}=4 \times 273$ $\mathrm{~T}_{1}=1092 \mathrm{~K}$ $\mathrm{~T}_{1}=1092-273=819^{\circ} \mathrm{C}$ $\mathrm{T}_{1}=819^{\circ} \mathrm{C}$
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 $\gamma$
2 $\sqrt{\frac{2 \gamma}{3}}$
3 $\gamma-1$
4 $\sqrt{\frac{\gamma}{3}}$
Explanation:
D Given, The root mean square velocity of molecule is, $\quad \mathrm{v}_{\mathrm{rms}}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}}}$ $\because \mathrm{PV}=\mathrm{RT} \text { and } \rho=\frac{\mathrm{M}}{\mathrm{V}}$ $\mathrm{v}_{\mathrm{rms}}=\sqrt{\frac{3 \mathrm{PM}}{\rho \mathrm{M}}}$ $\mathrm{v}_{\text {rms }}=\sqrt{\frac{3 \mathrm{P}}{\rho}}$ $\text { Speed of sound }(\mathrm{v})=\sqrt{\frac{\gamma \mathrm{P}}{\rho}}$ So, $\frac{\mathrm{v}_{\mathrm{s}}}{\mathrm{v}_{\mathrm{rms}}}=\frac{\sqrt{\frac{\gamma \mathrm{P}}{\rho}}}{\sqrt{\frac{3 \mathrm{P}}{\rho}}}=\sqrt{\frac{\gamma}{3}}$
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 $\left(C_{V}\right)$ and constant $(R)$
1 $\mathrm{C}_{\mathrm{V}}=\frac{\mathrm{R}}{2}$
2 $\mathrm{C}_{\mathrm{V}}=\mathrm{R}$
3 $\mathrm{C}_{\mathrm{V}}=2 \mathrm{R}$
4 $\mathrm{C}_{\mathrm{V}}=3 \mathrm{R}$
Explanation:
D Given, degrees of freedom (f) $=6$ We know that, $\mathrm{C}_{\mathrm{P}}-\mathrm{C}_{\mathrm{V}}=\mathrm{R}$ And, $\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}=\gamma=1+\frac{2}{\mathrm{f}}$ $\frac{C_{P}}{C_{V}}=1+\frac{2}{6}=\frac{4}{3}$ $C_{P}=\frac{4}{3} C_{V}$ 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 $0.0000893 \mathrm{gm} / \mathrm{cm}^{3}$, then the root mean square velocity of molecules of hydrogen molecules at NTP is
