139113
Consider a gas with density $\rho$ and $\bar{c}$ as the root mean square velocity of its molecules contained in a volume. If the system moves as whole with velocity $v$, then the pressure exerted by the gas is-
A Since we know that, $\mathrm{v}_{\mathrm{rms}}=\overline{\mathrm{c}}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}}} \ldots \text { (i) }$ For ideal gas $\mathrm{PV}=\mathrm{nRT}$ $\mathrm{PV}=\frac{\mathrm{m}}{\mathrm{M}} \mathrm{RT} \quad\left[\because \mathrm{n}=\frac{\mathrm{m}}{\mathrm{M}}\right]$ $\mathrm{P}=\frac{\mathrm{m}}{\mathrm{VM}} \mathrm{RT} \quad\left[\because \rho(\text { density })=\frac{\mathrm{m}}{\mathrm{V}}\right]$ $\mathrm{P}=\frac{\rho}{\mathrm{M}} \cdot \mathrm{RT}$ $\mathrm{P}=\rho \frac{\mathrm{RT}}{\mathrm{M}}$ $\therefore \quad \frac{\mathrm{RT}}{\mathrm{M}}=\frac{\mathrm{P}}{\rho}$ $\overline{\mathrm{c}}=\sqrt{\frac{3 \mathrm{P}}{\rho}}$ $\bar{c}^{2}=\frac{3 P}{\rho}$ $\mathrm{P}=\frac{1}{3} \rho \bar{c}^{2}$
BCECE-2011
Kinetic Theory of Gases
139114
A vessel contains a mixture of one mole of oxygen and two moles of nitrogen at $300 \mathrm{~K}$. The ratio of the average rotational $K E$ per oxygen molecule that per nitrogen molecule is-
1 $1: 1$
2 $1: 2$
3 $2: 1$
4 depends on the moments of inertia of the two molecules
Explanation:
A Here, the energy is equally distributed among its degree of freedom. $\because$ Average energy per degree of freedom per molecule $\mathrm{E}=\frac{1}{2} \mathrm{KT}$ Degree of freedom of $\mathrm{O}_{2}=3 \mathrm{~N}-1(\mathrm{~N}=2)$ $=3 \times 2-1$ $=5$ Degree of freedom of $\mathrm{N}_{2}=3 \mathrm{~N}-1(\mathrm{~N}=2)$ $=3 \times 2-1$ $=5$ Hence, it is clear that if the degree of freedom of both are same then the average kinetic energy of both will be same. Therefore, the ratio will be $=\frac{\mathrm{E}_{\mathrm{O}_{2}}}{\mathrm{E}_{\mathrm{N}_{2}}}=\frac{\mathrm{f}_{\mathrm{O}_{2}}}{\mathrm{f}_{\mathrm{N}_{2}}}=\frac{5}{5}=\frac{1}{1}$ $\mathrm{E}_{\mathrm{O}_{2}}: \mathrm{E}_{\mathrm{N}_{2}}=1: 1$
BCECE-2011
Kinetic Theory of Gases
139115
The average kinetic energy of the molecules of a low density gas at $27^{\circ} \mathrm{C}$ is
139116
To what temperature should the hydrogen at $327^{\circ} \mathrm{C}$ be cooled at constant pressure, so that the root mean square velocity of its molecules becomes half of its previous value?
1 $-123^{\circ} \mathrm{C}$
2 $123^{\circ} \mathrm{C}$
3 $-100^{\circ} \mathrm{C}$
4 $0^{\circ} \mathrm{C}$
Explanation:
A As we know that, According to root mean square velocity of its molecules is given by $\mathrm{V}_{\mathrm{rms}}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}}}$ Hence, we can write expression from above equation- $\mathrm{T} \propto \mathrm{V}_{\mathrm{rms}}^{2}$ $\therefore \quad \frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=\left[\frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}\right]^{2}$ $\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=\left(\frac{\mathrm{V}}{\frac{2}{\mathrm{~V}}}\right)^{2}=\frac{1}{4}$ $\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=\frac{1}{4}$ $\mathrm{~T}_{2} =\frac{\mathrm{T}_{1}}{4}=\frac{273+327}{4}$ $\mathrm{~T}_{2} =\frac{600 \mathrm{~K}}{4}$ $=150 \mathrm{~K}$ $=150-273$ $\mathrm{~T}_{2} =-123^{\circ} \mathrm{C}$
MHT-CET 2009
Kinetic Theory of Gases
139117
If $\gamma$ be the ratio of specific heats of a perfect gas, the number of degrees of freedom of a molecule of the gas is
1 $\frac{25}{2}(\gamma-1)$
2 $\frac{3 \gamma-1}{2 \gamma-1}$
3 $\frac{2}{\gamma-1}$
4 $\frac{9}{2}(\gamma-1)$
Explanation:
C Given that, $\gamma=$ Ratio of specific heat $\mathrm{f}=$ degree of freedom $\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}=\gamma$ $C_{P}=\left(1+\frac{f}{2}\right) R$ and $C_{V}=\frac{f}{2} R$ So, $\quad \gamma=\frac{\left(1+\frac{\mathrm{f}}{2}\right) \mathrm{R}}{\frac{\mathrm{f}}{2} \mathrm{R}}$ $\gamma=\frac{2}{f}+1$ $f=\frac{2}{\gamma-1}$
139113
Consider a gas with density $\rho$ and $\bar{c}$ as the root mean square velocity of its molecules contained in a volume. If the system moves as whole with velocity $v$, then the pressure exerted by the gas is-
A Since we know that, $\mathrm{v}_{\mathrm{rms}}=\overline{\mathrm{c}}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}}} \ldots \text { (i) }$ For ideal gas $\mathrm{PV}=\mathrm{nRT}$ $\mathrm{PV}=\frac{\mathrm{m}}{\mathrm{M}} \mathrm{RT} \quad\left[\because \mathrm{n}=\frac{\mathrm{m}}{\mathrm{M}}\right]$ $\mathrm{P}=\frac{\mathrm{m}}{\mathrm{VM}} \mathrm{RT} \quad\left[\because \rho(\text { density })=\frac{\mathrm{m}}{\mathrm{V}}\right]$ $\mathrm{P}=\frac{\rho}{\mathrm{M}} \cdot \mathrm{RT}$ $\mathrm{P}=\rho \frac{\mathrm{RT}}{\mathrm{M}}$ $\therefore \quad \frac{\mathrm{RT}}{\mathrm{M}}=\frac{\mathrm{P}}{\rho}$ $\overline{\mathrm{c}}=\sqrt{\frac{3 \mathrm{P}}{\rho}}$ $\bar{c}^{2}=\frac{3 P}{\rho}$ $\mathrm{P}=\frac{1}{3} \rho \bar{c}^{2}$
BCECE-2011
Kinetic Theory of Gases
139114
A vessel contains a mixture of one mole of oxygen and two moles of nitrogen at $300 \mathrm{~K}$. The ratio of the average rotational $K E$ per oxygen molecule that per nitrogen molecule is-
1 $1: 1$
2 $1: 2$
3 $2: 1$
4 depends on the moments of inertia of the two molecules
Explanation:
A Here, the energy is equally distributed among its degree of freedom. $\because$ Average energy per degree of freedom per molecule $\mathrm{E}=\frac{1}{2} \mathrm{KT}$ Degree of freedom of $\mathrm{O}_{2}=3 \mathrm{~N}-1(\mathrm{~N}=2)$ $=3 \times 2-1$ $=5$ Degree of freedom of $\mathrm{N}_{2}=3 \mathrm{~N}-1(\mathrm{~N}=2)$ $=3 \times 2-1$ $=5$ Hence, it is clear that if the degree of freedom of both are same then the average kinetic energy of both will be same. Therefore, the ratio will be $=\frac{\mathrm{E}_{\mathrm{O}_{2}}}{\mathrm{E}_{\mathrm{N}_{2}}}=\frac{\mathrm{f}_{\mathrm{O}_{2}}}{\mathrm{f}_{\mathrm{N}_{2}}}=\frac{5}{5}=\frac{1}{1}$ $\mathrm{E}_{\mathrm{O}_{2}}: \mathrm{E}_{\mathrm{N}_{2}}=1: 1$
BCECE-2011
Kinetic Theory of Gases
139115
The average kinetic energy of the molecules of a low density gas at $27^{\circ} \mathrm{C}$ is
139116
To what temperature should the hydrogen at $327^{\circ} \mathrm{C}$ be cooled at constant pressure, so that the root mean square velocity of its molecules becomes half of its previous value?
1 $-123^{\circ} \mathrm{C}$
2 $123^{\circ} \mathrm{C}$
3 $-100^{\circ} \mathrm{C}$
4 $0^{\circ} \mathrm{C}$
Explanation:
A As we know that, According to root mean square velocity of its molecules is given by $\mathrm{V}_{\mathrm{rms}}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}}}$ Hence, we can write expression from above equation- $\mathrm{T} \propto \mathrm{V}_{\mathrm{rms}}^{2}$ $\therefore \quad \frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=\left[\frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}\right]^{2}$ $\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=\left(\frac{\mathrm{V}}{\frac{2}{\mathrm{~V}}}\right)^{2}=\frac{1}{4}$ $\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=\frac{1}{4}$ $\mathrm{~T}_{2} =\frac{\mathrm{T}_{1}}{4}=\frac{273+327}{4}$ $\mathrm{~T}_{2} =\frac{600 \mathrm{~K}}{4}$ $=150 \mathrm{~K}$ $=150-273$ $\mathrm{~T}_{2} =-123^{\circ} \mathrm{C}$
MHT-CET 2009
Kinetic Theory of Gases
139117
If $\gamma$ be the ratio of specific heats of a perfect gas, the number of degrees of freedom of a molecule of the gas is
1 $\frac{25}{2}(\gamma-1)$
2 $\frac{3 \gamma-1}{2 \gamma-1}$
3 $\frac{2}{\gamma-1}$
4 $\frac{9}{2}(\gamma-1)$
Explanation:
C Given that, $\gamma=$ Ratio of specific heat $\mathrm{f}=$ degree of freedom $\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}=\gamma$ $C_{P}=\left(1+\frac{f}{2}\right) R$ and $C_{V}=\frac{f}{2} R$ So, $\quad \gamma=\frac{\left(1+\frac{\mathrm{f}}{2}\right) \mathrm{R}}{\frac{\mathrm{f}}{2} \mathrm{R}}$ $\gamma=\frac{2}{f}+1$ $f=\frac{2}{\gamma-1}$
139113
Consider a gas with density $\rho$ and $\bar{c}$ as the root mean square velocity of its molecules contained in a volume. If the system moves as whole with velocity $v$, then the pressure exerted by the gas is-
A Since we know that, $\mathrm{v}_{\mathrm{rms}}=\overline{\mathrm{c}}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}}} \ldots \text { (i) }$ For ideal gas $\mathrm{PV}=\mathrm{nRT}$ $\mathrm{PV}=\frac{\mathrm{m}}{\mathrm{M}} \mathrm{RT} \quad\left[\because \mathrm{n}=\frac{\mathrm{m}}{\mathrm{M}}\right]$ $\mathrm{P}=\frac{\mathrm{m}}{\mathrm{VM}} \mathrm{RT} \quad\left[\because \rho(\text { density })=\frac{\mathrm{m}}{\mathrm{V}}\right]$ $\mathrm{P}=\frac{\rho}{\mathrm{M}} \cdot \mathrm{RT}$ $\mathrm{P}=\rho \frac{\mathrm{RT}}{\mathrm{M}}$ $\therefore \quad \frac{\mathrm{RT}}{\mathrm{M}}=\frac{\mathrm{P}}{\rho}$ $\overline{\mathrm{c}}=\sqrt{\frac{3 \mathrm{P}}{\rho}}$ $\bar{c}^{2}=\frac{3 P}{\rho}$ $\mathrm{P}=\frac{1}{3} \rho \bar{c}^{2}$
BCECE-2011
Kinetic Theory of Gases
139114
A vessel contains a mixture of one mole of oxygen and two moles of nitrogen at $300 \mathrm{~K}$. The ratio of the average rotational $K E$ per oxygen molecule that per nitrogen molecule is-
1 $1: 1$
2 $1: 2$
3 $2: 1$
4 depends on the moments of inertia of the two molecules
Explanation:
A Here, the energy is equally distributed among its degree of freedom. $\because$ Average energy per degree of freedom per molecule $\mathrm{E}=\frac{1}{2} \mathrm{KT}$ Degree of freedom of $\mathrm{O}_{2}=3 \mathrm{~N}-1(\mathrm{~N}=2)$ $=3 \times 2-1$ $=5$ Degree of freedom of $\mathrm{N}_{2}=3 \mathrm{~N}-1(\mathrm{~N}=2)$ $=3 \times 2-1$ $=5$ Hence, it is clear that if the degree of freedom of both are same then the average kinetic energy of both will be same. Therefore, the ratio will be $=\frac{\mathrm{E}_{\mathrm{O}_{2}}}{\mathrm{E}_{\mathrm{N}_{2}}}=\frac{\mathrm{f}_{\mathrm{O}_{2}}}{\mathrm{f}_{\mathrm{N}_{2}}}=\frac{5}{5}=\frac{1}{1}$ $\mathrm{E}_{\mathrm{O}_{2}}: \mathrm{E}_{\mathrm{N}_{2}}=1: 1$
BCECE-2011
Kinetic Theory of Gases
139115
The average kinetic energy of the molecules of a low density gas at $27^{\circ} \mathrm{C}$ is
139116
To what temperature should the hydrogen at $327^{\circ} \mathrm{C}$ be cooled at constant pressure, so that the root mean square velocity of its molecules becomes half of its previous value?
1 $-123^{\circ} \mathrm{C}$
2 $123^{\circ} \mathrm{C}$
3 $-100^{\circ} \mathrm{C}$
4 $0^{\circ} \mathrm{C}$
Explanation:
A As we know that, According to root mean square velocity of its molecules is given by $\mathrm{V}_{\mathrm{rms}}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}}}$ Hence, we can write expression from above equation- $\mathrm{T} \propto \mathrm{V}_{\mathrm{rms}}^{2}$ $\therefore \quad \frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=\left[\frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}\right]^{2}$ $\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=\left(\frac{\mathrm{V}}{\frac{2}{\mathrm{~V}}}\right)^{2}=\frac{1}{4}$ $\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=\frac{1}{4}$ $\mathrm{~T}_{2} =\frac{\mathrm{T}_{1}}{4}=\frac{273+327}{4}$ $\mathrm{~T}_{2} =\frac{600 \mathrm{~K}}{4}$ $=150 \mathrm{~K}$ $=150-273$ $\mathrm{~T}_{2} =-123^{\circ} \mathrm{C}$
MHT-CET 2009
Kinetic Theory of Gases
139117
If $\gamma$ be the ratio of specific heats of a perfect gas, the number of degrees of freedom of a molecule of the gas is
1 $\frac{25}{2}(\gamma-1)$
2 $\frac{3 \gamma-1}{2 \gamma-1}$
3 $\frac{2}{\gamma-1}$
4 $\frac{9}{2}(\gamma-1)$
Explanation:
C Given that, $\gamma=$ Ratio of specific heat $\mathrm{f}=$ degree of freedom $\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}=\gamma$ $C_{P}=\left(1+\frac{f}{2}\right) R$ and $C_{V}=\frac{f}{2} R$ So, $\quad \gamma=\frac{\left(1+\frac{\mathrm{f}}{2}\right) \mathrm{R}}{\frac{\mathrm{f}}{2} \mathrm{R}}$ $\gamma=\frac{2}{f}+1$ $f=\frac{2}{\gamma-1}$
139113
Consider a gas with density $\rho$ and $\bar{c}$ as the root mean square velocity of its molecules contained in a volume. If the system moves as whole with velocity $v$, then the pressure exerted by the gas is-
A Since we know that, $\mathrm{v}_{\mathrm{rms}}=\overline{\mathrm{c}}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}}} \ldots \text { (i) }$ For ideal gas $\mathrm{PV}=\mathrm{nRT}$ $\mathrm{PV}=\frac{\mathrm{m}}{\mathrm{M}} \mathrm{RT} \quad\left[\because \mathrm{n}=\frac{\mathrm{m}}{\mathrm{M}}\right]$ $\mathrm{P}=\frac{\mathrm{m}}{\mathrm{VM}} \mathrm{RT} \quad\left[\because \rho(\text { density })=\frac{\mathrm{m}}{\mathrm{V}}\right]$ $\mathrm{P}=\frac{\rho}{\mathrm{M}} \cdot \mathrm{RT}$ $\mathrm{P}=\rho \frac{\mathrm{RT}}{\mathrm{M}}$ $\therefore \quad \frac{\mathrm{RT}}{\mathrm{M}}=\frac{\mathrm{P}}{\rho}$ $\overline{\mathrm{c}}=\sqrt{\frac{3 \mathrm{P}}{\rho}}$ $\bar{c}^{2}=\frac{3 P}{\rho}$ $\mathrm{P}=\frac{1}{3} \rho \bar{c}^{2}$
BCECE-2011
Kinetic Theory of Gases
139114
A vessel contains a mixture of one mole of oxygen and two moles of nitrogen at $300 \mathrm{~K}$. The ratio of the average rotational $K E$ per oxygen molecule that per nitrogen molecule is-
1 $1: 1$
2 $1: 2$
3 $2: 1$
4 depends on the moments of inertia of the two molecules
Explanation:
A Here, the energy is equally distributed among its degree of freedom. $\because$ Average energy per degree of freedom per molecule $\mathrm{E}=\frac{1}{2} \mathrm{KT}$ Degree of freedom of $\mathrm{O}_{2}=3 \mathrm{~N}-1(\mathrm{~N}=2)$ $=3 \times 2-1$ $=5$ Degree of freedom of $\mathrm{N}_{2}=3 \mathrm{~N}-1(\mathrm{~N}=2)$ $=3 \times 2-1$ $=5$ Hence, it is clear that if the degree of freedom of both are same then the average kinetic energy of both will be same. Therefore, the ratio will be $=\frac{\mathrm{E}_{\mathrm{O}_{2}}}{\mathrm{E}_{\mathrm{N}_{2}}}=\frac{\mathrm{f}_{\mathrm{O}_{2}}}{\mathrm{f}_{\mathrm{N}_{2}}}=\frac{5}{5}=\frac{1}{1}$ $\mathrm{E}_{\mathrm{O}_{2}}: \mathrm{E}_{\mathrm{N}_{2}}=1: 1$
BCECE-2011
Kinetic Theory of Gases
139115
The average kinetic energy of the molecules of a low density gas at $27^{\circ} \mathrm{C}$ is
139116
To what temperature should the hydrogen at $327^{\circ} \mathrm{C}$ be cooled at constant pressure, so that the root mean square velocity of its molecules becomes half of its previous value?
1 $-123^{\circ} \mathrm{C}$
2 $123^{\circ} \mathrm{C}$
3 $-100^{\circ} \mathrm{C}$
4 $0^{\circ} \mathrm{C}$
Explanation:
A As we know that, According to root mean square velocity of its molecules is given by $\mathrm{V}_{\mathrm{rms}}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}}}$ Hence, we can write expression from above equation- $\mathrm{T} \propto \mathrm{V}_{\mathrm{rms}}^{2}$ $\therefore \quad \frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=\left[\frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}\right]^{2}$ $\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=\left(\frac{\mathrm{V}}{\frac{2}{\mathrm{~V}}}\right)^{2}=\frac{1}{4}$ $\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=\frac{1}{4}$ $\mathrm{~T}_{2} =\frac{\mathrm{T}_{1}}{4}=\frac{273+327}{4}$ $\mathrm{~T}_{2} =\frac{600 \mathrm{~K}}{4}$ $=150 \mathrm{~K}$ $=150-273$ $\mathrm{~T}_{2} =-123^{\circ} \mathrm{C}$
MHT-CET 2009
Kinetic Theory of Gases
139117
If $\gamma$ be the ratio of specific heats of a perfect gas, the number of degrees of freedom of a molecule of the gas is
1 $\frac{25}{2}(\gamma-1)$
2 $\frac{3 \gamma-1}{2 \gamma-1}$
3 $\frac{2}{\gamma-1}$
4 $\frac{9}{2}(\gamma-1)$
Explanation:
C Given that, $\gamma=$ Ratio of specific heat $\mathrm{f}=$ degree of freedom $\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}=\gamma$ $C_{P}=\left(1+\frac{f}{2}\right) R$ and $C_{V}=\frac{f}{2} R$ So, $\quad \gamma=\frac{\left(1+\frac{\mathrm{f}}{2}\right) \mathrm{R}}{\frac{\mathrm{f}}{2} \mathrm{R}}$ $\gamma=\frac{2}{f}+1$ $f=\frac{2}{\gamma-1}$
139113
Consider a gas with density $\rho$ and $\bar{c}$ as the root mean square velocity of its molecules contained in a volume. If the system moves as whole with velocity $v$, then the pressure exerted by the gas is-
A Since we know that, $\mathrm{v}_{\mathrm{rms}}=\overline{\mathrm{c}}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}}} \ldots \text { (i) }$ For ideal gas $\mathrm{PV}=\mathrm{nRT}$ $\mathrm{PV}=\frac{\mathrm{m}}{\mathrm{M}} \mathrm{RT} \quad\left[\because \mathrm{n}=\frac{\mathrm{m}}{\mathrm{M}}\right]$ $\mathrm{P}=\frac{\mathrm{m}}{\mathrm{VM}} \mathrm{RT} \quad\left[\because \rho(\text { density })=\frac{\mathrm{m}}{\mathrm{V}}\right]$ $\mathrm{P}=\frac{\rho}{\mathrm{M}} \cdot \mathrm{RT}$ $\mathrm{P}=\rho \frac{\mathrm{RT}}{\mathrm{M}}$ $\therefore \quad \frac{\mathrm{RT}}{\mathrm{M}}=\frac{\mathrm{P}}{\rho}$ $\overline{\mathrm{c}}=\sqrt{\frac{3 \mathrm{P}}{\rho}}$ $\bar{c}^{2}=\frac{3 P}{\rho}$ $\mathrm{P}=\frac{1}{3} \rho \bar{c}^{2}$
BCECE-2011
Kinetic Theory of Gases
139114
A vessel contains a mixture of one mole of oxygen and two moles of nitrogen at $300 \mathrm{~K}$. The ratio of the average rotational $K E$ per oxygen molecule that per nitrogen molecule is-
1 $1: 1$
2 $1: 2$
3 $2: 1$
4 depends on the moments of inertia of the two molecules
Explanation:
A Here, the energy is equally distributed among its degree of freedom. $\because$ Average energy per degree of freedom per molecule $\mathrm{E}=\frac{1}{2} \mathrm{KT}$ Degree of freedom of $\mathrm{O}_{2}=3 \mathrm{~N}-1(\mathrm{~N}=2)$ $=3 \times 2-1$ $=5$ Degree of freedom of $\mathrm{N}_{2}=3 \mathrm{~N}-1(\mathrm{~N}=2)$ $=3 \times 2-1$ $=5$ Hence, it is clear that if the degree of freedom of both are same then the average kinetic energy of both will be same. Therefore, the ratio will be $=\frac{\mathrm{E}_{\mathrm{O}_{2}}}{\mathrm{E}_{\mathrm{N}_{2}}}=\frac{\mathrm{f}_{\mathrm{O}_{2}}}{\mathrm{f}_{\mathrm{N}_{2}}}=\frac{5}{5}=\frac{1}{1}$ $\mathrm{E}_{\mathrm{O}_{2}}: \mathrm{E}_{\mathrm{N}_{2}}=1: 1$
BCECE-2011
Kinetic Theory of Gases
139115
The average kinetic energy of the molecules of a low density gas at $27^{\circ} \mathrm{C}$ is
139116
To what temperature should the hydrogen at $327^{\circ} \mathrm{C}$ be cooled at constant pressure, so that the root mean square velocity of its molecules becomes half of its previous value?
1 $-123^{\circ} \mathrm{C}$
2 $123^{\circ} \mathrm{C}$
3 $-100^{\circ} \mathrm{C}$
4 $0^{\circ} \mathrm{C}$
Explanation:
A As we know that, According to root mean square velocity of its molecules is given by $\mathrm{V}_{\mathrm{rms}}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}}}$ Hence, we can write expression from above equation- $\mathrm{T} \propto \mathrm{V}_{\mathrm{rms}}^{2}$ $\therefore \quad \frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=\left[\frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}\right]^{2}$ $\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=\left(\frac{\mathrm{V}}{\frac{2}{\mathrm{~V}}}\right)^{2}=\frac{1}{4}$ $\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=\frac{1}{4}$ $\mathrm{~T}_{2} =\frac{\mathrm{T}_{1}}{4}=\frac{273+327}{4}$ $\mathrm{~T}_{2} =\frac{600 \mathrm{~K}}{4}$ $=150 \mathrm{~K}$ $=150-273$ $\mathrm{~T}_{2} =-123^{\circ} \mathrm{C}$
MHT-CET 2009
Kinetic Theory of Gases
139117
If $\gamma$ be the ratio of specific heats of a perfect gas, the number of degrees of freedom of a molecule of the gas is
1 $\frac{25}{2}(\gamma-1)$
2 $\frac{3 \gamma-1}{2 \gamma-1}$
3 $\frac{2}{\gamma-1}$
4 $\frac{9}{2}(\gamma-1)$
Explanation:
C Given that, $\gamma=$ Ratio of specific heat $\mathrm{f}=$ degree of freedom $\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}=\gamma$ $C_{P}=\left(1+\frac{f}{2}\right) R$ and $C_{V}=\frac{f}{2} R$ So, $\quad \gamma=\frac{\left(1+\frac{\mathrm{f}}{2}\right) \mathrm{R}}{\frac{\mathrm{f}}{2} \mathrm{R}}$ $\gamma=\frac{2}{f}+1$ $f=\frac{2}{\gamma-1}$
