Degree of Freedom, Various speeds of Gas Molecules
Kinetic Theory of Gases

139104 The root mean square speed of molecules of nitrogen gas at $27^{\circ} \mathrm{C}$ is approximately:
(Given mass of a nitrogen molecule $=4.6 \times 10^{-26} \mathrm{~kg}$ and take Boltzmann constant $\mathrm{k}_{\mathrm{B}}$ $=1.4 \times 10^{-23} \mathrm{JK}^{-1}$ )

1 $523 \mathrm{~m} / \mathrm{s}$
2 $1260 \mathrm{~m} / \mathrm{s}$
3 $91 \mathrm{~m} / \mathrm{s}$
4 $27.4 \mathrm{~m} / \mathrm{s}$
Kinetic Theory of Gases

139105 Three vessels of equal volume contain gases at the same temperature and pressure. The first vessel contains neon (monoatomic), the second contains chlorine (diatomic) and third contains uranium hexafluoride (polyatomic). Arrange these on the basis of their root mean square sped $\left(v_{\text {rms }}\right)$ and choose the correct answer from the options given below:

1 $\mathrm{v}_{\mathrm{rms}}$ (mono) $=\mathrm{v}_{\mathrm{rms}}$ (dia) $=\mathrm{v}_{\mathrm{rms}}$ (poly)
2 $\mathrm{V}_{\mathrm{rms}}\left(\right.$ mono) $>\mathrm{v}_{\mathrm{rms}}$ (dia) $>\mathrm{v}_{\mathrm{rms}}$ (poly)
3 $\mathrm{v}_{\text {rms }}$ (dia) $ \lt \mathrm{v}_{\text {rms }}$ (poly) $ \lt \mathrm{v}_{\text {rms }}$ (mono)
4 $\mathrm{v}_{\text {rms }}$ (mono) $ \lt \mathrm{v}_{\text {rms }}$ (dia) $ \lt \mathrm{v}_{\text {rms }}$ (poly)
Kinetic Theory of Gases

139106 The rms speed of oxygen molecule in a vessel at particular temperature is, $\left(1+\frac{5}{\mathrm{x}}\right)^{\frac{1}{2}} \mathrm{v}$, where $\mathrm{v}$ is the average speed of the molecule. The value of $x$ will be :
$\left(\text { Take } \pi=\frac{22}{7}\right)$

1 28
2 27
3 8
4 4
Kinetic Theory of Gases

139107 The mean free path of molecules of a certain gas at STP is 1500 d , where $d$ is the diameter of the gas molecules. While maintaining the standard pressure, the mean free path of the molecules at 373 K is approximately :

1 1098 d
2 2049 d
3 750 d
4 1500 d
Kinetic Theory of Gases

139104 The root mean square speed of molecules of nitrogen gas at $27^{\circ} \mathrm{C}$ is approximately:
(Given mass of a nitrogen molecule $=4.6 \times 10^{-26} \mathrm{~kg}$ and take Boltzmann constant $\mathrm{k}_{\mathrm{B}}$ $=1.4 \times 10^{-23} \mathrm{JK}^{-1}$ )

1 $523 \mathrm{~m} / \mathrm{s}$
2 $1260 \mathrm{~m} / \mathrm{s}$
3 $91 \mathrm{~m} / \mathrm{s}$
4 $27.4 \mathrm{~m} / \mathrm{s}$
Kinetic Theory of Gases

139105 Three vessels of equal volume contain gases at the same temperature and pressure. The first vessel contains neon (monoatomic), the second contains chlorine (diatomic) and third contains uranium hexafluoride (polyatomic). Arrange these on the basis of their root mean square sped $\left(v_{\text {rms }}\right)$ and choose the correct answer from the options given below:

1 $\mathrm{v}_{\mathrm{rms}}$ (mono) $=\mathrm{v}_{\mathrm{rms}}$ (dia) $=\mathrm{v}_{\mathrm{rms}}$ (poly)
2 $\mathrm{V}_{\mathrm{rms}}\left(\right.$ mono) $>\mathrm{v}_{\mathrm{rms}}$ (dia) $>\mathrm{v}_{\mathrm{rms}}$ (poly)
3 $\mathrm{v}_{\text {rms }}$ (dia) $ \lt \mathrm{v}_{\text {rms }}$ (poly) $ \lt \mathrm{v}_{\text {rms }}$ (mono)
4 $\mathrm{v}_{\text {rms }}$ (mono) $ \lt \mathrm{v}_{\text {rms }}$ (dia) $ \lt \mathrm{v}_{\text {rms }}$ (poly)
Kinetic Theory of Gases

139106 The rms speed of oxygen molecule in a vessel at particular temperature is, $\left(1+\frac{5}{\mathrm{x}}\right)^{\frac{1}{2}} \mathrm{v}$, where $\mathrm{v}$ is the average speed of the molecule. The value of $x$ will be :
$\left(\text { Take } \pi=\frac{22}{7}\right)$

1 28
2 27
3 8
4 4
Kinetic Theory of Gases

139107 The mean free path of molecules of a certain gas at STP is 1500 d , where $d$ is the diameter of the gas molecules. While maintaining the standard pressure, the mean free path of the molecules at 373 K is approximately :

1 1098 d
2 2049 d
3 750 d
4 1500 d
Kinetic Theory of Gases

139104 The root mean square speed of molecules of nitrogen gas at $27^{\circ} \mathrm{C}$ is approximately:
(Given mass of a nitrogen molecule $=4.6 \times 10^{-26} \mathrm{~kg}$ and take Boltzmann constant $\mathrm{k}_{\mathrm{B}}$ $=1.4 \times 10^{-23} \mathrm{JK}^{-1}$ )

1 $523 \mathrm{~m} / \mathrm{s}$
2 $1260 \mathrm{~m} / \mathrm{s}$
3 $91 \mathrm{~m} / \mathrm{s}$
4 $27.4 \mathrm{~m} / \mathrm{s}$
Kinetic Theory of Gases

139105 Three vessels of equal volume contain gases at the same temperature and pressure. The first vessel contains neon (monoatomic), the second contains chlorine (diatomic) and third contains uranium hexafluoride (polyatomic). Arrange these on the basis of their root mean square sped $\left(v_{\text {rms }}\right)$ and choose the correct answer from the options given below:

1 $\mathrm{v}_{\mathrm{rms}}$ (mono) $=\mathrm{v}_{\mathrm{rms}}$ (dia) $=\mathrm{v}_{\mathrm{rms}}$ (poly)
2 $\mathrm{V}_{\mathrm{rms}}\left(\right.$ mono) $>\mathrm{v}_{\mathrm{rms}}$ (dia) $>\mathrm{v}_{\mathrm{rms}}$ (poly)
3 $\mathrm{v}_{\text {rms }}$ (dia) $ \lt \mathrm{v}_{\text {rms }}$ (poly) $ \lt \mathrm{v}_{\text {rms }}$ (mono)
4 $\mathrm{v}_{\text {rms }}$ (mono) $ \lt \mathrm{v}_{\text {rms }}$ (dia) $ \lt \mathrm{v}_{\text {rms }}$ (poly)
Kinetic Theory of Gases

139106 The rms speed of oxygen molecule in a vessel at particular temperature is, $\left(1+\frac{5}{\mathrm{x}}\right)^{\frac{1}{2}} \mathrm{v}$, where $\mathrm{v}$ is the average speed of the molecule. The value of $x$ will be :
$\left(\text { Take } \pi=\frac{22}{7}\right)$

1 28
2 27
3 8
4 4
Kinetic Theory of Gases

139107 The mean free path of molecules of a certain gas at STP is 1500 d , where $d$ is the diameter of the gas molecules. While maintaining the standard pressure, the mean free path of the molecules at 373 K is approximately :

1 1098 d
2 2049 d
3 750 d
4 1500 d
Kinetic Theory of Gases

139104 The root mean square speed of molecules of nitrogen gas at $27^{\circ} \mathrm{C}$ is approximately:
(Given mass of a nitrogen molecule $=4.6 \times 10^{-26} \mathrm{~kg}$ and take Boltzmann constant $\mathrm{k}_{\mathrm{B}}$ $=1.4 \times 10^{-23} \mathrm{JK}^{-1}$ )

1 $523 \mathrm{~m} / \mathrm{s}$
2 $1260 \mathrm{~m} / \mathrm{s}$
3 $91 \mathrm{~m} / \mathrm{s}$
4 $27.4 \mathrm{~m} / \mathrm{s}$
Kinetic Theory of Gases

139105 Three vessels of equal volume contain gases at the same temperature and pressure. The first vessel contains neon (monoatomic), the second contains chlorine (diatomic) and third contains uranium hexafluoride (polyatomic). Arrange these on the basis of their root mean square sped $\left(v_{\text {rms }}\right)$ and choose the correct answer from the options given below:

1 $\mathrm{v}_{\mathrm{rms}}$ (mono) $=\mathrm{v}_{\mathrm{rms}}$ (dia) $=\mathrm{v}_{\mathrm{rms}}$ (poly)
2 $\mathrm{V}_{\mathrm{rms}}\left(\right.$ mono) $>\mathrm{v}_{\mathrm{rms}}$ (dia) $>\mathrm{v}_{\mathrm{rms}}$ (poly)
3 $\mathrm{v}_{\text {rms }}$ (dia) $ \lt \mathrm{v}_{\text {rms }}$ (poly) $ \lt \mathrm{v}_{\text {rms }}$ (mono)
4 $\mathrm{v}_{\text {rms }}$ (mono) $ \lt \mathrm{v}_{\text {rms }}$ (dia) $ \lt \mathrm{v}_{\text {rms }}$ (poly)
Kinetic Theory of Gases

139106 The rms speed of oxygen molecule in a vessel at particular temperature is, $\left(1+\frac{5}{\mathrm{x}}\right)^{\frac{1}{2}} \mathrm{v}$, where $\mathrm{v}$ is the average speed of the molecule. The value of $x$ will be :
$\left(\text { Take } \pi=\frac{22}{7}\right)$

1 28
2 27
3 8
4 4
Kinetic Theory of Gases

139107 The mean free path of molecules of a certain gas at STP is 1500 d , where $d$ is the diameter of the gas molecules. While maintaining the standard pressure, the mean free path of the molecules at 373 K is approximately :

1 1098 d
2 2049 d
3 750 d
4 1500 d