139244
The ratio of rms speed of $\mathrm{O}_{2}$ to $\mathrm{H}_{2}$ is-
1 $\frac{1}{4}$
2 4
3 2
4 $\frac{1}{2}$
Explanation:
A Given, molecular weight of $\mathrm{O}_{2}\left(\mathrm{M}_{\mathrm{O}_{2}}\right)=32$ Molecular weight of $\left(\mathrm{M}_{\mathrm{H}_{2}}\right)=2$ We know, Root mean square velocity $\left(\mathrm{v}_{\mathrm{rms}}\right)=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{m}}}$ $\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{o}_{2}}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}_{\mathrm{O}_{2}}}}$ $\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{H}_{2}}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}_{\mathrm{H}_{2}}}}$ Taking ratio of equation (i) \& (ii), we get - $\frac{\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{O}_{2}}}{\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{H}_{2}}}=\sqrt{\frac{\mathrm{M}_{\mathrm{H}_{2}}}{\mathrm{M}_{\mathrm{O}_{2}}}}$ $=\sqrt{\frac{2}{32}}=\frac{1}{4}$
Karnataka CET-2002
Kinetic Theory of Gases
139245
If mass of $\mathrm{He}$ is 4 times that of hydrogen, then mean velocity of $\mathrm{He}$ is :
1 2 times of $\mathrm{H}$-mean value
2 $\frac{1}{2}$ times of H-mean value
3 4 times of H-mean value
4 same as H-mean value
Explanation:
B Given, $\mathrm{M}_{\mathrm{He}}=4 \mathrm{M}_{\mathrm{H}}$ Mean velocity $\left(\mathrm{v}_{\text {avg }}\right)=\sqrt{\frac{8 \mathrm{RT}}{\pi \mathrm{M}}}$ For hydrogen gas, $\left(\mathrm{v}_{\text {avg }}\right)_{\mathrm{H}}=\sqrt{\frac{8 \mathrm{RT}}{\pi \mathrm{M}_{\mathrm{H}}}}$ For helium gas, $\left(\mathrm{v}_{\mathrm{avg}}\right)_{\mathrm{He}}=\sqrt{\frac{8 \mathrm{RT}}{\pi \mathrm{M}_{\mathrm{He}}}}=\sqrt{\frac{8 \mathrm{RT}}{\pi 4 \mathrm{M}_{\mathrm{H}}}}$ So, $\frac{\left(\mathrm{v}_{\text {avg }}\right)_{\mathrm{H}}}{\left(\mathrm{v}_{\text {avg }}\right)_{\mathrm{He}}}=\sqrt{\frac{4 \mathrm{M}_{\mathrm{H}}}{\mathrm{M}_{\mathrm{H}}}}=\sqrt{\frac{4}{1}}=2$ Or $\left(\mathrm{v}_{\text {avg }}\right)_{\mathrm{He}}=\frac{1}{2}\left(\mathrm{v}_{\text {avg }}\right)_{\mathrm{H}}$
BCECE-2004
Kinetic Theory of Gases
139246
The ratio of root mean square velocities of $\mathrm{O}_{3}$ and $\mathrm{O}_{2}$ is:
1 $1: 1$
2 $2: 3$
3 $3: 2$
4 $\sqrt{2}: \sqrt{3}$
Explanation:
D Given, molecular weight of $\mathrm{O}_{3}\left(\mathrm{M}_{\mathrm{O}_{3}}\right)=48$ Molecular weight of $\mathrm{O}_{2}\left(\mathrm{M}_{\mathrm{O}_{2}}\right)=32$ We know, root mean square velocities. $\mathrm{v}_{\mathrm{rms}}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}}}$ Root mean square velocities of $\mathrm{O}_{3}$ and $\mathrm{O}_{2}$ are as under, $\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{O}_{2}}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}_{\mathrm{O}_{2}}}}$ $\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{O}_{3}}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}_{\mathrm{O}_{3}}}}$ $\therefore \quad \frac{\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{O}_{3}}}{\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{O}_{2}}}=\sqrt{\frac{\mathrm{M}_{\mathrm{O}_{2}}}{\mathrm{M}_{\mathrm{O}_{3}}}}$ $=\sqrt{\frac{32}{48}}=\frac{\sqrt{2}}{\sqrt{3}}=\sqrt{2}: \sqrt{3}$
BCECE-2004
Kinetic Theory of Gases
139247
Nitrogen $\left(\mathbf{N}_{2}\right)$ is in equilibrium state at $T=$ $421 \mathrm{~K}$. The value of most probable speed, $v_{\mathrm{mp}}$ is-
1 $400 \mathrm{~m} / \mathrm{s}$
2 $421 \mathrm{~m} / \mathrm{s}$
3 $500 \mathrm{~m} / \mathrm{s}$
4 $600 \mathrm{~m} / \mathrm{s}$
Explanation:
C Given, molecular weight of $\mathrm{N}_{2}\left(\mathrm{M}_{\mathrm{N}_{2}}\right)=28$ Temperature $(\mathrm{T})=421 \mathrm{~K}$ We know, Most probable velocity $\left(\mathrm{v}_{\mathrm{mp}}\right)=\sqrt{\frac{2 \mathrm{RT}}{\mathrm{M}}}$ $\mathrm{v}_{\mathrm{mp}} =\sqrt{\frac{2 \times 8.31 \times 421}{28 \times 10^{-3}}}$ $\mathrm{v}_{\mathrm{mp}} =499.89 \mathrm{~m} / \mathrm{s} = 500 \mathrm{~m} / \mathrm{s}$
139244
The ratio of rms speed of $\mathrm{O}_{2}$ to $\mathrm{H}_{2}$ is-
1 $\frac{1}{4}$
2 4
3 2
4 $\frac{1}{2}$
Explanation:
A Given, molecular weight of $\mathrm{O}_{2}\left(\mathrm{M}_{\mathrm{O}_{2}}\right)=32$ Molecular weight of $\left(\mathrm{M}_{\mathrm{H}_{2}}\right)=2$ We know, Root mean square velocity $\left(\mathrm{v}_{\mathrm{rms}}\right)=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{m}}}$ $\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{o}_{2}}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}_{\mathrm{O}_{2}}}}$ $\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{H}_{2}}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}_{\mathrm{H}_{2}}}}$ Taking ratio of equation (i) \& (ii), we get - $\frac{\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{O}_{2}}}{\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{H}_{2}}}=\sqrt{\frac{\mathrm{M}_{\mathrm{H}_{2}}}{\mathrm{M}_{\mathrm{O}_{2}}}}$ $=\sqrt{\frac{2}{32}}=\frac{1}{4}$
Karnataka CET-2002
Kinetic Theory of Gases
139245
If mass of $\mathrm{He}$ is 4 times that of hydrogen, then mean velocity of $\mathrm{He}$ is :
1 2 times of $\mathrm{H}$-mean value
2 $\frac{1}{2}$ times of H-mean value
3 4 times of H-mean value
4 same as H-mean value
Explanation:
B Given, $\mathrm{M}_{\mathrm{He}}=4 \mathrm{M}_{\mathrm{H}}$ Mean velocity $\left(\mathrm{v}_{\text {avg }}\right)=\sqrt{\frac{8 \mathrm{RT}}{\pi \mathrm{M}}}$ For hydrogen gas, $\left(\mathrm{v}_{\text {avg }}\right)_{\mathrm{H}}=\sqrt{\frac{8 \mathrm{RT}}{\pi \mathrm{M}_{\mathrm{H}}}}$ For helium gas, $\left(\mathrm{v}_{\mathrm{avg}}\right)_{\mathrm{He}}=\sqrt{\frac{8 \mathrm{RT}}{\pi \mathrm{M}_{\mathrm{He}}}}=\sqrt{\frac{8 \mathrm{RT}}{\pi 4 \mathrm{M}_{\mathrm{H}}}}$ So, $\frac{\left(\mathrm{v}_{\text {avg }}\right)_{\mathrm{H}}}{\left(\mathrm{v}_{\text {avg }}\right)_{\mathrm{He}}}=\sqrt{\frac{4 \mathrm{M}_{\mathrm{H}}}{\mathrm{M}_{\mathrm{H}}}}=\sqrt{\frac{4}{1}}=2$ Or $\left(\mathrm{v}_{\text {avg }}\right)_{\mathrm{He}}=\frac{1}{2}\left(\mathrm{v}_{\text {avg }}\right)_{\mathrm{H}}$
BCECE-2004
Kinetic Theory of Gases
139246
The ratio of root mean square velocities of $\mathrm{O}_{3}$ and $\mathrm{O}_{2}$ is:
1 $1: 1$
2 $2: 3$
3 $3: 2$
4 $\sqrt{2}: \sqrt{3}$
Explanation:
D Given, molecular weight of $\mathrm{O}_{3}\left(\mathrm{M}_{\mathrm{O}_{3}}\right)=48$ Molecular weight of $\mathrm{O}_{2}\left(\mathrm{M}_{\mathrm{O}_{2}}\right)=32$ We know, root mean square velocities. $\mathrm{v}_{\mathrm{rms}}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}}}$ Root mean square velocities of $\mathrm{O}_{3}$ and $\mathrm{O}_{2}$ are as under, $\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{O}_{2}}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}_{\mathrm{O}_{2}}}}$ $\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{O}_{3}}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}_{\mathrm{O}_{3}}}}$ $\therefore \quad \frac{\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{O}_{3}}}{\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{O}_{2}}}=\sqrt{\frac{\mathrm{M}_{\mathrm{O}_{2}}}{\mathrm{M}_{\mathrm{O}_{3}}}}$ $=\sqrt{\frac{32}{48}}=\frac{\sqrt{2}}{\sqrt{3}}=\sqrt{2}: \sqrt{3}$
BCECE-2004
Kinetic Theory of Gases
139247
Nitrogen $\left(\mathbf{N}_{2}\right)$ is in equilibrium state at $T=$ $421 \mathrm{~K}$. The value of most probable speed, $v_{\mathrm{mp}}$ is-
1 $400 \mathrm{~m} / \mathrm{s}$
2 $421 \mathrm{~m} / \mathrm{s}$
3 $500 \mathrm{~m} / \mathrm{s}$
4 $600 \mathrm{~m} / \mathrm{s}$
Explanation:
C Given, molecular weight of $\mathrm{N}_{2}\left(\mathrm{M}_{\mathrm{N}_{2}}\right)=28$ Temperature $(\mathrm{T})=421 \mathrm{~K}$ We know, Most probable velocity $\left(\mathrm{v}_{\mathrm{mp}}\right)=\sqrt{\frac{2 \mathrm{RT}}{\mathrm{M}}}$ $\mathrm{v}_{\mathrm{mp}} =\sqrt{\frac{2 \times 8.31 \times 421}{28 \times 10^{-3}}}$ $\mathrm{v}_{\mathrm{mp}} =499.89 \mathrm{~m} / \mathrm{s} = 500 \mathrm{~m} / \mathrm{s}$
139244
The ratio of rms speed of $\mathrm{O}_{2}$ to $\mathrm{H}_{2}$ is-
1 $\frac{1}{4}$
2 4
3 2
4 $\frac{1}{2}$
Explanation:
A Given, molecular weight of $\mathrm{O}_{2}\left(\mathrm{M}_{\mathrm{O}_{2}}\right)=32$ Molecular weight of $\left(\mathrm{M}_{\mathrm{H}_{2}}\right)=2$ We know, Root mean square velocity $\left(\mathrm{v}_{\mathrm{rms}}\right)=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{m}}}$ $\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{o}_{2}}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}_{\mathrm{O}_{2}}}}$ $\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{H}_{2}}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}_{\mathrm{H}_{2}}}}$ Taking ratio of equation (i) \& (ii), we get - $\frac{\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{O}_{2}}}{\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{H}_{2}}}=\sqrt{\frac{\mathrm{M}_{\mathrm{H}_{2}}}{\mathrm{M}_{\mathrm{O}_{2}}}}$ $=\sqrt{\frac{2}{32}}=\frac{1}{4}$
Karnataka CET-2002
Kinetic Theory of Gases
139245
If mass of $\mathrm{He}$ is 4 times that of hydrogen, then mean velocity of $\mathrm{He}$ is :
1 2 times of $\mathrm{H}$-mean value
2 $\frac{1}{2}$ times of H-mean value
3 4 times of H-mean value
4 same as H-mean value
Explanation:
B Given, $\mathrm{M}_{\mathrm{He}}=4 \mathrm{M}_{\mathrm{H}}$ Mean velocity $\left(\mathrm{v}_{\text {avg }}\right)=\sqrt{\frac{8 \mathrm{RT}}{\pi \mathrm{M}}}$ For hydrogen gas, $\left(\mathrm{v}_{\text {avg }}\right)_{\mathrm{H}}=\sqrt{\frac{8 \mathrm{RT}}{\pi \mathrm{M}_{\mathrm{H}}}}$ For helium gas, $\left(\mathrm{v}_{\mathrm{avg}}\right)_{\mathrm{He}}=\sqrt{\frac{8 \mathrm{RT}}{\pi \mathrm{M}_{\mathrm{He}}}}=\sqrt{\frac{8 \mathrm{RT}}{\pi 4 \mathrm{M}_{\mathrm{H}}}}$ So, $\frac{\left(\mathrm{v}_{\text {avg }}\right)_{\mathrm{H}}}{\left(\mathrm{v}_{\text {avg }}\right)_{\mathrm{He}}}=\sqrt{\frac{4 \mathrm{M}_{\mathrm{H}}}{\mathrm{M}_{\mathrm{H}}}}=\sqrt{\frac{4}{1}}=2$ Or $\left(\mathrm{v}_{\text {avg }}\right)_{\mathrm{He}}=\frac{1}{2}\left(\mathrm{v}_{\text {avg }}\right)_{\mathrm{H}}$
BCECE-2004
Kinetic Theory of Gases
139246
The ratio of root mean square velocities of $\mathrm{O}_{3}$ and $\mathrm{O}_{2}$ is:
1 $1: 1$
2 $2: 3$
3 $3: 2$
4 $\sqrt{2}: \sqrt{3}$
Explanation:
D Given, molecular weight of $\mathrm{O}_{3}\left(\mathrm{M}_{\mathrm{O}_{3}}\right)=48$ Molecular weight of $\mathrm{O}_{2}\left(\mathrm{M}_{\mathrm{O}_{2}}\right)=32$ We know, root mean square velocities. $\mathrm{v}_{\mathrm{rms}}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}}}$ Root mean square velocities of $\mathrm{O}_{3}$ and $\mathrm{O}_{2}$ are as under, $\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{O}_{2}}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}_{\mathrm{O}_{2}}}}$ $\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{O}_{3}}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}_{\mathrm{O}_{3}}}}$ $\therefore \quad \frac{\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{O}_{3}}}{\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{O}_{2}}}=\sqrt{\frac{\mathrm{M}_{\mathrm{O}_{2}}}{\mathrm{M}_{\mathrm{O}_{3}}}}$ $=\sqrt{\frac{32}{48}}=\frac{\sqrt{2}}{\sqrt{3}}=\sqrt{2}: \sqrt{3}$
BCECE-2004
Kinetic Theory of Gases
139247
Nitrogen $\left(\mathbf{N}_{2}\right)$ is in equilibrium state at $T=$ $421 \mathrm{~K}$. The value of most probable speed, $v_{\mathrm{mp}}$ is-
1 $400 \mathrm{~m} / \mathrm{s}$
2 $421 \mathrm{~m} / \mathrm{s}$
3 $500 \mathrm{~m} / \mathrm{s}$
4 $600 \mathrm{~m} / \mathrm{s}$
Explanation:
C Given, molecular weight of $\mathrm{N}_{2}\left(\mathrm{M}_{\mathrm{N}_{2}}\right)=28$ Temperature $(\mathrm{T})=421 \mathrm{~K}$ We know, Most probable velocity $\left(\mathrm{v}_{\mathrm{mp}}\right)=\sqrt{\frac{2 \mathrm{RT}}{\mathrm{M}}}$ $\mathrm{v}_{\mathrm{mp}} =\sqrt{\frac{2 \times 8.31 \times 421}{28 \times 10^{-3}}}$ $\mathrm{v}_{\mathrm{mp}} =499.89 \mathrm{~m} / \mathrm{s} = 500 \mathrm{~m} / \mathrm{s}$
139244
The ratio of rms speed of $\mathrm{O}_{2}$ to $\mathrm{H}_{2}$ is-
1 $\frac{1}{4}$
2 4
3 2
4 $\frac{1}{2}$
Explanation:
A Given, molecular weight of $\mathrm{O}_{2}\left(\mathrm{M}_{\mathrm{O}_{2}}\right)=32$ Molecular weight of $\left(\mathrm{M}_{\mathrm{H}_{2}}\right)=2$ We know, Root mean square velocity $\left(\mathrm{v}_{\mathrm{rms}}\right)=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{m}}}$ $\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{o}_{2}}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}_{\mathrm{O}_{2}}}}$ $\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{H}_{2}}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}_{\mathrm{H}_{2}}}}$ Taking ratio of equation (i) \& (ii), we get - $\frac{\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{O}_{2}}}{\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{H}_{2}}}=\sqrt{\frac{\mathrm{M}_{\mathrm{H}_{2}}}{\mathrm{M}_{\mathrm{O}_{2}}}}$ $=\sqrt{\frac{2}{32}}=\frac{1}{4}$
Karnataka CET-2002
Kinetic Theory of Gases
139245
If mass of $\mathrm{He}$ is 4 times that of hydrogen, then mean velocity of $\mathrm{He}$ is :
1 2 times of $\mathrm{H}$-mean value
2 $\frac{1}{2}$ times of H-mean value
3 4 times of H-mean value
4 same as H-mean value
Explanation:
B Given, $\mathrm{M}_{\mathrm{He}}=4 \mathrm{M}_{\mathrm{H}}$ Mean velocity $\left(\mathrm{v}_{\text {avg }}\right)=\sqrt{\frac{8 \mathrm{RT}}{\pi \mathrm{M}}}$ For hydrogen gas, $\left(\mathrm{v}_{\text {avg }}\right)_{\mathrm{H}}=\sqrt{\frac{8 \mathrm{RT}}{\pi \mathrm{M}_{\mathrm{H}}}}$ For helium gas, $\left(\mathrm{v}_{\mathrm{avg}}\right)_{\mathrm{He}}=\sqrt{\frac{8 \mathrm{RT}}{\pi \mathrm{M}_{\mathrm{He}}}}=\sqrt{\frac{8 \mathrm{RT}}{\pi 4 \mathrm{M}_{\mathrm{H}}}}$ So, $\frac{\left(\mathrm{v}_{\text {avg }}\right)_{\mathrm{H}}}{\left(\mathrm{v}_{\text {avg }}\right)_{\mathrm{He}}}=\sqrt{\frac{4 \mathrm{M}_{\mathrm{H}}}{\mathrm{M}_{\mathrm{H}}}}=\sqrt{\frac{4}{1}}=2$ Or $\left(\mathrm{v}_{\text {avg }}\right)_{\mathrm{He}}=\frac{1}{2}\left(\mathrm{v}_{\text {avg }}\right)_{\mathrm{H}}$
BCECE-2004
Kinetic Theory of Gases
139246
The ratio of root mean square velocities of $\mathrm{O}_{3}$ and $\mathrm{O}_{2}$ is:
1 $1: 1$
2 $2: 3$
3 $3: 2$
4 $\sqrt{2}: \sqrt{3}$
Explanation:
D Given, molecular weight of $\mathrm{O}_{3}\left(\mathrm{M}_{\mathrm{O}_{3}}\right)=48$ Molecular weight of $\mathrm{O}_{2}\left(\mathrm{M}_{\mathrm{O}_{2}}\right)=32$ We know, root mean square velocities. $\mathrm{v}_{\mathrm{rms}}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}}}$ Root mean square velocities of $\mathrm{O}_{3}$ and $\mathrm{O}_{2}$ are as under, $\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{O}_{2}}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}_{\mathrm{O}_{2}}}}$ $\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{O}_{3}}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}_{\mathrm{O}_{3}}}}$ $\therefore \quad \frac{\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{O}_{3}}}{\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{O}_{2}}}=\sqrt{\frac{\mathrm{M}_{\mathrm{O}_{2}}}{\mathrm{M}_{\mathrm{O}_{3}}}}$ $=\sqrt{\frac{32}{48}}=\frac{\sqrt{2}}{\sqrt{3}}=\sqrt{2}: \sqrt{3}$
BCECE-2004
Kinetic Theory of Gases
139247
Nitrogen $\left(\mathbf{N}_{2}\right)$ is in equilibrium state at $T=$ $421 \mathrm{~K}$. The value of most probable speed, $v_{\mathrm{mp}}$ is-
1 $400 \mathrm{~m} / \mathrm{s}$
2 $421 \mathrm{~m} / \mathrm{s}$
3 $500 \mathrm{~m} / \mathrm{s}$
4 $600 \mathrm{~m} / \mathrm{s}$
Explanation:
C Given, molecular weight of $\mathrm{N}_{2}\left(\mathrm{M}_{\mathrm{N}_{2}}\right)=28$ Temperature $(\mathrm{T})=421 \mathrm{~K}$ We know, Most probable velocity $\left(\mathrm{v}_{\mathrm{mp}}\right)=\sqrt{\frac{2 \mathrm{RT}}{\mathrm{M}}}$ $\mathrm{v}_{\mathrm{mp}} =\sqrt{\frac{2 \times 8.31 \times 421}{28 \times 10^{-3}}}$ $\mathrm{v}_{\mathrm{mp}} =499.89 \mathrm{~m} / \mathrm{s} = 500 \mathrm{~m} / \mathrm{s}$
