139234
One $\mathrm{kg}$ of a diatomic gas is at a pressure of $8 \times$ $10^{4} \mathrm{~N} / \mathrm{m}^{2}$. The density of the gas is $4 \mathrm{~kg} / \mathrm{m}^{3}$. What is the energy of the gas due to its thermal motion?
1 $5 \times 10^{4} \mathrm{~J}$
2 $6 \times 10^{4} \mathrm{~J}$
3 $7 \times 10^{4} \mathrm{~J}$
4 $3 \times 10^{4} \mathrm{~J}$
Explanation:
A Given that, $\text { Pressure }(P)=8 \times 10^{4} \mathrm{~N} / \mathrm{m}^{2}$ $\text { Density }(\rho)=4 \mathrm{~kg} / \mathrm{m}^{3}$ $\text { Mass of gas }=1 \mathrm{~kg}$ $\text { Volume } =\frac{\text { mass }}{\text { Density }}$ $=\frac{1}{4}$ $\mathrm{KE} =\frac{5}{2} \mathrm{P} \cdot \mathrm{V}$ $\mathrm{KE} =\frac{5}{2} \times 8 \times 10^4 \times \frac{1}{4}$ $\mathrm{KE} =\frac{5 \times 8 \times 10^4}{8}$ $\mathrm{KE} =5 \times 10^4 \text { Joule. }$
BITSAT-2016
Kinetic Theory of Gases
139235
Which of the following relation correct? $\left(\mathrm{V}_{\mathrm{rms}^{-}}\right.$ root mean square velocity, $\overrightarrow{\mathbf{u}}$-mean velocity and $\mathbf{u}_{\mathrm{mp}}$-most probable velocity)
C Ratio of root mean square velocity, mean velocity and most probable velocity are given as, Root mean square velocity $\left(\mathrm{v}_{\mathrm{rms}}\right)=\sqrt{\frac{3 R T}{\mathrm{M}}}$ Average velocity, $\overline{\mathrm{V}}=\sqrt{\frac{8 \mathrm{RT}}{\pi \mathrm{M}}}$ Most probable velocity, $\mathrm{V}_{\mathrm{mp}}=\sqrt{\frac{2 \mathrm{RT}}{\mathrm{M}}}$ $\mathrm{V}_{\mathrm{rms}}: \overline{\mathrm{V}}: \mathrm{V}_{\mathrm{mp}}=\sqrt{3}: \sqrt{\frac{8}{\pi}}: \sqrt{2}$ $\mathrm{V}_{\mathrm{rms}}: \overline{\mathrm{V}}: \mathrm{V}_{\mathrm{mp}}=1.732: 1.59: 1.41$ $\therefore \mathrm{V}_{\mathrm{rms}}>\overline{\mathrm{V}}>\mathrm{V}_{\mathrm{mp}}$.
CG PET- 2014]
Kinetic Theory of Gases
139237
The root mean square velocity of gas molecules at $27^{\circ} \mathrm{C}$ is $1365 \mathrm{~m} / \mathrm{s}$. The gas is
1 $\mathrm{O}_{2}$
2 $\mathrm{He}$
3 $\mathrm{N}_{2}$
4 $\mathrm{CO}_{2}$
Explanation:
B Given that, $\mathrm{T}= 27^{\circ} \mathrm{C}=273+27=300 \mathrm{~K}$ $\mathrm{v}_{\mathrm{rms}}=1365 \mathrm{~m} / \mathrm{sec} .$ Root mean square velocity, $\mathrm{v}_{\mathrm{rms}}=\sqrt{\frac{3 R T}{\mathrm{M}}}$ $1365=\sqrt{\frac{3 R T}{M}}$ On squaring both side, we get- $(1365)^{2}=\frac{3 \mathrm{RT}}{\mathrm{M}}$ $\mathrm{M}=\frac{3 \times 8.314 \times 300}{(1365)^{2}}$ $\mathrm{M}=\frac{7482.62}{(1365)^{2}}$ $\mathrm{M}=0.0040 \mathrm{~kg}$ $\mathrm{M}=4 \mathrm{gm} .$
CG PET- 2008
Kinetic Theory of Gases
139238
At which temperature the velocity of $\mathrm{O}_{2}$ molecules will be equal to the velocity of $\mathrm{N}_{2}$ molecules at $0^{\circ} \mathrm{C}$ ?
1 $40^{\circ} \mathrm{C}$
2 $93^{\circ} \mathrm{C}$
3 $39^{\circ} \mathrm{C}$
4 cannot be calculated
Explanation:
C Given that, Temperature, $\mathrm{T}=0^{\circ} \mathrm{C}+273 \mathrm{~K}=273 \mathrm{~K}$ Root mean square velocity of oxygen $\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{O}_{2}}$ is equal to $\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{N}_{2}}$ $\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{O}_{2}}=\mathrm{v}_{\mathrm{N}_{2}}$ $\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}_{\mathrm{O}_{2}}}}=\sqrt{\frac{3 \mathrm{R} \times 273}{\mathrm{M}_{\mathrm{N}_{2}}}}\left(\because \mathrm{M}_{\mathrm{O}_{2}}=32 \& \mathrm{M}_{\mathrm{N}_{2}}=28\right)$ $\sqrt{\frac{\mathrm{T}}{32}}=\sqrt{\frac{273}{28}}$ Squaring on both side, we get - $\frac{\mathrm{T}}{32} =\frac{273}{28}$ $\mathrm{~T} =312 \mathrm{~K}$ $\mathrm{~T} =(312-273)^{\circ} \mathrm{C}$ $\mathrm{T} =39^{\circ} \mathrm{C}$
JCECE-2016
Kinetic Theory of Gases
139239
Two gases are at absolute temperatures $300 \mathrm{~K}$ and $350 \mathrm{~K}$ respectively. The ratio of average kinetic energy of their molecules is
1 $6: 7$
2 $36: 49$
3 $7: 6$
4 $343: 216$
Explanation:
A Given that, $\mathrm{T}_{1}=300 \mathrm{~K}$ $\mathrm{~T}_{2}=350 \mathrm{~K}$ Average kinetic energy of gases, $\mathrm{E}=\frac{3}{2} \mathrm{kT}$ $\mathrm{E} \propto \mathrm{T}$ $\therefore \frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{300}{350}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{6}{7}$ $\mathrm{E}_{1}: \mathrm{E}_{2}=6: 7$
139234
One $\mathrm{kg}$ of a diatomic gas is at a pressure of $8 \times$ $10^{4} \mathrm{~N} / \mathrm{m}^{2}$. The density of the gas is $4 \mathrm{~kg} / \mathrm{m}^{3}$. What is the energy of the gas due to its thermal motion?
1 $5 \times 10^{4} \mathrm{~J}$
2 $6 \times 10^{4} \mathrm{~J}$
3 $7 \times 10^{4} \mathrm{~J}$
4 $3 \times 10^{4} \mathrm{~J}$
Explanation:
A Given that, $\text { Pressure }(P)=8 \times 10^{4} \mathrm{~N} / \mathrm{m}^{2}$ $\text { Density }(\rho)=4 \mathrm{~kg} / \mathrm{m}^{3}$ $\text { Mass of gas }=1 \mathrm{~kg}$ $\text { Volume } =\frac{\text { mass }}{\text { Density }}$ $=\frac{1}{4}$ $\mathrm{KE} =\frac{5}{2} \mathrm{P} \cdot \mathrm{V}$ $\mathrm{KE} =\frac{5}{2} \times 8 \times 10^4 \times \frac{1}{4}$ $\mathrm{KE} =\frac{5 \times 8 \times 10^4}{8}$ $\mathrm{KE} =5 \times 10^4 \text { Joule. }$
BITSAT-2016
Kinetic Theory of Gases
139235
Which of the following relation correct? $\left(\mathrm{V}_{\mathrm{rms}^{-}}\right.$ root mean square velocity, $\overrightarrow{\mathbf{u}}$-mean velocity and $\mathbf{u}_{\mathrm{mp}}$-most probable velocity)
C Ratio of root mean square velocity, mean velocity and most probable velocity are given as, Root mean square velocity $\left(\mathrm{v}_{\mathrm{rms}}\right)=\sqrt{\frac{3 R T}{\mathrm{M}}}$ Average velocity, $\overline{\mathrm{V}}=\sqrt{\frac{8 \mathrm{RT}}{\pi \mathrm{M}}}$ Most probable velocity, $\mathrm{V}_{\mathrm{mp}}=\sqrt{\frac{2 \mathrm{RT}}{\mathrm{M}}}$ $\mathrm{V}_{\mathrm{rms}}: \overline{\mathrm{V}}: \mathrm{V}_{\mathrm{mp}}=\sqrt{3}: \sqrt{\frac{8}{\pi}}: \sqrt{2}$ $\mathrm{V}_{\mathrm{rms}}: \overline{\mathrm{V}}: \mathrm{V}_{\mathrm{mp}}=1.732: 1.59: 1.41$ $\therefore \mathrm{V}_{\mathrm{rms}}>\overline{\mathrm{V}}>\mathrm{V}_{\mathrm{mp}}$.
CG PET- 2014]
Kinetic Theory of Gases
139237
The root mean square velocity of gas molecules at $27^{\circ} \mathrm{C}$ is $1365 \mathrm{~m} / \mathrm{s}$. The gas is
1 $\mathrm{O}_{2}$
2 $\mathrm{He}$
3 $\mathrm{N}_{2}$
4 $\mathrm{CO}_{2}$
Explanation:
B Given that, $\mathrm{T}= 27^{\circ} \mathrm{C}=273+27=300 \mathrm{~K}$ $\mathrm{v}_{\mathrm{rms}}=1365 \mathrm{~m} / \mathrm{sec} .$ Root mean square velocity, $\mathrm{v}_{\mathrm{rms}}=\sqrt{\frac{3 R T}{\mathrm{M}}}$ $1365=\sqrt{\frac{3 R T}{M}}$ On squaring both side, we get- $(1365)^{2}=\frac{3 \mathrm{RT}}{\mathrm{M}}$ $\mathrm{M}=\frac{3 \times 8.314 \times 300}{(1365)^{2}}$ $\mathrm{M}=\frac{7482.62}{(1365)^{2}}$ $\mathrm{M}=0.0040 \mathrm{~kg}$ $\mathrm{M}=4 \mathrm{gm} .$
CG PET- 2008
Kinetic Theory of Gases
139238
At which temperature the velocity of $\mathrm{O}_{2}$ molecules will be equal to the velocity of $\mathrm{N}_{2}$ molecules at $0^{\circ} \mathrm{C}$ ?
1 $40^{\circ} \mathrm{C}$
2 $93^{\circ} \mathrm{C}$
3 $39^{\circ} \mathrm{C}$
4 cannot be calculated
Explanation:
C Given that, Temperature, $\mathrm{T}=0^{\circ} \mathrm{C}+273 \mathrm{~K}=273 \mathrm{~K}$ Root mean square velocity of oxygen $\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{O}_{2}}$ is equal to $\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{N}_{2}}$ $\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{O}_{2}}=\mathrm{v}_{\mathrm{N}_{2}}$ $\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}_{\mathrm{O}_{2}}}}=\sqrt{\frac{3 \mathrm{R} \times 273}{\mathrm{M}_{\mathrm{N}_{2}}}}\left(\because \mathrm{M}_{\mathrm{O}_{2}}=32 \& \mathrm{M}_{\mathrm{N}_{2}}=28\right)$ $\sqrt{\frac{\mathrm{T}}{32}}=\sqrt{\frac{273}{28}}$ Squaring on both side, we get - $\frac{\mathrm{T}}{32} =\frac{273}{28}$ $\mathrm{~T} =312 \mathrm{~K}$ $\mathrm{~T} =(312-273)^{\circ} \mathrm{C}$ $\mathrm{T} =39^{\circ} \mathrm{C}$
JCECE-2016
Kinetic Theory of Gases
139239
Two gases are at absolute temperatures $300 \mathrm{~K}$ and $350 \mathrm{~K}$ respectively. The ratio of average kinetic energy of their molecules is
1 $6: 7$
2 $36: 49$
3 $7: 6$
4 $343: 216$
Explanation:
A Given that, $\mathrm{T}_{1}=300 \mathrm{~K}$ $\mathrm{~T}_{2}=350 \mathrm{~K}$ Average kinetic energy of gases, $\mathrm{E}=\frac{3}{2} \mathrm{kT}$ $\mathrm{E} \propto \mathrm{T}$ $\therefore \frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{300}{350}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{6}{7}$ $\mathrm{E}_{1}: \mathrm{E}_{2}=6: 7$
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Kinetic Theory of Gases
139234
One $\mathrm{kg}$ of a diatomic gas is at a pressure of $8 \times$ $10^{4} \mathrm{~N} / \mathrm{m}^{2}$. The density of the gas is $4 \mathrm{~kg} / \mathrm{m}^{3}$. What is the energy of the gas due to its thermal motion?
1 $5 \times 10^{4} \mathrm{~J}$
2 $6 \times 10^{4} \mathrm{~J}$
3 $7 \times 10^{4} \mathrm{~J}$
4 $3 \times 10^{4} \mathrm{~J}$
Explanation:
A Given that, $\text { Pressure }(P)=8 \times 10^{4} \mathrm{~N} / \mathrm{m}^{2}$ $\text { Density }(\rho)=4 \mathrm{~kg} / \mathrm{m}^{3}$ $\text { Mass of gas }=1 \mathrm{~kg}$ $\text { Volume } =\frac{\text { mass }}{\text { Density }}$ $=\frac{1}{4}$ $\mathrm{KE} =\frac{5}{2} \mathrm{P} \cdot \mathrm{V}$ $\mathrm{KE} =\frac{5}{2} \times 8 \times 10^4 \times \frac{1}{4}$ $\mathrm{KE} =\frac{5 \times 8 \times 10^4}{8}$ $\mathrm{KE} =5 \times 10^4 \text { Joule. }$
BITSAT-2016
Kinetic Theory of Gases
139235
Which of the following relation correct? $\left(\mathrm{V}_{\mathrm{rms}^{-}}\right.$ root mean square velocity, $\overrightarrow{\mathbf{u}}$-mean velocity and $\mathbf{u}_{\mathrm{mp}}$-most probable velocity)
C Ratio of root mean square velocity, mean velocity and most probable velocity are given as, Root mean square velocity $\left(\mathrm{v}_{\mathrm{rms}}\right)=\sqrt{\frac{3 R T}{\mathrm{M}}}$ Average velocity, $\overline{\mathrm{V}}=\sqrt{\frac{8 \mathrm{RT}}{\pi \mathrm{M}}}$ Most probable velocity, $\mathrm{V}_{\mathrm{mp}}=\sqrt{\frac{2 \mathrm{RT}}{\mathrm{M}}}$ $\mathrm{V}_{\mathrm{rms}}: \overline{\mathrm{V}}: \mathrm{V}_{\mathrm{mp}}=\sqrt{3}: \sqrt{\frac{8}{\pi}}: \sqrt{2}$ $\mathrm{V}_{\mathrm{rms}}: \overline{\mathrm{V}}: \mathrm{V}_{\mathrm{mp}}=1.732: 1.59: 1.41$ $\therefore \mathrm{V}_{\mathrm{rms}}>\overline{\mathrm{V}}>\mathrm{V}_{\mathrm{mp}}$.
CG PET- 2014]
Kinetic Theory of Gases
139237
The root mean square velocity of gas molecules at $27^{\circ} \mathrm{C}$ is $1365 \mathrm{~m} / \mathrm{s}$. The gas is
1 $\mathrm{O}_{2}$
2 $\mathrm{He}$
3 $\mathrm{N}_{2}$
4 $\mathrm{CO}_{2}$
Explanation:
B Given that, $\mathrm{T}= 27^{\circ} \mathrm{C}=273+27=300 \mathrm{~K}$ $\mathrm{v}_{\mathrm{rms}}=1365 \mathrm{~m} / \mathrm{sec} .$ Root mean square velocity, $\mathrm{v}_{\mathrm{rms}}=\sqrt{\frac{3 R T}{\mathrm{M}}}$ $1365=\sqrt{\frac{3 R T}{M}}$ On squaring both side, we get- $(1365)^{2}=\frac{3 \mathrm{RT}}{\mathrm{M}}$ $\mathrm{M}=\frac{3 \times 8.314 \times 300}{(1365)^{2}}$ $\mathrm{M}=\frac{7482.62}{(1365)^{2}}$ $\mathrm{M}=0.0040 \mathrm{~kg}$ $\mathrm{M}=4 \mathrm{gm} .$
CG PET- 2008
Kinetic Theory of Gases
139238
At which temperature the velocity of $\mathrm{O}_{2}$ molecules will be equal to the velocity of $\mathrm{N}_{2}$ molecules at $0^{\circ} \mathrm{C}$ ?
1 $40^{\circ} \mathrm{C}$
2 $93^{\circ} \mathrm{C}$
3 $39^{\circ} \mathrm{C}$
4 cannot be calculated
Explanation:
C Given that, Temperature, $\mathrm{T}=0^{\circ} \mathrm{C}+273 \mathrm{~K}=273 \mathrm{~K}$ Root mean square velocity of oxygen $\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{O}_{2}}$ is equal to $\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{N}_{2}}$ $\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{O}_{2}}=\mathrm{v}_{\mathrm{N}_{2}}$ $\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}_{\mathrm{O}_{2}}}}=\sqrt{\frac{3 \mathrm{R} \times 273}{\mathrm{M}_{\mathrm{N}_{2}}}}\left(\because \mathrm{M}_{\mathrm{O}_{2}}=32 \& \mathrm{M}_{\mathrm{N}_{2}}=28\right)$ $\sqrt{\frac{\mathrm{T}}{32}}=\sqrt{\frac{273}{28}}$ Squaring on both side, we get - $\frac{\mathrm{T}}{32} =\frac{273}{28}$ $\mathrm{~T} =312 \mathrm{~K}$ $\mathrm{~T} =(312-273)^{\circ} \mathrm{C}$ $\mathrm{T} =39^{\circ} \mathrm{C}$
JCECE-2016
Kinetic Theory of Gases
139239
Two gases are at absolute temperatures $300 \mathrm{~K}$ and $350 \mathrm{~K}$ respectively. The ratio of average kinetic energy of their molecules is
1 $6: 7$
2 $36: 49$
3 $7: 6$
4 $343: 216$
Explanation:
A Given that, $\mathrm{T}_{1}=300 \mathrm{~K}$ $\mathrm{~T}_{2}=350 \mathrm{~K}$ Average kinetic energy of gases, $\mathrm{E}=\frac{3}{2} \mathrm{kT}$ $\mathrm{E} \propto \mathrm{T}$ $\therefore \frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{300}{350}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{6}{7}$ $\mathrm{E}_{1}: \mathrm{E}_{2}=6: 7$
139234
One $\mathrm{kg}$ of a diatomic gas is at a pressure of $8 \times$ $10^{4} \mathrm{~N} / \mathrm{m}^{2}$. The density of the gas is $4 \mathrm{~kg} / \mathrm{m}^{3}$. What is the energy of the gas due to its thermal motion?
1 $5 \times 10^{4} \mathrm{~J}$
2 $6 \times 10^{4} \mathrm{~J}$
3 $7 \times 10^{4} \mathrm{~J}$
4 $3 \times 10^{4} \mathrm{~J}$
Explanation:
A Given that, $\text { Pressure }(P)=8 \times 10^{4} \mathrm{~N} / \mathrm{m}^{2}$ $\text { Density }(\rho)=4 \mathrm{~kg} / \mathrm{m}^{3}$ $\text { Mass of gas }=1 \mathrm{~kg}$ $\text { Volume } =\frac{\text { mass }}{\text { Density }}$ $=\frac{1}{4}$ $\mathrm{KE} =\frac{5}{2} \mathrm{P} \cdot \mathrm{V}$ $\mathrm{KE} =\frac{5}{2} \times 8 \times 10^4 \times \frac{1}{4}$ $\mathrm{KE} =\frac{5 \times 8 \times 10^4}{8}$ $\mathrm{KE} =5 \times 10^4 \text { Joule. }$
BITSAT-2016
Kinetic Theory of Gases
139235
Which of the following relation correct? $\left(\mathrm{V}_{\mathrm{rms}^{-}}\right.$ root mean square velocity, $\overrightarrow{\mathbf{u}}$-mean velocity and $\mathbf{u}_{\mathrm{mp}}$-most probable velocity)
C Ratio of root mean square velocity, mean velocity and most probable velocity are given as, Root mean square velocity $\left(\mathrm{v}_{\mathrm{rms}}\right)=\sqrt{\frac{3 R T}{\mathrm{M}}}$ Average velocity, $\overline{\mathrm{V}}=\sqrt{\frac{8 \mathrm{RT}}{\pi \mathrm{M}}}$ Most probable velocity, $\mathrm{V}_{\mathrm{mp}}=\sqrt{\frac{2 \mathrm{RT}}{\mathrm{M}}}$ $\mathrm{V}_{\mathrm{rms}}: \overline{\mathrm{V}}: \mathrm{V}_{\mathrm{mp}}=\sqrt{3}: \sqrt{\frac{8}{\pi}}: \sqrt{2}$ $\mathrm{V}_{\mathrm{rms}}: \overline{\mathrm{V}}: \mathrm{V}_{\mathrm{mp}}=1.732: 1.59: 1.41$ $\therefore \mathrm{V}_{\mathrm{rms}}>\overline{\mathrm{V}}>\mathrm{V}_{\mathrm{mp}}$.
CG PET- 2014]
Kinetic Theory of Gases
139237
The root mean square velocity of gas molecules at $27^{\circ} \mathrm{C}$ is $1365 \mathrm{~m} / \mathrm{s}$. The gas is
1 $\mathrm{O}_{2}$
2 $\mathrm{He}$
3 $\mathrm{N}_{2}$
4 $\mathrm{CO}_{2}$
Explanation:
B Given that, $\mathrm{T}= 27^{\circ} \mathrm{C}=273+27=300 \mathrm{~K}$ $\mathrm{v}_{\mathrm{rms}}=1365 \mathrm{~m} / \mathrm{sec} .$ Root mean square velocity, $\mathrm{v}_{\mathrm{rms}}=\sqrt{\frac{3 R T}{\mathrm{M}}}$ $1365=\sqrt{\frac{3 R T}{M}}$ On squaring both side, we get- $(1365)^{2}=\frac{3 \mathrm{RT}}{\mathrm{M}}$ $\mathrm{M}=\frac{3 \times 8.314 \times 300}{(1365)^{2}}$ $\mathrm{M}=\frac{7482.62}{(1365)^{2}}$ $\mathrm{M}=0.0040 \mathrm{~kg}$ $\mathrm{M}=4 \mathrm{gm} .$
CG PET- 2008
Kinetic Theory of Gases
139238
At which temperature the velocity of $\mathrm{O}_{2}$ molecules will be equal to the velocity of $\mathrm{N}_{2}$ molecules at $0^{\circ} \mathrm{C}$ ?
1 $40^{\circ} \mathrm{C}$
2 $93^{\circ} \mathrm{C}$
3 $39^{\circ} \mathrm{C}$
4 cannot be calculated
Explanation:
C Given that, Temperature, $\mathrm{T}=0^{\circ} \mathrm{C}+273 \mathrm{~K}=273 \mathrm{~K}$ Root mean square velocity of oxygen $\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{O}_{2}}$ is equal to $\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{N}_{2}}$ $\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{O}_{2}}=\mathrm{v}_{\mathrm{N}_{2}}$ $\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}_{\mathrm{O}_{2}}}}=\sqrt{\frac{3 \mathrm{R} \times 273}{\mathrm{M}_{\mathrm{N}_{2}}}}\left(\because \mathrm{M}_{\mathrm{O}_{2}}=32 \& \mathrm{M}_{\mathrm{N}_{2}}=28\right)$ $\sqrt{\frac{\mathrm{T}}{32}}=\sqrt{\frac{273}{28}}$ Squaring on both side, we get - $\frac{\mathrm{T}}{32} =\frac{273}{28}$ $\mathrm{~T} =312 \mathrm{~K}$ $\mathrm{~T} =(312-273)^{\circ} \mathrm{C}$ $\mathrm{T} =39^{\circ} \mathrm{C}$
JCECE-2016
Kinetic Theory of Gases
139239
Two gases are at absolute temperatures $300 \mathrm{~K}$ and $350 \mathrm{~K}$ respectively. The ratio of average kinetic energy of their molecules is
1 $6: 7$
2 $36: 49$
3 $7: 6$
4 $343: 216$
Explanation:
A Given that, $\mathrm{T}_{1}=300 \mathrm{~K}$ $\mathrm{~T}_{2}=350 \mathrm{~K}$ Average kinetic energy of gases, $\mathrm{E}=\frac{3}{2} \mathrm{kT}$ $\mathrm{E} \propto \mathrm{T}$ $\therefore \frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{300}{350}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{6}{7}$ $\mathrm{E}_{1}: \mathrm{E}_{2}=6: 7$
139234
One $\mathrm{kg}$ of a diatomic gas is at a pressure of $8 \times$ $10^{4} \mathrm{~N} / \mathrm{m}^{2}$. The density of the gas is $4 \mathrm{~kg} / \mathrm{m}^{3}$. What is the energy of the gas due to its thermal motion?
1 $5 \times 10^{4} \mathrm{~J}$
2 $6 \times 10^{4} \mathrm{~J}$
3 $7 \times 10^{4} \mathrm{~J}$
4 $3 \times 10^{4} \mathrm{~J}$
Explanation:
A Given that, $\text { Pressure }(P)=8 \times 10^{4} \mathrm{~N} / \mathrm{m}^{2}$ $\text { Density }(\rho)=4 \mathrm{~kg} / \mathrm{m}^{3}$ $\text { Mass of gas }=1 \mathrm{~kg}$ $\text { Volume } =\frac{\text { mass }}{\text { Density }}$ $=\frac{1}{4}$ $\mathrm{KE} =\frac{5}{2} \mathrm{P} \cdot \mathrm{V}$ $\mathrm{KE} =\frac{5}{2} \times 8 \times 10^4 \times \frac{1}{4}$ $\mathrm{KE} =\frac{5 \times 8 \times 10^4}{8}$ $\mathrm{KE} =5 \times 10^4 \text { Joule. }$
BITSAT-2016
Kinetic Theory of Gases
139235
Which of the following relation correct? $\left(\mathrm{V}_{\mathrm{rms}^{-}}\right.$ root mean square velocity, $\overrightarrow{\mathbf{u}}$-mean velocity and $\mathbf{u}_{\mathrm{mp}}$-most probable velocity)
C Ratio of root mean square velocity, mean velocity and most probable velocity are given as, Root mean square velocity $\left(\mathrm{v}_{\mathrm{rms}}\right)=\sqrt{\frac{3 R T}{\mathrm{M}}}$ Average velocity, $\overline{\mathrm{V}}=\sqrt{\frac{8 \mathrm{RT}}{\pi \mathrm{M}}}$ Most probable velocity, $\mathrm{V}_{\mathrm{mp}}=\sqrt{\frac{2 \mathrm{RT}}{\mathrm{M}}}$ $\mathrm{V}_{\mathrm{rms}}: \overline{\mathrm{V}}: \mathrm{V}_{\mathrm{mp}}=\sqrt{3}: \sqrt{\frac{8}{\pi}}: \sqrt{2}$ $\mathrm{V}_{\mathrm{rms}}: \overline{\mathrm{V}}: \mathrm{V}_{\mathrm{mp}}=1.732: 1.59: 1.41$ $\therefore \mathrm{V}_{\mathrm{rms}}>\overline{\mathrm{V}}>\mathrm{V}_{\mathrm{mp}}$.
CG PET- 2014]
Kinetic Theory of Gases
139237
The root mean square velocity of gas molecules at $27^{\circ} \mathrm{C}$ is $1365 \mathrm{~m} / \mathrm{s}$. The gas is
1 $\mathrm{O}_{2}$
2 $\mathrm{He}$
3 $\mathrm{N}_{2}$
4 $\mathrm{CO}_{2}$
Explanation:
B Given that, $\mathrm{T}= 27^{\circ} \mathrm{C}=273+27=300 \mathrm{~K}$ $\mathrm{v}_{\mathrm{rms}}=1365 \mathrm{~m} / \mathrm{sec} .$ Root mean square velocity, $\mathrm{v}_{\mathrm{rms}}=\sqrt{\frac{3 R T}{\mathrm{M}}}$ $1365=\sqrt{\frac{3 R T}{M}}$ On squaring both side, we get- $(1365)^{2}=\frac{3 \mathrm{RT}}{\mathrm{M}}$ $\mathrm{M}=\frac{3 \times 8.314 \times 300}{(1365)^{2}}$ $\mathrm{M}=\frac{7482.62}{(1365)^{2}}$ $\mathrm{M}=0.0040 \mathrm{~kg}$ $\mathrm{M}=4 \mathrm{gm} .$
CG PET- 2008
Kinetic Theory of Gases
139238
At which temperature the velocity of $\mathrm{O}_{2}$ molecules will be equal to the velocity of $\mathrm{N}_{2}$ molecules at $0^{\circ} \mathrm{C}$ ?
1 $40^{\circ} \mathrm{C}$
2 $93^{\circ} \mathrm{C}$
3 $39^{\circ} \mathrm{C}$
4 cannot be calculated
Explanation:
C Given that, Temperature, $\mathrm{T}=0^{\circ} \mathrm{C}+273 \mathrm{~K}=273 \mathrm{~K}$ Root mean square velocity of oxygen $\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{O}_{2}}$ is equal to $\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{N}_{2}}$ $\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{O}_{2}}=\mathrm{v}_{\mathrm{N}_{2}}$ $\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}_{\mathrm{O}_{2}}}}=\sqrt{\frac{3 \mathrm{R} \times 273}{\mathrm{M}_{\mathrm{N}_{2}}}}\left(\because \mathrm{M}_{\mathrm{O}_{2}}=32 \& \mathrm{M}_{\mathrm{N}_{2}}=28\right)$ $\sqrt{\frac{\mathrm{T}}{32}}=\sqrt{\frac{273}{28}}$ Squaring on both side, we get - $\frac{\mathrm{T}}{32} =\frac{273}{28}$ $\mathrm{~T} =312 \mathrm{~K}$ $\mathrm{~T} =(312-273)^{\circ} \mathrm{C}$ $\mathrm{T} =39^{\circ} \mathrm{C}$
JCECE-2016
Kinetic Theory of Gases
139239
Two gases are at absolute temperatures $300 \mathrm{~K}$ and $350 \mathrm{~K}$ respectively. The ratio of average kinetic energy of their molecules is
1 $6: 7$
2 $36: 49$
3 $7: 6$
4 $343: 216$
Explanation:
A Given that, $\mathrm{T}_{1}=300 \mathrm{~K}$ $\mathrm{~T}_{2}=350 \mathrm{~K}$ Average kinetic energy of gases, $\mathrm{E}=\frac{3}{2} \mathrm{kT}$ $\mathrm{E} \propto \mathrm{T}$ $\therefore \frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{300}{350}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{6}{7}$ $\mathrm{E}_{1}: \mathrm{E}_{2}=6: 7$