NEET Test Series from KOTA - 10 Papers In MS WORD
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Kinetic Theory of Gases
139282
The kinetic energy of one molecule of a gas at normal temperature and pressure will be $(\mathrm{k}=8.31 \mathrm{~J} / \mathrm{mole} \mathrm{K})$
1 $1.7 \times 10^{3} \mathrm{~J}$
2 $10.2 \times 10^{3} \mathrm{~J}$
3 $3.4 \times 10^{3} \mathrm{~J}$
4 $6.8 \times 10^{3} \mathrm{~J}$
Explanation:
C Given, Normal Temperature $=273^{\circ} \mathrm{C}$ $\mathrm{R}=8.31 \mathrm{~J} / \mathrm{mol} \mathrm{K}$ Kinetic energy of one molecule of gas, $\text { K.E } =\frac{3}{2} \mathrm{RT}$ $=\frac{3}{2} \times 8.31 \times 273=3.4 \times 10^{3} \mathrm{~J}$
JIPMER-2004
Kinetic Theory of Gases
139286
Calculate the ratio of rms speed of oxygen gas molecules to that of hydrogen gas molecules kept at the same temperature
1 $1: 4$
2 $1: 8$
3 $1: 2$
4 $1: 6$
Explanation:
A Given, Molecular mass of oxygen $\left(\mathrm{O}_{2}\right)=32$ Molecular mass of Hydrogen $\left(\mathrm{H}_{2}\right)=2$ We know, $\mathrm{v}_{\mathrm{rms}}=\sqrt{\frac{3 R T}{M}}$ $\frac{\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{O}_{2}}}{\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{H}_{2}}}=\frac{\sqrt{\frac{3 \mathrm{RT}}{32}}}{\sqrt{\frac{3 R T}{2}}}$ $\frac{\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{O}_{2}}}{\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{H}_{2}}} \frac{1}{4}$
AMU-2010
Kinetic Theory of Gases
139294
Four particles have speeds $1 \mathrm{~m} / \mathrm{s}, 2 \mathrm{~m} / \mathrm{s}, 3 \mathrm{~m} / \mathrm{s}$ and $4 \mathrm{~m} / \mathrm{s}$. The root mean square velocity of the molecule
1 $2.74 \mathrm{~m} / \mathrm{s}$
2 $2 \mathrm{~m} / \mathrm{s}$
3 $1.37 \mathrm{~m} / \mathrm{s}$
4 $5.48 \mathrm{~m} / \mathrm{s}$
Explanation:
A We know that, $\mathrm{v}_{\mathrm{rms}}=\sqrt{\frac{\mathrm{v}_{1}^{2}+\mathrm{v}_{2}^{2}+\mathrm{v}_{3}^{2}+\mathrm{v}_{4}^{2}}{\mathrm{n}}}$ $\mathrm{v}_{\mathrm{rms}}=\sqrt{\frac{(1)^{2}+(2)^{2}+(3)^{2}+(4)^{2}}{4}}$ $\mathrm{v}_{\mathrm{rms}}=\sqrt{\frac{1+4+9+16}{4}}$ $\mathrm{v}_{\mathrm{rms}}=\sqrt{\frac{30}{4}}=2.74 \mathrm{~m} / \mathrm{s}$
AMU-2001
Kinetic Theory of Gases
139297
According to kinetic theory of gases, at absolute zero temperature
1 Water freezes
2 Liquid helium freezes
3 Molecular motion stops
4 Liquid hydrogen freezes
Explanation:
C We know that, $\mathrm{v}_{\mathrm{rms}}=\sqrt{\frac{3 R T}{\mathrm{M}}}$ The root mean square velocity is directly proportional to the square root of the absolute temperature. Hence, at zero absolute temperature, the velocity turn out to be zero, So molecular motion would stop.
139282
The kinetic energy of one molecule of a gas at normal temperature and pressure will be $(\mathrm{k}=8.31 \mathrm{~J} / \mathrm{mole} \mathrm{K})$
1 $1.7 \times 10^{3} \mathrm{~J}$
2 $10.2 \times 10^{3} \mathrm{~J}$
3 $3.4 \times 10^{3} \mathrm{~J}$
4 $6.8 \times 10^{3} \mathrm{~J}$
Explanation:
C Given, Normal Temperature $=273^{\circ} \mathrm{C}$ $\mathrm{R}=8.31 \mathrm{~J} / \mathrm{mol} \mathrm{K}$ Kinetic energy of one molecule of gas, $\text { K.E } =\frac{3}{2} \mathrm{RT}$ $=\frac{3}{2} \times 8.31 \times 273=3.4 \times 10^{3} \mathrm{~J}$
JIPMER-2004
Kinetic Theory of Gases
139286
Calculate the ratio of rms speed of oxygen gas molecules to that of hydrogen gas molecules kept at the same temperature
1 $1: 4$
2 $1: 8$
3 $1: 2$
4 $1: 6$
Explanation:
A Given, Molecular mass of oxygen $\left(\mathrm{O}_{2}\right)=32$ Molecular mass of Hydrogen $\left(\mathrm{H}_{2}\right)=2$ We know, $\mathrm{v}_{\mathrm{rms}}=\sqrt{\frac{3 R T}{M}}$ $\frac{\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{O}_{2}}}{\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{H}_{2}}}=\frac{\sqrt{\frac{3 \mathrm{RT}}{32}}}{\sqrt{\frac{3 R T}{2}}}$ $\frac{\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{O}_{2}}}{\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{H}_{2}}} \frac{1}{4}$
AMU-2010
Kinetic Theory of Gases
139294
Four particles have speeds $1 \mathrm{~m} / \mathrm{s}, 2 \mathrm{~m} / \mathrm{s}, 3 \mathrm{~m} / \mathrm{s}$ and $4 \mathrm{~m} / \mathrm{s}$. The root mean square velocity of the molecule
1 $2.74 \mathrm{~m} / \mathrm{s}$
2 $2 \mathrm{~m} / \mathrm{s}$
3 $1.37 \mathrm{~m} / \mathrm{s}$
4 $5.48 \mathrm{~m} / \mathrm{s}$
Explanation:
A We know that, $\mathrm{v}_{\mathrm{rms}}=\sqrt{\frac{\mathrm{v}_{1}^{2}+\mathrm{v}_{2}^{2}+\mathrm{v}_{3}^{2}+\mathrm{v}_{4}^{2}}{\mathrm{n}}}$ $\mathrm{v}_{\mathrm{rms}}=\sqrt{\frac{(1)^{2}+(2)^{2}+(3)^{2}+(4)^{2}}{4}}$ $\mathrm{v}_{\mathrm{rms}}=\sqrt{\frac{1+4+9+16}{4}}$ $\mathrm{v}_{\mathrm{rms}}=\sqrt{\frac{30}{4}}=2.74 \mathrm{~m} / \mathrm{s}$
AMU-2001
Kinetic Theory of Gases
139297
According to kinetic theory of gases, at absolute zero temperature
1 Water freezes
2 Liquid helium freezes
3 Molecular motion stops
4 Liquid hydrogen freezes
Explanation:
C We know that, $\mathrm{v}_{\mathrm{rms}}=\sqrt{\frac{3 R T}{\mathrm{M}}}$ The root mean square velocity is directly proportional to the square root of the absolute temperature. Hence, at zero absolute temperature, the velocity turn out to be zero, So molecular motion would stop.
139282
The kinetic energy of one molecule of a gas at normal temperature and pressure will be $(\mathrm{k}=8.31 \mathrm{~J} / \mathrm{mole} \mathrm{K})$
1 $1.7 \times 10^{3} \mathrm{~J}$
2 $10.2 \times 10^{3} \mathrm{~J}$
3 $3.4 \times 10^{3} \mathrm{~J}$
4 $6.8 \times 10^{3} \mathrm{~J}$
Explanation:
C Given, Normal Temperature $=273^{\circ} \mathrm{C}$ $\mathrm{R}=8.31 \mathrm{~J} / \mathrm{mol} \mathrm{K}$ Kinetic energy of one molecule of gas, $\text { K.E } =\frac{3}{2} \mathrm{RT}$ $=\frac{3}{2} \times 8.31 \times 273=3.4 \times 10^{3} \mathrm{~J}$
JIPMER-2004
Kinetic Theory of Gases
139286
Calculate the ratio of rms speed of oxygen gas molecules to that of hydrogen gas molecules kept at the same temperature
1 $1: 4$
2 $1: 8$
3 $1: 2$
4 $1: 6$
Explanation:
A Given, Molecular mass of oxygen $\left(\mathrm{O}_{2}\right)=32$ Molecular mass of Hydrogen $\left(\mathrm{H}_{2}\right)=2$ We know, $\mathrm{v}_{\mathrm{rms}}=\sqrt{\frac{3 R T}{M}}$ $\frac{\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{O}_{2}}}{\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{H}_{2}}}=\frac{\sqrt{\frac{3 \mathrm{RT}}{32}}}{\sqrt{\frac{3 R T}{2}}}$ $\frac{\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{O}_{2}}}{\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{H}_{2}}} \frac{1}{4}$
AMU-2010
Kinetic Theory of Gases
139294
Four particles have speeds $1 \mathrm{~m} / \mathrm{s}, 2 \mathrm{~m} / \mathrm{s}, 3 \mathrm{~m} / \mathrm{s}$ and $4 \mathrm{~m} / \mathrm{s}$. The root mean square velocity of the molecule
1 $2.74 \mathrm{~m} / \mathrm{s}$
2 $2 \mathrm{~m} / \mathrm{s}$
3 $1.37 \mathrm{~m} / \mathrm{s}$
4 $5.48 \mathrm{~m} / \mathrm{s}$
Explanation:
A We know that, $\mathrm{v}_{\mathrm{rms}}=\sqrt{\frac{\mathrm{v}_{1}^{2}+\mathrm{v}_{2}^{2}+\mathrm{v}_{3}^{2}+\mathrm{v}_{4}^{2}}{\mathrm{n}}}$ $\mathrm{v}_{\mathrm{rms}}=\sqrt{\frac{(1)^{2}+(2)^{2}+(3)^{2}+(4)^{2}}{4}}$ $\mathrm{v}_{\mathrm{rms}}=\sqrt{\frac{1+4+9+16}{4}}$ $\mathrm{v}_{\mathrm{rms}}=\sqrt{\frac{30}{4}}=2.74 \mathrm{~m} / \mathrm{s}$
AMU-2001
Kinetic Theory of Gases
139297
According to kinetic theory of gases, at absolute zero temperature
1 Water freezes
2 Liquid helium freezes
3 Molecular motion stops
4 Liquid hydrogen freezes
Explanation:
C We know that, $\mathrm{v}_{\mathrm{rms}}=\sqrt{\frac{3 R T}{\mathrm{M}}}$ The root mean square velocity is directly proportional to the square root of the absolute temperature. Hence, at zero absolute temperature, the velocity turn out to be zero, So molecular motion would stop.
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Kinetic Theory of Gases
139282
The kinetic energy of one molecule of a gas at normal temperature and pressure will be $(\mathrm{k}=8.31 \mathrm{~J} / \mathrm{mole} \mathrm{K})$
1 $1.7 \times 10^{3} \mathrm{~J}$
2 $10.2 \times 10^{3} \mathrm{~J}$
3 $3.4 \times 10^{3} \mathrm{~J}$
4 $6.8 \times 10^{3} \mathrm{~J}$
Explanation:
C Given, Normal Temperature $=273^{\circ} \mathrm{C}$ $\mathrm{R}=8.31 \mathrm{~J} / \mathrm{mol} \mathrm{K}$ Kinetic energy of one molecule of gas, $\text { K.E } =\frac{3}{2} \mathrm{RT}$ $=\frac{3}{2} \times 8.31 \times 273=3.4 \times 10^{3} \mathrm{~J}$
JIPMER-2004
Kinetic Theory of Gases
139286
Calculate the ratio of rms speed of oxygen gas molecules to that of hydrogen gas molecules kept at the same temperature
1 $1: 4$
2 $1: 8$
3 $1: 2$
4 $1: 6$
Explanation:
A Given, Molecular mass of oxygen $\left(\mathrm{O}_{2}\right)=32$ Molecular mass of Hydrogen $\left(\mathrm{H}_{2}\right)=2$ We know, $\mathrm{v}_{\mathrm{rms}}=\sqrt{\frac{3 R T}{M}}$ $\frac{\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{O}_{2}}}{\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{H}_{2}}}=\frac{\sqrt{\frac{3 \mathrm{RT}}{32}}}{\sqrt{\frac{3 R T}{2}}}$ $\frac{\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{O}_{2}}}{\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{H}_{2}}} \frac{1}{4}$
AMU-2010
Kinetic Theory of Gases
139294
Four particles have speeds $1 \mathrm{~m} / \mathrm{s}, 2 \mathrm{~m} / \mathrm{s}, 3 \mathrm{~m} / \mathrm{s}$ and $4 \mathrm{~m} / \mathrm{s}$. The root mean square velocity of the molecule
1 $2.74 \mathrm{~m} / \mathrm{s}$
2 $2 \mathrm{~m} / \mathrm{s}$
3 $1.37 \mathrm{~m} / \mathrm{s}$
4 $5.48 \mathrm{~m} / \mathrm{s}$
Explanation:
A We know that, $\mathrm{v}_{\mathrm{rms}}=\sqrt{\frac{\mathrm{v}_{1}^{2}+\mathrm{v}_{2}^{2}+\mathrm{v}_{3}^{2}+\mathrm{v}_{4}^{2}}{\mathrm{n}}}$ $\mathrm{v}_{\mathrm{rms}}=\sqrt{\frac{(1)^{2}+(2)^{2}+(3)^{2}+(4)^{2}}{4}}$ $\mathrm{v}_{\mathrm{rms}}=\sqrt{\frac{1+4+9+16}{4}}$ $\mathrm{v}_{\mathrm{rms}}=\sqrt{\frac{30}{4}}=2.74 \mathrm{~m} / \mathrm{s}$
AMU-2001
Kinetic Theory of Gases
139297
According to kinetic theory of gases, at absolute zero temperature
1 Water freezes
2 Liquid helium freezes
3 Molecular motion stops
4 Liquid hydrogen freezes
Explanation:
C We know that, $\mathrm{v}_{\mathrm{rms}}=\sqrt{\frac{3 R T}{\mathrm{M}}}$ The root mean square velocity is directly proportional to the square root of the absolute temperature. Hence, at zero absolute temperature, the velocity turn out to be zero, So molecular motion would stop.