QBManager

Kinetic Theory of Gases

139282 The kinetic energy of one molecule of a gas at normal temperature and pressure will be $(k = 8.31 J / mole K)$

1

1.7 \times 10^{3} J

2

10.2 \times 10^{3} J

3

3.4 \times 10^{3} J

4

6.8 \times 10^{3} J

Explanation:

C Given,
Normal Temperature $= 273^{\circ} C$
$R = 8.31 J / mol K$ Kinetic energy of one molecule of gas,
$K.E = \frac{3}{2} RT$
$= \frac{3}{2} \times 8.31 \times 273 = 3.4 \times 10^{3} J$

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 $(O_{2}) = 32$ Molecular mass of Hydrogen $(H_{2}) = 2$ We know,
$v_{rms} = \sqrt{\frac{3 R T}{M}}$
$\frac{{(v_{rms})}_{O_{2}}}{{(v_{rms})}_{H_{2}}} = \frac{\sqrt{\frac{3 RT}{32}}}{\sqrt{\frac{3 R T}{2}}}$
$\frac{{(v_{rms})}_{O_{2}}}{{(v_{rms})}_{H_{2}}} \frac{1}{4}$

AMU-2010

Kinetic Theory of Gases

139294 Four particles have speeds $1 m / s, 2 m / s, 3 m / s$ and $4 m / s$ . The root mean square velocity of the molecule

1

2.74 m / s

2

2 m / s

3

1.37 m / s

4

5.48 m / s

Explanation:

A We know that,
$v_{rms} = \sqrt{\frac{v_{1}^{2} + v_{2}^{2} + v_{3}^{2} + v_{4}^{2}}{n}}$
$v_{rms} = \sqrt{\frac{(1)^{2} + (2)^{2} + (3)^{2} + (4)^{2}}{4}}$
$v_{rms} = \sqrt{\frac{1 + 4 + 9 + 16}{4}}$
$v_{rms} = \sqrt{\frac{30}{4}} = 2.74 m / s$

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,
$v_{rms} = \sqrt{\frac{3 R T}{M}}$
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.

AIPMT- 1990

Kinetic Theory of Gases

139282 The kinetic energy of one molecule of a gas at normal temperature and pressure will be $(k = 8.31 J / mole K)$

1

1.7 \times 10^{3} J

2

10.2 \times 10^{3} J

3

3.4 \times 10^{3} J

4

6.8 \times 10^{3} J

Explanation:

C Given,
Normal Temperature $= 273^{\circ} C$
$R = 8.31 J / mol K$ Kinetic energy of one molecule of gas,
$K.E = \frac{3}{2} RT$
$= \frac{3}{2} \times 8.31 \times 273 = 3.4 \times 10^{3} J$

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 $(O_{2}) = 32$ Molecular mass of Hydrogen $(H_{2}) = 2$ We know,
$v_{rms} = \sqrt{\frac{3 R T}{M}}$
$\frac{{(v_{rms})}_{O_{2}}}{{(v_{rms})}_{H_{2}}} = \frac{\sqrt{\frac{3 RT}{32}}}{\sqrt{\frac{3 R T}{2}}}$
$\frac{{(v_{rms})}_{O_{2}}}{{(v_{rms})}_{H_{2}}} \frac{1}{4}$

AMU-2010

Kinetic Theory of Gases

139294 Four particles have speeds $1 m / s, 2 m / s, 3 m / s$ and $4 m / s$ . The root mean square velocity of the molecule

1

2.74 m / s

2

2 m / s

3

1.37 m / s

4

5.48 m / s

Explanation:

A We know that,
$v_{rms} = \sqrt{\frac{v_{1}^{2} + v_{2}^{2} + v_{3}^{2} + v_{4}^{2}}{n}}$
$v_{rms} = \sqrt{\frac{(1)^{2} + (2)^{2} + (3)^{2} + (4)^{2}}{4}}$
$v_{rms} = \sqrt{\frac{1 + 4 + 9 + 16}{4}}$
$v_{rms} = \sqrt{\frac{30}{4}} = 2.74 m / s$

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,
$v_{rms} = \sqrt{\frac{3 R T}{M}}$
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.

AIPMT- 1990

Kinetic Theory of Gases

139282 The kinetic energy of one molecule of a gas at normal temperature and pressure will be $(k = 8.31 J / mole K)$

1

1.7 \times 10^{3} J

2

10.2 \times 10^{3} J

3

3.4 \times 10^{3} J

4

6.8 \times 10^{3} J

Explanation:

C Given,
Normal Temperature $= 273^{\circ} C$
$R = 8.31 J / mol K$ Kinetic energy of one molecule of gas,
$K.E = \frac{3}{2} RT$
$= \frac{3}{2} \times 8.31 \times 273 = 3.4 \times 10^{3} J$

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 $(O_{2}) = 32$ Molecular mass of Hydrogen $(H_{2}) = 2$ We know,
$v_{rms} = \sqrt{\frac{3 R T}{M}}$
$\frac{{(v_{rms})}_{O_{2}}}{{(v_{rms})}_{H_{2}}} = \frac{\sqrt{\frac{3 RT}{32}}}{\sqrt{\frac{3 R T}{2}}}$
$\frac{{(v_{rms})}_{O_{2}}}{{(v_{rms})}_{H_{2}}} \frac{1}{4}$

AMU-2010

Kinetic Theory of Gases

139294 Four particles have speeds $1 m / s, 2 m / s, 3 m / s$ and $4 m / s$ . The root mean square velocity of the molecule

1

2.74 m / s

2

2 m / s

3

1.37 m / s

4

5.48 m / s

Explanation:

A We know that,
$v_{rms} = \sqrt{\frac{v_{1}^{2} + v_{2}^{2} + v_{3}^{2} + v_{4}^{2}}{n}}$
$v_{rms} = \sqrt{\frac{(1)^{2} + (2)^{2} + (3)^{2} + (4)^{2}}{4}}$
$v_{rms} = \sqrt{\frac{1 + 4 + 9 + 16}{4}}$
$v_{rms} = \sqrt{\frac{30}{4}} = 2.74 m / s$

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,
$v_{rms} = \sqrt{\frac{3 R T}{M}}$
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.

AIPMT- 1990

Kinetic Theory of Gases

139282 The kinetic energy of one molecule of a gas at normal temperature and pressure will be $(k = 8.31 J / mole K)$

1

1.7 \times 10^{3} J

2

10.2 \times 10^{3} J

3

3.4 \times 10^{3} J

4

6.8 \times 10^{3} J

Explanation:

C Given,
Normal Temperature $= 273^{\circ} C$
$R = 8.31 J / mol K$ Kinetic energy of one molecule of gas,
$K.E = \frac{3}{2} RT$
$= \frac{3}{2} \times 8.31 \times 273 = 3.4 \times 10^{3} J$

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 $(O_{2}) = 32$ Molecular mass of Hydrogen $(H_{2}) = 2$ We know,
$v_{rms} = \sqrt{\frac{3 R T}{M}}$
$\frac{{(v_{rms})}_{O_{2}}}{{(v_{rms})}_{H_{2}}} = \frac{\sqrt{\frac{3 RT}{32}}}{\sqrt{\frac{3 R T}{2}}}$
$\frac{{(v_{rms})}_{O_{2}}}{{(v_{rms})}_{H_{2}}} \frac{1}{4}$

AMU-2010

Kinetic Theory of Gases

139294 Four particles have speeds $1 m / s, 2 m / s, 3 m / s$ and $4 m / s$ . The root mean square velocity of the molecule

1

2.74 m / s

2

2 m / s

3

1.37 m / s

4

5.48 m / s

Explanation:

A We know that,
$v_{rms} = \sqrt{\frac{v_{1}^{2} + v_{2}^{2} + v_{3}^{2} + v_{4}^{2}}{n}}$
$v_{rms} = \sqrt{\frac{(1)^{2} + (2)^{2} + (3)^{2} + (4)^{2}}{4}}$
$v_{rms} = \sqrt{\frac{1 + 4 + 9 + 16}{4}}$
$v_{rms} = \sqrt{\frac{30}{4}} = 2.74 m / s$

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,
$v_{rms} = \sqrt{\frac{3 R T}{M}}$
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.

AIPMT- 1990

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