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The rate of diffusion of methane at a given temperature is twice that of gas X. The molecular weight of X is

64.0

32.0

4.0

8.0

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Explanation

$\frac{r_{C H_{4}}}{r_{X}} = \sqrt{\frac{M_{X}}{M_{C H_{4}}}} \Rightarrow 2 = \sqrt{\frac{M_{X}}{16}} \Rightarrow M_{X} = 64$ .

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Density ratio of O₂ andH₂ is 16 : 1. The ratio of their r.m.s. velocities will be [AIIMS 2000]

4 : 1

1 : 16

1 : 4

16 : 1

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Explanation

$\frac{r_{1}}{r_{2}} = \frac{v_{1}}{v_{2}} = \sqrt{\frac{d_{2}}{d_{1}}} = \sqrt{\frac{1}{16}} = 1 : 4$ .

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The rate of diffusion of a gas having molecular weight just double of nitrogen gas is 56 mls^–1. The rate of diffusion of nitrogen will be [CPMT 2000]

79.19 ml s^–1

112.0 ml s^–1

56 ml s^–1

90.0 ml s^–1

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Explanation

$\frac{r_{X}}{r_{N_{2}}} = \sqrt{\frac{M_{N_{2}}}{M_{X}}} = \sqrt{\frac{28}{56}} = \sqrt{\frac{1}{2}}$ ;

$\frac{56}{r_{N_{2}}} = \frac{1}{\sqrt{2}}$ or $r_{N_{2}} = 56 \sqrt{2} = 79 .19 m s^{- 1}$

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50 ml of gas A diffuse through a membrane in the same time as 40 ml of a gas B under identical pressure-temperature conditions. If the molecular weight of A is 64, that of B would be [CBSE PMT 1992]

100

250

200

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Explanation

$\frac{r_{1}}{r_{2}} = \sqrt{\frac{M_{2}}{M_{1}}} \Rightarrow \frac{50 / t}{40 / t} = \sqrt{\frac{M_{2}}{64}}$

$\frac{50}{40} = \sqrt{\frac{M_{2}}{64}}$ or $\frac{5}{4} = \sqrt{\frac{M_{2}}{64}}$ or $\frac{25}{16} = \frac{M_{2}}{64}$ or M₂ = 100

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The kinetic energy for 14 grams of nitrogen gas at 127°C is nearly (mol. mass of nitrogen = 28 and gas constant = 8.31JK^–1mol^–1)

1.0 J

4.15 J

2493 J

3.3 J

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Explanation

K.E. $= \frac{3}{2} R T m o l^{- 1}$

or K.E. $= \frac{3}{2} n R T = \frac{3}{2} \times \frac{14}{28} \times 8 .31 \times 400 J = 2493 J$

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The rms velocity of CO₂ at a temperature T (in kelvin) is x cms^–1. At what temperature (in kelvin) the rms velocity of nitrous oxide would be 4x cms^–1 [EAMCET 2001]

16 T

2 T

4 T

32 T

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Explanation

$u = \sqrt{\frac{3 R T}{M}}$

∴ $\frac{u_{C O_{2}}}{u_{N_{2} O}} = \sqrt{\frac{T_{C O_{2}}}{M_{C O_{2}}} \times \frac{M_{N_{2} O}}{T_{N_{2} O}}}$

i.e., $\frac{x}{4 x} = \sqrt{\frac{T}{44} \times \frac{44}{T_{N_{2} O}}}$ or $\frac{1}{4} = \sqrt{\frac{T}{T_{N_{2} O}}}$ or $T_{N_{2} O} = 16 T$

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The rms velocity of an ideal gas at 27°C is 0.3 ms^–1. Its rms velocity at 927°C (in ms^–1) is [IIT 1996; EAMCET 1991]

3.0

2.4

0.9

0.6

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Explanation

$u = \sqrt{\frac{3 R T}{M}}$

For the same gas at two different temperatures,

$\frac{u_{1}}{u_{2}} = \sqrt{\frac{T_{1}}{T_{2}}}$ ; $\frac{0 .3}{u_{2}} = \sqrt{\frac{300}{1200}} = \frac{1}{2}$ , u₂ = 0.6 ms^–1

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The rms velocity of hydrogen is $\sqrt{7}$ times the rms velocity of nitrogen. If T is the temperature of the gas [IIT 2000]

$T (H_{2}) = T (N_{2})$

$T (H_{2}) > T (N_{2})$

$T (H_{2}) < T (N_{2})$

$T (H_{2}) = \sqrt{7} T (N_{2})$

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Explanation

$u = \sqrt{\frac{3 R T}{M}}$ ;

∴ $\frac{u (H_{2})}{u (N_{2})} = \sqrt{\frac{T (H_{2})}{M (H_{2})} \times \frac{M (N_{2})}{T (N_{2})}}$ or $\sqrt{7} = \sqrt{\frac{T (H_{2})}{T (N_{2})} \times \frac{28}{2}}$

or $7 = \frac{T (H_{2})}{T (N_{2})} \times 14$ or $\frac{T (H_{2})}{T (N_{2})} = \frac{1}{2}$

or $T (N_{2}) = 2 \times T (H_{2})$ i.e., $T (N_{2}) > T (H_{2})$

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If the average velocity of N₂ molecules is 0.3 m/s at 27°C, then the velocity of 0.6 m/s will take place at

273 K

927 K

1000 K

1200 K

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Explanation

$v = 0 .921 u$

∴ $\frac{v_{1}}{v_{2}} = \frac{u_{1}}{u_{2}} = \sqrt{\frac{T_{1}}{T_{2}}}$

∴ $\frac{0 .3}{0 .6} = \sqrt{\frac{300}{T_{2}}}$ or $\frac{1}{2} = \sqrt{\frac{300}{T_{2}}}$

or T₂ = 300 × 4 = 1200 K

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Equal moles of hydrogen and oxygen gases are placed in container with a pin-hole through which both can escape. What fraction of the oxygen escapes in the time required for one-half of the hydrogen to escape?

1/4

3/8

1/2

1/8

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Explanation

(d) Given, number of moles of hydrogen ( $n_{H_{2}}$ ) and that of oxygen ( $n_{O_{2}}$ ) are equal.

.^.. We have, the relation between ratio of number of moles escaped and ratio of molecular mass.

$\frac{n_{H_{2}}}{n_{O_{2}}} = \sqrt{\frac{M_{H_{2}}}{M_{O_{2}}}}$

where, M = Molecular mass of the molecule

$\frac{n_{O_{2}}}{n_{H_{2}}} = \sqrt{\frac{2}{32}} = \sqrt{\frac{1}{16}}$

$\frac{n_{O_{2}}}{0.5}$ = $\frac{1}{4}$

$n_{O_{2}}$ = 0.5/4 =1/8

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