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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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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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A gas such as carbon monoxide would be most likely to obey the ideal gas law at

high temperatures and low pressures

low temperatures and high pressures

high temperatures and high pressures

low temperatures and low pressures

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Explanation

Real gases show ideal gas behaviour at high temperature and low pressures.

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Equal masses of H₂, O₂ and methane have been taken in a container of volume V at temperature 27°C in identical conditions. The ratio of the volumes of gases H₂: O₂ : CH₄ would be

8:16:1

16:8:1

16:1:2

8:1:2

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Explanation

Therefore, the ratio of the volumes of gases can be determined in terms of their moles. The ratio of volumes of H₂: O₂:methane (CH₄)
is given by

$V_{H_{2}} : V_{O_{2}} : V_{{CH}_{4}} = n_{H_{2}} : n_{O_{2}} : n_{{CH}_{4}} \Rightarrow V_{H_{2}} : V_{O_{2}} : V_{{CH}_{4}} = \frac{m_{H_{2}}}{M_{H_{2}}} : \frac{m_{O_{2}}}{M_{O_{2}}} : \frac{m_{{CH}_{4}}}{M_{{CH}_{4}}} But m_{H_{2}} = m_{O_{2}} = m_{{CH}_{4}} = \frac{m}{2} : \frac{m}{32} : \frac{m}{16} = 16 : 1 : 2$

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Maximum deviation from ideal gas is expected from:

H₂(g)

N₂(g)

CH₂(g)

NH₃(g)

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Explanation

(d) Easily liquefiable gases like NH₃, SO₂ etc. exhibit maximum deviation from ideal gas as for them Z<<< 1 .

${CH}_{4}$ also exhibits deviation but it is less as compared to NH₃.

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A gaseous mixture was prepared by taking equal moles of CO and N₂. If the total pressure of the mixture was found 1 atmosphere, the partial pressure of the nitrogen (N₂) in the mixture is

0.8 atm

0.9 aim

1 atm

0.5 atm

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Explanation

"CO="N₂

P_co= P_N2

Given, P_co + P_N₂ = 1atm

or 2P_N2 = 1 atm

or P_N2 = 0.5atm

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The pressure-volume plot for an ideal gas at a given temperature has the form of a

Straight line

Exponential curve

Rectangular hyperbola

U-shaped curve

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Yes. You can attempt every Physics question on this page for free without logging in, and check the correct answer with a detailed explanation instantly.

No account is required to attempt questions and view answers. A free account adds bookmarks, personal notes, and progress tracking.

The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.