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The average, RMS and most probable velocities of gas molecules at STP increase in the order

RMS < average velocity < most probable velocity

Most probable velocity < average velocity < RMS

Average velocity < RMS < most probable velocity

RMS < most probable velocity < average velocity

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Explanation

(B). Molecular speed of gases are of velocity

(1) The RMS root mean square velocity

C = $\sqrt{(\frac{3 RT}{M})}$

(2) Average velocity = $\sqrt{\frac{8 RT}{πM}}$

RMS=1.085 × Average velocity

(3) Most probable velocity = $\sqrt{\frac{2 RT}{M}}$

The ratio is

Most probable velocity : Average velocity : RMS

= $\sqrt{2} : \sqrt{\frac{8}{π}} : \sqrt{3} = 1 : 1.128 : 1.224$

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Equal volumes of ${SO}_{2}$ and $He$ at a temperature T and pressure P are allowed to effuse through a hole. The rate of effusion of helium is

Equal to the rate of effusion of ${SO}_{2}$

Four times the rate of effusion of ${SO}_{2}$

Half of the rate of effusion of ${SO}_{2}$

Twice the rate of effusion of ${SO}_{2}$

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Explanation

(B). The relative rates of diffusion of gases with respect to molecular weights is given by the expression

$\frac{r_{1}}{r_{2}} = \sqrt{\frac{M_{2}}{M_{1}}} = 4; M_{2} = 64 and M_{1} = 4$

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Flask X is filled with 20g of ${CH}_{4}$ gas at 100°C and another identical flask Y with 40g $O_{2}$ gas at the same temperature. Which one of the following statements is correct – [Molar masses $(g {mol}^{- 1})$ ${CH}_{4}$ = 16.0, $O_{2}$ = 32.0]

The pressure of the gases in the two flasks are identical

The pressure of ${CH}_{4}$ in flask X is higher than that of $O_{2}$ in flask Y

The pressure of ${CH}_{4}$ in flask X is lower than that of $O_{2}$ in flask Y

The pressure of ${CH}_{4}$ in flask X is half that of $O_{2}$ in flask Y

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Explanation

$(A) . n_{{CH}_{4}} = \frac{20}{16} = \frac{5}{4} n_{O_{2}} = \frac{40}{32} = \frac{5}{4} n_{{CH}_{4}} = n_{O_{2}} hence pressure is equal$

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A certain volume of argon gas (Mol. wt. = 40) requires 45 s to effuse through a hole at a certain pressure and temperature. The same volume of another gas of unknown molecular weight requires 60s to pass through the same hole under the same conditions of temperature and pressure. The molecular weight of the gas is

120

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Explanation

(C). This based on Graham's law of Diffusion.

$\frac{r_{1}}{r_{2}} = \sqrt{(\frac{M_{2}}{M_{1}})}$

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The rms speed of a gas molecules at temperature $27 K$ and pressure $1.5$ bar is $1 \times 10^{4}$ cm/sec. If both temperature and pressure are raised three time, the rms speed of the gas will be

$9 \times 10^{4}$ cm/sec

$3 \times 10^{4}$ cm/sec

$\sqrt{3} \times 10^{4}$ cm/sec

$1 \times 10^{4}$ cm/sec

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Explanation

(C). Do remember rms speed does not depend upon the pressure.

$C_{2} = \sqrt{\frac{3 R \times 3 \times 27}{M}} ∴ \frac{\bar{C_{2}}}{C_{1}} = \sqrt{3} C_{2} = \sqrt{\frac{3 R \times 27}{M}} or C_{2} = \sqrt{3} C_{1} = \sqrt{3} \times 10^{4} cm / \sec$

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The rate of effusion of helium gas at a pressure of 1000 torr is 10 torr min–1 What will be the rate of effusion of hydrogen gas at a pressure of 2000 torr at the same temperature?

$20 torr \min^{- 1}$

$40 torr \min^{- 1}$

$20 \sqrt{2} torr \min^{- 1}$

$10 torr \min^{- 1}$

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Explanation

$(C) . At a given temperature, rate of effusion \propto P and \propto \frac{1}{\sqrt{M}} ∴ rate of effusion of hydrogen gas = 10 torr \min^{- 1} \times \frac{2000}{1000} \times \sqrt{\frac{4}{2}} = 20 \sqrt{2} torr \min^{- 1}$

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The van der waals' constants for a gas are : $a = 4 {lit}^{2} atm {mol}^{- 2}, b = 0.04 lit {mol}^{- 1}$ . lts Boyle temperature is roughly

$100^{0} C$

$1220 K$

$1220^{0} C$

$1600 K$

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Explanation

(B). Boyle temperature,

$T_{B} = \frac{a}{Rb} = \frac{4 {lit}^{2} atm . {mol}^{- 2}}{0.082 lit atm K^{- 1} {mol}^{- 1} x 0.04 lit {mol}^{- 1}} = 1219.5 = 1220 K .$

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Since the atomic weights of C, N and O are 12, 14 and 16 respectively, among the following pair, the pair that will diffuse at the same rate is $-$

Carbon dioxide and nitrous oxide

Carbon dioxide and nitrogen peroxide

Carbon dioxide and carbon monoxide

Nitrous oxide and nitrogen peroxide

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Explanation

(A). Mol. wt. of $CO$ = 28 Mol wt. of ${CO}_{2}$ = 44

Mol. wt. of $N_{2} O$ = 28 + 16 = 44

Mol. wt. of ${NO}_{2}$ = 14 + 32 = 46

$∵ {CO}_{2}$ and $N_{2} O$ having same mol. wt. therefore, rate of diffusion for both the gases are same.

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Oxygen is present in 1-litre flask at a pressure of $7.6 \times 10^{- 10}$ mmHg. Calculate the number of oxygen molecules in the flask at $0 º C$ .

$2.686 \times 10^{10}$

$26.86 \times 10^{10}$

$0.626 \times 10^{12}$

$4.123 \times 10^{8}$

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Explanation

(A). Using the expression $pV = nRT,$ we have

$n = \frac{pV}{RT} = \frac{\{(7.6 \times 10^{- 10} / 760) atm\} (1 L)}{(0.0821 atm L K^{- 1} {mol}^{- 1}) (273.15 K)} = 4.459 \times 10^{- 14} mol N = {nN}_{A} = (4.459 \times 10^{- 14} mol) (6.023 \times 10^{23} {mol}^{- 1}) = 2.686 \times 10^{10} .$

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The critical temperature and critical pressure of a gas obeying van der Waals’ equation are 30ºC and 73 atm respectively. Its van der Waals’ constant, b $(in litres {mol}^{- 1})$ is, therefore

0.500

0.060

0.265

0.128

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Explanation

(D). The vander Waals constant $T_{C}, P_{C} and V_{C}$ are related by the expression

$V_{C} = 3 b; T_{C} = \frac{8 a}{27 Rb} and P_{C} = \frac{a}{27 b^{2}}$

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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.

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