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NEET Practice Questions (MCQs) with Answers & Solutions

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The average kinetic energy of two moles of ${CO}_{2}$ at a certain temperature is 1800 cal. The temperature of the gas is $-$

300 K

150 K

200 K

400 K

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Explanation

(A). The average kinetic energy of molecule is given by $\frac{3}{2} \frac{RT}{N}$ where $\frac{R}{N}$ is Boltzmann constant.

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For a real gas, PV is a constant over a small range of pressures, at

Boyle’s temperature

Critical temperature

Inversion temperature

Ordinary temperature

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Explanation

(A). This is referred as Boyle’s temperature.

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A vessel of volume $0.02 m^{3}$ contains a mixture of hydrogen and helium at 20°C and 2 atm pressure. The mass of mixture is 5 gms. Find the ratio of mass of hydrogen to that of helium in the mixture (At. wt. He = 4)

1 : 2

1 : 3

2 : 3

3 : 2

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Explanation

$(A) . n = \frac{2 \times 20}{293 \times 0.0821} = 1.67 = \frac{5}{3} \Rightarrow \frac{x}{2} + \frac{5 - x}{4} = \frac{5}{3} \Rightarrow 15 + 3 x = 20 so x = \frac{5}{3} Required ratio = \frac{x}{5 - x} = \frac{5 / 3}{5 - 5 / 3} = \frac{1}{2}$

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Four particles have velocities 1, 0, 2, 3 m/s. The root mean square of the particles is (in m/s)

$3.5$

$\sqrt{3.5}$

$1.5$

$\sqrt{14 / 3}$

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Explanation

(B). $V_{rms} = \sqrt{\frac{1^{2} + 0^{2} + 2^{2} + 3^{2}}{4}} = \sqrt{3.5}$

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Joule-Thomson expansion of an ideal gas is an

Isothermal process

Isobaric process

Isoenthalpic process

Ideal process

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Explanation

(C). Joule-Thomson effect occurs at constant enthalpy.

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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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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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Yes. You can attempt every NEET 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.