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Kinetic Theory of Gases MCQs for NEET — Physics Questions with Answers

Practice free Kinetic Theory of Gases (Physics) NEET multiple-choice questions online with instant answers and detailed explanations. No login required.

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The rate of diffusion is :

Faster in solids than in liquids and gases

Faster in liquids than in solids and gases

Equal to solids, liquids, and gases

Faster in gases than in liquids and solids

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Explanation

4. It is because of their low densities.

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Two thermally insulated vessels 1 and 2 are filled with air at temperatures $(T_{1}, T_{2}),$ volume $(V_{1}, V_{2})$ and pressure $(P_{1}, P_{2})$ respectively. If the valve joining the two vessels is opened, the temperature inside the vessel at equilibrium will be

$T_{1} + T_{2}$

$(T_{1} + T_{2}) / 2$

$\frac{T_{1} T_{2} (P_{1} V_{1} + P_{2} V_{2})}{P_{1} V_{1} T_{2} + P_{2} V_{2} T_{1}}$

$\frac{T_{1} T_{2} (P_{1} V_{1} + P_{2} V_{2})}{P_{1} V_{1} T_{2} + P_{2} V_{2} T_{2}}$

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Explanation

3. The number of moles of the system remains same,

$\frac{P_{1} V_{1}}{{RT}_{1}} + \frac{P_{2} V_{2}}{{RT}_{2}} = \frac{P (V_{1} + V_{2})}{RT} \Rightarrow T = \frac{P (V_{1} + V_{2}) T_{1} T_{2}}{(P_{1} V_{1} T_{2} + P_{2} V_{2} T_{1})}$

According to Boyle's law,

$P_{1} V_{1} + P_{2} V_{2} = P (V_{1} + V_{2}) ∴ T = \frac{(P_{1} V_{1} + P_{2} V_{2}) T_{1} T_{2}}{(P_{1} V_{1} T_{2} + P_{2} V_{2} T_{1})}$

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For matter to exist simultaneously in gas and liquid phases :

The temperature must be 0 K.

The temperature must be less than $0 ° C$ .

The temperature must be less than the critical temperature.

The temperature must be less than the reduced temperature.

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Explanation

3. The temperature must be less than the critical temperature.

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The value of critical temperature in terms of Vander Waal’s constant a and b is

$T_{c} = \frac{8 a}{27 Rb}$

$T_{c} = \frac{a}{2 Rb}$

$T_{c} = \frac{8}{27 Rb}$

$T_{c} = \frac{27 a}{8 Rb}$

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Explanation

1. $T_{c} = \frac{8 a}{27 Rb}$

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In Vander Waal’s equation, a and b represent $(P + \frac{a}{V^{2}}) (v - b) = RT$

Both a and b represent correction in volume

Both a and b represent adhesive force between molecules

a represents adhesive force between molecules and b correction in volume

a represents correction in volume and b represents adhesive force between molecules

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Explanation

3. In Vander Wall's equation, $(P + \frac{a}{V^{2}}) (V - b) = RT$ a represents intermolecular attractive force and b represents

volume correction.

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At 0°C the density of a fixed mass of a gas divided by pressure is x. At 100°C, the ratio will be:

$x$

$\frac{273}{373} x$

$\frac{373}{273} x$

$\frac{100}{273} x$

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Explanation

$2 . PV = μRT = \frac{m}{M} RT \Rightarrow \frac{m}{VP} = \frac{density}{P} = \frac{M}{RT} {(\frac{density}{P})}_{At 0 ° C} = \frac{M}{R (273)} = x . . . . . . . . . . (i) {(\frac{density}{P})}_{At 100 ° C} = \frac{M}{R (373)} . . . . . . . . . . (ii) \Rightarrow {(\frac{density}{P})}_{At 100 ° C} = \frac{273}{373} x$

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At what temperature volume of an ideal gas at 0°C becomes triple :

546°C

182°C

819°C

646°C

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Explanation

$1 . At constant pressure V \propto T \Rightarrow \frac{V_{2}}{V_{1}} = \frac{T_{2}}{T_{1}} \Rightarrow T_{2} = (\frac{V_{2}}{V_{1}}) T_{1} \Rightarrow T_{2} = (\frac{3 V}{V}) \times 273 = 819 K = 546 ° C$

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If an ideal gas has volume V at 27°C and it is heated at a constant pressure so that its volume becomes 1.5V. Then the value of final temperature will be

600°C

177°C

817°C

None of these

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Explanation

2. $T_{2} = (\frac{V_{2}}{V_{1}}) T_{1} = (\frac{1.5 V}{V}) \times (273 + 27) = 450 K \Rightarrow 177 ° C$

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A tyre kept outside in sunlight bursts off after sometime because of :

Increase in pressure

Increases in volume

Both (a) and (b)

None of these

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Explanation

1. Increase in pressure

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In Boyle's law what remains constant :

$PV$

$TV$

$\frac{V}{T}$

$\frac{P}{T}$

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Explanation

1. According to Boyle’s law PV = constant.

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