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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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Which is lighter than dry air?

Moist air

SO₂

Cl₂

O₂

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Explanation

(a) Molar mass of moist air is lesser than dry air.

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The ratio a/b (the terms used in van der Waals' equation) has the unit:

atm lire mol^-1

atm dm³ mol^-1

dyne cm mol^-1

all of these

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Explanation

(d) The units of 'a; are: atm litre² mol^-2

= atm dm⁶mol^-2 = dyne cm²mol^-2

The units of 'b' are: litre mol^-1 = dm³mol^-1 = cm³mol^-1

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In the vander Waals' equation, the constant 'a' and 'b' with temperature shows which trend:

both remains same

'a' remains same, 'b' varies

'a' varies, 'b' remains same

both varies

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Explanation

(d) This is one of the limitation of vander Waal's equation.

'a' is measure of force, which is sensitive for distance, with increase in temp, the distance between molecule increase,so magnitude of force decrease. 'b' is measure of size, which is not so sensitive as 'a' . but it also changes.

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At relatively high pressure, van der Waals' equation reduces to:

PV = RT

PV = RT + a/v

PV = RT + Pb

PV = RT - a/V²

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Explanation

(c) At high pressure, volume of molecules should not be neglected in comparison to volume of gas. Also experimental to volume of gas. Also, experimental studies reveals PV > RT at high P.

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Joule-Thomson coefficient ( $\partial T / \partial P$ )_H for an ideal gas is:

zero

+ve

-ve

none of these

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Explanation

(a) Ideal gas does not show Joule-Thomson effect.

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The unit of van der Waals' constant 'a' is:

atm litre² mol-²

dyne cm⁴ mol^-2

newton m⁴ mol^-2

all of these

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Explanation

(d) a=P x V² = atm litre² mol-² = dyne cm⁴ mol^-2 = Newton m⁴ mol^-2 = atm dm⁶ mol^-2

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The unit of van der Waals' constant 'b' is :

cm³ mol^-1

litre mol^-1

m³ mol^-1

all of these

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Explanation

(d) b = 4Nv; unit of b = litre mol^-1 = cm³ mol^-1 = m³ mol^-1

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At STP, 0.50 mole H₂ gas and 1.0 mole He gas

have equal average kinetic energies

have equal molecular speeds

occupy equal volumes

have equal effusion rates

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An ideal gas cannot be liquefied because

its critical temperature is always above 0 $°$ C

its molecules are relatively smaller in size

it solidifies before becoming a liquid

forces operating between its molecules are negligible

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Explanation

(d) Gases can be liquefied by lowering the temperature and increasing the pressure. An ideal gas have no intermolecular force of attraction, so it cannot be liquefied by applying high pressure and decreasing temperature.

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The compressibility factor for H₂ and He is usually:

either of these

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Explanation

(a) For H₂ and He, PV > nRT; Also Z = PV/nRT .

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