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

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Volume, pressure and temperature of an ideal gas are V, P and T respectively.
If mass of molecule is m, then its density is [ $k_B = Boltzmann`s constant$]

A. $ { P \over k_B T } $ B. $ { Pm \over k_B T } $ C. $ { P \over k_B TV } $ D. $ m k_B T $

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Explanation

$ PV = RT \Rightarrow PV = { M \over M_o } RT $ $ \therefore {PMo \over RT } ={ M \over V} = \rho $
$ \rho = { PMo \over RT} = {P \times m \times N_A \over RT} = { Pm \over \left( { R \over N_A} \right)} T = { Pm \over k_B T } $

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The temperature of an ideal gas at atmospheric pressure is 300 K and volume $1m^3$. If temperature and volume become double, then pressure will be

A. $ 4 \times 10^5 Nm^{-2} $ ` B. $ 2 \\times 10^5 Nm^{-2} $ C. $ 1 \times 10^5 Nm^{-2 } $ D. $ 0.5 \times 10^5 Nm^{-2 } $

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Explanation

$ PV = \mu RT \Rightarrow P \alpha { T \over V } $ If T and V both doubled, then pressure remains same

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At 100 K and 0.1 atmospheric pressure, the volume of helium gas is 10 litres. If volume and pressure are doubled, its temperature will change to

A. 127 K B. 400 K C. 25 K D. 200 K

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Explanation

$ PV = \mu RT \Rightarrow PV \alpha T $ If P and V doubled, then T becomes four times,

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What is the mass of 2 litres of nitrogen at 22.4 atmospheric pressure and
273 K. $ (R = 8.314 Jmol k^{-1})$

A. $ 14 \times 22.4 g $ B. 56 g C. 28 g D. None of these

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Explanation

$ PV = RT \Rightarrow = {PV \over RT } $ $ Mass of \mu litre nitrogen = \times M_o $

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A vessel contains 1 mole of $O_2$ gas (relative molar mass 32) at a temperature T. The pressure of the gas is P. An identical vessel containing 1 mole of He gas (relative molar mass 4) at a temperature 2T has pressure of ..........

A. 8 P B. p/8 C. 2 P D. P

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Explanation

$PV = RT \Rightarrow P = { RT \over V} $ $ For same , R and V, P \alpha T $

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The equation of state for 5 g of oxygen at a pressure P and temperature T, when occupying a volume V, will be

A. $ PV = \left( { 5 \over 32} \right) RT $ B. $ PV = \left( { 5 \over 16} \right) RT $ C. $ PV = \left( { 5 \over 2} \right) RT $ D. $ PV = 5 RT $

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Explanation

molecular weight of oxygen = 32 (g) number of moles in 5 g of oxygen = 5 /32 $ \therefore Equation of state is PV = RT$ $ PV = { 5 \over 32 } RT$

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A gas at 1 atmosphere and having volume 100 ml is mixed with another gas of equal moles at 0.5 atm and having volume 50 ml in flask of one litre, what is the final pressure?

A. 0.125 atm B. 0.75 atm C. 1 atm D. 0.5 atm

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Explanation

Total number of moles is conserved, $ \therefore { P_1 V_1 \over RT} + { P_2 V_2 \over RT } = { PV \over RT} $ $ \therefore { 1 \times 100 \over RT } + { 0.5 \times 50 \over RT} = { P \times 1000 \over RT} $ ( 1 L = 1000 ml ) P = 0.125 atm

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The quantity $ { PV \over k_E T } $ represents

A. mass of gas B. number of moles of gas C. number of molecules in gas D. K. E. of gas

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Explanation

The ideal gas equation is PV = RT PV = RT ( = 1 ) $ \ PV = k_B N_A T $ $ \left( { R \over N_A} = k_B \right) $ $ \therefore {PV \over k_B T } = N_A = Avogadro's number $

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Equation of gas in terms of pressure (P), absolute temperature (T) and density ($ \rho $) is

A. $ { P_1 T_1 \over \rho_1} = { P_2 T_2 \over \rho_2 } $ B. $ { P_1 \over T_1 \rho_1} = { P_2 \over T_2 \rho_2 } $ C. $ { P_1 \rho_2 \over T_1 } = { P_2 \rho_1 \over T_2 } $ D. $ { P_1 \rho_1 \over T_1 } = { P_2 \rho_2 \over T_2 } $

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Explanation

PV = RT $ \therefore Pv = \left( { M \over M_0 } \right) RT $ $ = \left( { M \over V } \right) {RT \over M_o } $ $ { \rho RT \over M_o } $ $ \left( {M \over V } = \rho \right) $ $ \Rightarrow { P \over \rho T } = { R \over M_o } = const $

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$O_2$ gas is filled in a vessel. If pressure is doubled, temperature becomes four times, how many times its density will become.

A. 4 B. 1/4 C. 2 D. 1/2

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Explanation

PV = RT $ \therefore p \left( { M \over \rho} \right) = RT $ $ \therefore P \left( {M \over \rho} \right) = \left( { M \over M_o} \right) RT \Rightarrow \alpha { P \over T } $

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