JEE Physics Kinetic Theory MCQs Set 19

Download JEE MCQs for JEE Physics: Kinetic Theory of gases

Access targeted multiple-choice questions for Kinetic Theory of gases designed to align with the latest JEE academic syllabus for JEE Physics. These objective practice sets help students evaluate their conceptual understanding and improve exam readiness.

Chapter-wise Objective Questions: Kinetic Theory of gases

Navigate directly to the 50 objective questions for Kinetic Theory of gases using the digital viewer below. Each practice set includes verified answer keys, allowing students to instantly cross-check their work and identify areas requiring further revision.

Multiple Choice Questions

Question 1: Two identical containers each of volume \( V_0 \) are joined by a small pipe. They contain identical gases at temperature \( T_0 \) and pressure \( p_0 \). One container is heated to temperature \( 2T_0 \) while the other is maintained at \( T_0 \). The common pressure of the gas is
(a) \( 2p_0 \)
(b) \( \dfrac43p_0 \)
(c) \( p_0 \)
(d) \( \dfrac32p_0 \)
Show Answer & Explanation

Answer: (b) \( \dfrac43p_0 \)

Explanation:
1. Total moles \( =2n_0 \) with \( n_0=\dfrac{p_0V_0}{RT_0} \). With equal pressure, \( n_1(2T_0)=n_2T_0\Rightarrow n_2=2n_1 \).
2. \( n_1=\dfrac{2n_0}3 \), so \( p=\dfrac{n_1R(2T_0)}{V_0}=\dfrac43p_0 \).

Teacher's Note:
1. The hot side holds fewer moles at the same pressure.

Question 2: The volumes of containers A and B, connected by a tube and a closed valve, are \( V \) and \( 4V \) respectively. Both contain the same ideal gas at pressures \( 5.0\times10^5 \) Pa (300 K) and \( 1.0\times10^5 \) Pa (400 K). The valve is opened to equalise the pressure, but the temperature of each container is kept at its initial value. The common pressure is
(a) \( 2.5\times10^5 \) Pa
(b) \( 2.0\times10^5 \) Pa
(c) \( 3.0\times10^5 \) Pa
(d) \( 1.5\times10^5 \) Pa
Show Answer & Explanation

Answer: (b) \( 2.0\times10^5 \) Pa

Explanation:
1. Moles are conserved: \( \dfrac{PV}{300}+\dfrac{P\,4V}{400}=\dfrac{5\times10^5V}{300}+\dfrac{10^5\,4V}{400} \).
2. \( P\left(\dfrac1{300}+\dfrac1{100}\right)=1666.7+1000 \Rightarrow P=2\times10^5 \) Pa.

Teacher's Note:
1. Use \( n=pV/RT \) for each container.

Question 3: Two non-reactive monatomic ideal gases have their atomic masses in the ratio 2 : 3. The ratio of their partial pressures, when enclosed in a vessel kept at a constant temperature, is 4 : 3. The ratio of their densities is
(a) 1 : 4
(b) 1 : 2
(c) 6 : 9
(d) 8 : 9
Show Answer & Explanation

Answer: (d) 8 : 9

Explanation:
1. \( \rho=\dfrac{pM}{RT} \) at the same \( T \).
2. \( \dfrac{\rho_1}{\rho_2}=\dfrac{4\times2}{3\times3}=\dfrac89 \).

Teacher's Note:
1. Density is proportional to partial pressure times molar mass.

Question 4: A container with insulating walls is divided into two equal parts by a partition fitted with a valve. One part is filled with an ideal gas at pressure \( p \) and temperature \( T \), whereas the other part is completely evacuated. If the valve is suddenly opened, the pressure and temperature of the gas will be
(a) \( \dfrac p2,\ T \)
(b) \( \dfrac p2,\ \dfrac T2 \)
(c) \( p,\ T \)
(d) \( p,\ \dfrac T2 \)
Show Answer & Explanation

Answer: (a) \( \dfrac p2,\ T \)

Explanation:
1. Free expansion: \( Q=0 \), \( W=0 \), so \( \Delta U=0 \) and the temperature of an ideal gas is unchanged.
2. Volume doubles at constant \( T \), so the pressure halves.

Teacher's Note:
1. Free expansion does no work and leaves \( T \) unchanged.

Question 5: A perfect gas at 27\(^\circ\)C is heated at constant pressure so as to double its volume. The increase in temperature of the gas will be
(a) 300\(^\circ\)C
(b) 54\(^\circ\)C
(c) 327\(^\circ\)C
(d) 600\(^\circ\)C
Show Answer & Explanation

Answer: (a) 300\(^\circ\)C

Explanation:
1. \( V\propto T \): \( T_2=2\times300=600 \) K.
2. Increase \( =300 \) K \( =300^\circ \)C.

Teacher's Note:
1. Use kelvin for gas laws.

Question 6: A real gas behaves like an ideal gas if its
(a) pressure and temperature are both high
(b) pressure and temperature are both low
(c) pressure is high and temperature is low
(d) pressure is low and temperature is high
Show Answer & Explanation

Answer: (d) pressure is low and temperature is high

Explanation:
1. Real gases approach ideal behaviour at low pressure (molecules far apart) and high temperature (weak effect of attraction).

Teacher's Note:
1. Deviations are largest near liquefaction.

Question 7: The temperature of a gas contained in a closed vessel of constant volume increases by 1\(^\circ\)C when the pressure of the gas is increased by 1%. The initial temperature of the gas is
(a) 100 K
(b) 273\(^\circ\)C
(c) 100\(^\circ\)C
(d) 200 K
Show Answer & Explanation

Answer: (a) 100 K

Explanation:
1. At constant volume \( \dfrac{\Delta p}{p}=\dfrac{\Delta T}{T} \), so \( T=\dfrac{1}{0.01}=100 \) K.

Teacher's Note:
1. Pressure is proportional to absolute temperature.

Question 8: 1 mole of a gas occupies a volume of 200 mL at 100 mm pressure. What is the volume occupied by 2 moles of the gas at 400 mm pressure and at the same temperature?
(a) 50 mL
(b) 100 mL
(c) 200 mL
(d) 400 mL
Show Answer & Explanation

Answer: (b) 100 mL

Explanation:
1. \( V\propto\dfrac np \): \( V_2=200\times\dfrac21\times\dfrac{100}{400}=100 \) mL.

Teacher's Note:
1. Volume rises with the number of moles and falls with pressure.

Question 9: Air is pumped into an automobile tube up to a pressure of 200 kPa in the morning when the air temperature is 22\(^\circ\)C. During the day, the temperature rises to 42\(^\circ\)C and the tube expands by 2%. The pressure of the air in the tube at this temperature will be approximately
(a) 212 kPa
(b) 209 kPa
(c) 206 kPa
(d) 200 kPa
Show Answer & Explanation

Answer: (b) 209 kPa

Explanation:
1. \( \dfrac{p_1V_1}{T_1}=\dfrac{p_2V_2}{T_2} \) with \( T_1=295 \) K, \( T_2=315 \) K, \( V_2=1.02V_1 \).
2. \( p_2=200\times\dfrac{315}{295}\times\dfrac1{1.02}=209 \) kPa.

Teacher's Note:
1. Convert temperatures to kelvin.

Question 10: An ideal gas is expanding such that \( pT^2=\) constant. The coefficient of volume expansion of the gas is
(a) \( \dfrac1T \)
(b) \( \dfrac2T \)
(c) \( \dfrac3T \)
(d) \( \dfrac4T \)
Show Answer & Explanation

Answer: (c) \( \dfrac3T \)

Explanation:
1. \( p=\dfrac{nRT}{V} \) gives \( \dfrac{nRT^3}{V}=\) constant, so \( V\propto T^3 \).
2. \( \gamma_V=\dfrac1V\dfrac{dV}{dT}=\dfrac3T \).

Teacher's Note:
1. If \( V\propto T^k \) the coefficient of expansion is \( k/T \).

Question 11: 1 L of oxygen at a pressure of 1 atm and 2 L of nitrogen at a pressure of 0.5 atm are introduced into a vessel of volume 1 L. If there is no change in temperature, the final pressure of the mixture of gas (in atm) is
(a) 1.5
(b) 2.5
(c) 2
(d) 4
Show Answer & Explanation

Answer: (c) 2

Explanation:
1. \( p_{O_2}=1\times\dfrac11=1 \) atm and \( p_{N_2}=0.5\times\dfrac21=1 \) atm.
2. Total \( =2 \) atm.

Teacher's Note:
1. Each gas obeys Boyle's law; add the partial pressures.

Question 12: What is an ideal gas?
(a) One that consists of molecules
(b) A gas satisfying the assumptions of kinetic theory
(c) A gas having Maxwellian distribution of speed
(d) A gas consisting of massless particles
Show Answer & Explanation

Answer: (b) A gas satisfying the assumptions of kinetic theory

Explanation:
1. An ideal gas is one whose molecules obey the assumptions of the kinetic theory: point-like, no mutual forces, elastic collisions.

Teacher's Note:
1. Real gases approximate this at low pressure.

Question 13: When the temperature of a gas increases by 1\(^\circ\)C, its pressure increases by 0.4% (constant volume). What is its initial temperature?
(a) 250 K
(b) 125 K
(c) 195 K
(d) 329 K
Show Answer & Explanation

Answer: (a) 250 K

Explanation:
1. \( \dfrac{\Delta p}{p}=\dfrac{\Delta T}{T}\Rightarrow T=\dfrac1{0.004}=250 \) K.

Teacher's Note:
1. Pressure is proportional to \( T \) at constant volume.

Question 14: Two gases A and B having the same pressure \( p \), volume \( V \) and absolute temperature \( T \) are mixed. If the mixture has the volume and temperature as \( V \) and \( T \) respectively, then the pressure of the mixture is
(a) \( 2p \)
(b) \( p \)
(c) \( p/2 \)
(d) \( 4p \)
Show Answer & Explanation

Answer: (a) \( 2p \)

Explanation:
1. The mixture has \( n_A+n_B=\dfrac{2pV}{RT} \) moles in volume \( V \) at \( T \), so the pressure is \( 2p \).

Teacher's Note:
1. Dalton's law: partial pressures add.

Question 15: In a certain region of space, there are only 5 molecules per cm\(^3\) on an average and the temperature is 3 K. The pressure of this dilute gas is (\( k=1.38\times10^{-23} \) J K\(^{-1}\))
(a) \( 20.7\times10^{-17} \) N m\(^{-2}\)
(b) \( 1.53\times10^{-15} \) N m\(^{-2}\)
(c) \( 2.3\times10^{-10} \) N m\(^{-2}\)
(d) \( 5.3\times10^{-5} \) N m\(^{-2}\)
Show Answer & Explanation

Answer: (a) \( 20.7\times10^{-17} \) N m\(^{-2}\)

Explanation:
1. \( n=5\times10^6 \) m\(^{-3}\).
2. \( p=nkT=5\times10^6\times1.38\times10^{-23}\times3=2.07\times10^{-16}=20.7\times10^{-17} \) N m\(^{-2}\).

Teacher's Note:
1. Convert cm\(^{-3}\) to m\(^{-3}\) first.

Question 16: Two balloons are filled, one with pure He gas and the other with air, respectively. If the pressure and temperature of these balloons are the same, then the number of molecules per unit volume is
(a) more in the He filled balloon
(b) the same in both balloons
(c) more in the air filled balloon
(d) in the ratio of 1 : 4
Show Answer & Explanation

Answer: (b) the same in both balloons

Explanation:
1. \( p=nkT \): equal \( p \) and \( T \) give equal number density, whatever the gas.

Teacher's Note:
1. Avogadro's law.

Question 17: Which one of the following graphs represents the behaviour of an ideal gas (at constant temperature) for the variation of \( pV \) with \( V \)?
(a) a straight line with negative slope
(b) a rectangular hyperbola falling with \( V \)
(c) a straight line rising with \( V \)
(d) a horizontal straight line
Show Answer & Explanation

Answer: (d) a horizontal straight line

Explanation:
1. At constant temperature \( pV=nRT \) is constant, so the plot of \( pV \) against \( V \) is a horizontal line.

Teacher's Note:
1. Boyle's law.

Question 18: The temperature remaining constant, the pressure of a gas is decreased by 20%. The percentage change in volume is
(a) increases by 20%
(b) decreases by 20%
(c) increases by 25%
(d) decreases by 25%
Show Answer & Explanation

Answer: (c) increases by 25%

Explanation:
1. \( pV \) constant: \( V_2=\dfrac{V}{0.8}=1.25V \).
2. Volume increases by 25%.

Teacher's Note:
1. The percentage change in volume is not equal to that in pressure.

Question 19: The equation of state for 5 g of oxygen at a pressure \( p \) and temperature \( T \), when occupying a volume \( V \), will be
(a) \( pV=\dfrac5{32}RT \)
(b) \( pV=5RT \)
(c) \( pV=\dfrac52RT \)
(d) \( pV=\dfrac5{16}RT \)
Show Answer & Explanation

Answer: (a) \( pV=\dfrac5{32}RT \)

Explanation:
1. Moles \( n=\dfrac{5}{32} \) (molar mass of O\(_2\) is 32 g), so \( pV=\dfrac5{32}RT \).

Teacher's Note:
1. \( n=m/M \).

Question 20: A bubble is at the bottom of a lake of depth \( h \). As the bubble comes to sea level, its radius increases three times. If atmospheric pressure is equal to \( l \) metre of water column, then \( h \) is equal to
(a) \( 26l \)
(b) \( l \)
(c) \( 25l \)
(d) \( 30l \)
Show Answer & Explanation

Answer: (a) \( 26l \)

Explanation:
1. Volume becomes \( 3^3=27 \) times, so \( p_{bottom}=27p_{atm}=27l \).
2. \( h+l=27l\Rightarrow h=26l \).

Teacher's Note:
1. Radius ratio cubed gives the volume ratio.

Question 21: If \( m \) represents the mass of each molecule of a gas and \( T \) its absolute temperature, then the root mean square velocity of the gaseous molecule is proportional to
(a) \( mT \)
(b) \( m^{1/2}T^{1/2} \)
(c) \( m^{-1/2}T \)
(d) \( m^{-1/2}T^{1/2} \)
Show Answer & Explanation

Answer: (d) \( m^{-1/2}T^{1/2} \)

Explanation:
1. \( v_{rms}=\sqrt{\dfrac{3kT}{m}}\propto m^{-1/2}T^{1/2} \).

Teacher's Note:
1. Lighter molecules and higher temperature give larger speeds.

Question 22: The rms velocity of gas molecules is 300 m s\(^{-1}\). The rms velocity of molecules of gas with twice the molecular weight and half the absolute temperature is
(a) 300 m s\(^{-1}\)
(b) 600 m s\(^{-1}\)
(c) 75 m s\(^{-1}\)
(d) 150 m s\(^{-1}\)
Show Answer & Explanation

Answer: (d) 150 m s\(^{-1}\)

Explanation:
1. \( v\propto\sqrt{T/M} \) changes by \( \sqrt{\tfrac{1/2}{2}}=\tfrac12 \), so \( v=150 \) m s\(^{-1}\).

Teacher's Note:
1. Both changes lower the speed.

Question 23: The velocities of 4 gas molecules are given by 1 km/s, 3 km/s, 5 km/s and 7 km/s. The difference between the rms velocity and the average velocity is
(a) 0.338 km/s
(b) 0.438 km/s
(c) 0.583 km/s
(d) 0.683 km/s
Show Answer & Explanation

Answer: (c) 0.583 km/s

Explanation:
1. Average \( =\dfrac{1+3+5+7}4=4 \) km/s.
2. \( v_{rms}=\sqrt{\dfrac{1+9+25+49}{4}}=\sqrt{21}=4.583 \) km/s.
3. Difference \( =0.583 \) km/s.

Teacher's Note:
1. The rms speed always exceeds the mean speed.

Question 24: An ideal gas (\( \gamma=1.5 \)) is expanded adiabatically. How many times has the gas to be expanded to reduce the root mean square velocity of its molecules 2.0 times?
(a) 4 times
(b) 16 times
(c) 8 times
(d) 2 times
Show Answer & Explanation

Answer: (b) 16 times

Explanation:
1. \( v_{rms}\propto\sqrt T \), so \( T \) falls to \( \tfrac14 \).
2. \( TV^{\gamma-1}=\) constant: \( V^{0.5}\) must rise 4 times, so \( V \) rises 16 times.

Teacher's Note:
1. Adiabatic relation with \( \gamma-1=0.5 \).

Question 25: The root mean square speed of the molecules of an ideal gas is \( v \). If the pressure is increased two times at constant temperature, the rms speed will become
(a) \( v/2 \)
(b) \( v \)
(c) \( 2v \)
(d) \( 4v \)
Show Answer & Explanation

Answer: (b) \( v \)

Explanation:
1. \( v_{rms}=\sqrt{3RT/M} \) depends only on \( T \) and \( M \), so it is unchanged.

Teacher's Note:
1. Pressure does not enter the formula.

Question 26: The molecules of a given mass of a gas have an rms velocity of 200 m/s at 27\(^\circ\)C and \( 1.0\times10^5 \) N/m\(^2\) pressure. When the temperature is 127\(^\circ\)C and pressure is \( 0.5\times10^5 \) N/m\(^2\), the rms velocity in m/s will be
(a) \( \dfrac{100\sqrt2}{3} \)
(b) \( 100\sqrt2 \)
(c) \( \dfrac{400}{\sqrt3} \)
(d) none of these
Show Answer & Explanation

Answer: (c) \( \dfrac{400}{\sqrt3} \)

Explanation:
1. \( v\propto\sqrt T \): \( v=200\sqrt{\dfrac{400}{300}}=\dfrac{400}{\sqrt3} \) m/s.

Teacher's Note:
1. Pressure has no effect.

Question 27: According to Maxwell's law of distribution of velocities of molecules, the most probable velocity is
(a) greater than the mean velocity
(b) equal to the mean velocity
(c) equal to the root mean square velocity
(d) less than the root mean square velocity
Show Answer & Explanation

Answer: (d) less than the root mean square velocity

Explanation:
1. \( v_p=\sqrt{2kT/m} \), \( \bar v=\sqrt{8kT/\pi m} \), \( v_{rms}=\sqrt{3kT/m} \), so \( v_p<\bar v<v_{rms} \).

Teacher's Note:
1. Ratio \( v_p:\bar v:v_{rms}=1:1.13:1.22 \).

Question 28: To what temperature should hydrogen at 327\(^\circ\)C be cooled at constant pressure, so that the root mean square velocity of its molecules becomes half of its previous value?
(a) \( -123^\circ \)C
(b) 123\(^\circ\)C
(c) \( -100^\circ \)C
(d) 0\(^\circ\)C
Show Answer & Explanation

Answer: (a) \( -123^\circ \)C

Explanation:
1. \( v\propto\sqrt T \): half the speed needs \( T=\dfrac{600}4=150 \) K \( =-123^\circ \)C.

Teacher's Note:
1. Start from 327 + 273 = 600 K.

Question 29: At what temperature is the rms speed of air molecules doubled of that at NTP?
(a) 819\(^\circ\)C
(b) 719\(^\circ\)C
(c) 909\(^\circ\)C
(d) none of these
Show Answer & Explanation

Answer: (a) 819\(^\circ\)C

Explanation:
1. Doubling \( v_{rms} \) needs \( T=4\times273=1092 \) K \( =819^\circ \)C.

Teacher's Note:
1. NTP is 273 K.

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Kinetic Theory of gases Objective Questions & Solutions for JEE Physics

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