JEE Physics Kinetic Theory MCQs Set 05

Answer: (c) 114

Explanation:
1. At constant volume and temperature, pressure is proportional to the number of molecules, that is to the mass of gas.
2. New pressure \( =1.5\times76=114 \) cm of Hg.

Teacher's Note:
1. \( P=\dfrac{nRT}{V} \), so for fixed \( V \) and \( T \) the pressure is proportional to \( n \).
2. A 50% increase in mass is a factor of 1.5.

Answer: (d) \( P=\dfrac23E \)

Explanation:
1. Kinetic theory gives \( P=\dfrac13\rho\overline{v^{2}} \).
2. The kinetic energy per unit volume is \( E=\dfrac12\rho\overline{v^{2}} \), so \( \rho\overline{v^{2}}=2E \).
3. Hence \( P=\dfrac23E \).

Teacher's Note:
1. Pressure equals two-thirds of the translational kinetic energy density.
2. Remember the factor 1/3 from three directions in space.

Answer: (c) \( \dfrac{273}{373}x \)

Explanation:
1. \( PV=nRT \) gives \( \dfrac{\rho}{P}=\dfrac{M}{RT} \), which is inversely proportional to the absolute temperature.
2. \( \dfrac{(\rho/P)_{373}}{(\rho/P)_{273}}=\dfrac{273}{373} \).
3. So the quotient becomes \( \dfrac{273}{373}x \).

Teacher's Note:
1. Always convert Celsius to kelvin for gas laws.
2. At fixed pressure, a hotter gas has a lower density.

Answer: (a) \( 3.5\times10^{11} \)

Explanation:
1. \( P=nkT \), so number density \( n=\dfrac{P}{kT}=\dfrac{10^{-5}\times133.3}{1.38\times10^{-23}\times273}\approx3.5\times10^{17}\ \text{m}^{-3} \).
2. Per cm³ (\( 10^{-6} \) m³): \( 3.5\times10^{11} \).

Teacher's Note:
1. Use \( P=nkT \) for number density.
2. Convert mm of Hg to pascals and m³ to cm³ carefully.

Answer: (c) \( 2v \)

Explanation:
1. \( v_{rms}\propto\sqrt{T/M} \).
2. Doubling \( T \) gives a factor \( \sqrt2 \); the molar mass halves (O to \( \text{O}_2 \)), giving another \( \sqrt2 \).
3. New speed \( =\sqrt2\times\sqrt2\,v=2v \).

Teacher's Note:
1. Dissociation halves the molecular mass.
2. Both changes raise the speed, so multiply the two factors.

Answer: (a) \( V_1 \)

Explanation:
1. The average speed of a gas depends only on its temperature and its molecular mass.
2. The \( \text{O}_2 \) molecules in vessel C are at the same temperature \( T \) as in A, and the presence of \( \text{N}_2 \) does not change their speed distribution.
3. So the average speed is \( V_1 \).

Teacher's Note:
1. In a mixture each gas keeps its own speed distribution at the common temperature.
2. Molecules of different gases only exchange energy, they do not change the average speed of each type.

Answer: (c) the ratio of rms speed to average speed of the molecules is \( \dfrac{2}{\sqrt6}\sqrt{\dfrac{2N+1}{N+1}} \)

Explanation:
1. Average speed \( =\dfrac{1+2+\dots+N}{N}=\dfrac{N+1}{2} \).
2. \( v_{rms}^{2}=\dfrac{1^{2}+2^{2}+\dots+N^{2}}{N}=\dfrac{(N+1)(2N+1)}{6} \).
3. Ratio \( =\dfrac{\sqrt{(N+1)(2N+1)/6}}{(N+1)/2}=\dfrac{2}{\sqrt6}\sqrt{\dfrac{2N+1}{N+1}} \).

Teacher's Note:
1. Rms speed is always larger than the average speed unless all speeds are equal.
2. Use the sum formulas for \( n \) and \( n^{2} \).

Answer: (b) collide with the walls and transfer momentum to the walls of the container

Explanation:
1. Each collision reverses the normal component of a molecule's momentum.
2. The momentum transferred to the wall per second is a force, and force per unit area is pressure.
3. So pressure arises from momentum transfer in collisions with the walls.

Teacher's Note:
1. Pressure is momentum transferred per unit area per unit time.
2. Collisions between molecules do not change the force on the walls on average.

Answer: (d) 5 and 6

Explanation:
1. Hydrogen is a diatomic molecule: 3 translational and 2 rotational degrees of freedom, giving 5.
2. Ozone is a non-linear triatomic molecule: 3 translational and 3 rotational degrees of freedom, giving 6 (at ordinary temperatures, ignoring vibration).

Teacher's Note:
1. Linear molecules have 2 rotational degrees of freedom, non-linear ones have 3.
2. Vibrational modes are frozen at room temperature.

Answer: (b) 1 : 1

Explanation:
1. Both molecules are diatomic with 2 rotational degrees of freedom.
2. By equipartition each carries \( \tfrac12kT \) per degree, so the average rotational energy is \( kT \) for each molecule at the same temperature.
3. The ratio is 1 : 1, independent of moments of inertia.

Teacher's Note:
1. Equipartition of energy does not depend on mass or moment of inertia.
2. Only the temperature and the number of degrees of freedom matter.

Answer: (b) 1870 m/s

Explanation:
1. \( \dfrac{v_{H_2}}{v_{N_2}}=\sqrt{\dfrac{M_{N_2}}{M_{H_2}}}=\sqrt{\dfrac{28}{2}}=\sqrt{14}\approx3.74 \).
2. \( v_{H_2}\approx500\times3.74\approx1870 \) m/s.

Teacher's Note:
1. The ratio of speeds is the square root of the inverse ratio of molar masses.
2. Use 28 for nitrogen and 2 for hydrogen.

Answer: (b) nature of the gas

Explanation:
1. For \( N \) molecules, \( PV=NkT \), so \( \dfrac{kT}{PV}=\dfrac{1}{N} \).
2. The quantity \( \dfrac{M}{N}=m_0 \) is the mass of one molecule.
3. It depends only on the nature of the gas.

Teacher's Note:
1. Reduce the expression using \( PV=NkT \).
2. The result is a property of the gas, not of its state.

Answer: (a) pressure

Explanation:
1. \( P=\dfrac{nRT}{V} \) depends only on \( n \), \( T \) and \( V \), so it is the same for all ideal gases.
2. Rms speed, density and momentum depend on the molar mass.

Teacher's Note:
1. Equal moles, volume and temperature fix the pressure.
2. Quantities that involve mass differ from gas to gas.

Answer: (d) 150 cm

Explanation:
1. At fixed \( V \) and \( T \), \( P\propto n \), which is proportional to mass.
2. Doubling the mass doubles the pressure: 150 cm of Hg.

Teacher's Note:
1. Pressure is proportional to the number of molecules at fixed volume and temperature.
2. Do not square the factor.

Answer: (b) 4

Explanation:
1. Average speed \( \propto\sqrt T \), so doubling it makes \( T \) four times larger.
2. The volume is fixed, so \( P\propto T \): the pressure increases by a factor 4.

Teacher's Note:
1. Speed doubles means temperature quadruples.
2. A closed container means constant volume.

Answer: (b) \( E \)

Explanation:
1. The mean translational kinetic energy per molecule is \( \tfrac32kT \), which depends only on temperature.
2. At the same temperature, \( \text{H}_2 \) molecules have the same mean kinetic energy \( E \) as \( \text{O}_2 \).

Teacher's Note:
1. Equal temperature means equal mean kinetic energy per molecule.
2. Lighter molecules compensate with higher speed.

Answer: (d) 1.0 g

Explanation:
1. At fixed \( P \) and \( V \), \( nT=\) constant, so \( m_1T_1=m_2T_2 \).
2. \( m_2=13\times\dfrac{300}{325}=12 \) g.
3. Mass released \( =13-12=1 \) g.

Teacher's Note:
1. Use \( PV=\dfrac mM RT \) with \( P \) and \( V \) fixed.
2. Convert to kelvin: 300 K and 325 K.

Answer: (d) the mean translational kinetic energy of a diatomic gas increases with an increase in absolute temperature

Explanation:
1. Mean translational kinetic energy \( =\tfrac32kT \), proportional to the absolute temperature, so (d) is correct.
2. Collisions are elastic, so the average speed does not fall with each collision.
3. The pressure is proportional to the mean square speed, and the kinetic energy of an ideal gas at constant temperature is unchanged on expansion.

Teacher's Note:
1. For an ideal gas the internal energy depends only on temperature.
2. Pressure involves \( v^{2} \), not \( v \).

Answer: (a) (1) and (4) only

Explanation:
1. At the same temperature the average kinetic energy per molecule is the same (2).
2. At the same pressure and temperature the kinetic energy per unit volume is \( \tfrac32P \), the same (3).
3. Average momentum per molecule \( \propto\sqrt{M} \) differs (1), and kinetic energy per unit mass \( \propto\dfrac1M \) differs (4).

Teacher's Note:
1. Same temperature gives the same energy per molecule; mass-dependent quantities differ.
2. Check each quantity separately.

Answer: (a) (1) and (3) only

Explanation:
1. \( PV=N_AkT \), so \( kT=\dfrac{10^{6}\times2\times10^{4}}{6\times10^{23}}=\dfrac13\times10^{-13} \) erg: statement (1).
2. Mean translational KE \( =\tfrac32kT=\tfrac32\times\dfrac13\times10^{-13}=5\times10^{-14} \) erg: statement (3).
3. Statements (2) and (4) are false.

Teacher's Note:
1. One gram molecule means one mole, so use \( N_A \) molecules.
2. Translational energy per molecule is \( \tfrac32kT \).

Answer: (c) 2 : 3

Explanation:
1. Isothermal expansion to double volume halves the pressure, so \( \Delta P=\tfrac12P_{initial} \) for each gas.
2. \( P_A=2\Delta p \) and \( P_B=2(1.5\Delta p)=3\Delta p \).
3. At the same \( V \) and \( T \), \( P\propto m \): \( \dfrac{m_A}{m_B}=\dfrac{2}{3} \).

Teacher's Note:
1. The pressure drop is half the initial pressure.
2. At fixed \( V \) and \( T \) the pressure is proportional to the mass.

Answer: (b) \( 2.9\times10^{4} \) s

Explanation:
1. Average speed \( \bar v=\sqrt{\dfrac{8RT}{\pi M}}=\sqrt{\dfrac{8\times8.314\times300}{\pi\times0.032}}\approx445\ \text{m/s} \).
2. Time \( =\dfrac{1.28\times10^{7}}{445}\approx2.9\times10^{4} \) s (about 8 hours).

Teacher's Note:
1. Use the average speed formula with the molar mass in kg.
2. Molecules take a long time to travel in a straight line only because of their frequent collisions in reality.

Answer: (d) \( 8.0\times10^{-24} \) kg m/s

Explanation:
1. Average speed \( \bar v=\sqrt{\dfrac{8kT}{\pi m}}=\sqrt{\dfrac{8\times1.38\times10^{-23}\times273}{\pi\times6.64\times10^{-27}}}\approx1.20\times10^{3}\ \text{m/s} \).
2. Average momentum magnitude \( =m\bar v=6.64\times10^{-27}\times1.2\times10^{3}\approx8.0\times10^{-24} \) kg m/s.

Teacher's Note:
1. The average of the vector momentum is zero; the question asks for the average magnitude.
2. Use the mean speed, not the rms speed.

Answer: (a) 4 atm

Explanation:
1. \( P=\dfrac{nRT}{V} \): equal \( n \), \( T \) and \( V \) give equal pressures.
2. The molecular weights do not matter.
3. He is at 4 atm.

Teacher's Note:
1. Pressure depends on the number of moles, not on the type of gas.
2. Avoid being distracted by the molecular weights.

Answer: (a) \( \sqrt{14} \)

Explanation:
1. In a mixture both gases are at the same temperature.
2. Mean speed \( \propto\dfrac{1}{\sqrt M} \): \( \dfrac{\bar v_{H_2}}{\bar v_{N_2}}=\sqrt{\dfrac{28}{2}}=\sqrt{14} \).

Teacher's Note:
1. A mixture is in thermal equilibrium, so all components share one temperature.
2. The lighter molecules are faster.

Answer: (d) \( -2.5\times10^{25} \)

Explanation:
1. \( PV=NkT \), so \( N=\dfrac{PV}{kT} \).
2. \( n_f-n_i=\dfrac{PV}{k}\left(\dfrac{1}{T_f}-\dfrac{1}{T_i}\right)=\dfrac{10^{5}\times30}{1.38\times10^{-23}}\left(\dfrac{1}{300}-\dfrac{1}{290}\right) \).
3. \( \approx-2.5\times10^{25} \) (molecules leave the open room as it warms).

Teacher's Note:
1. In an open room the pressure and volume are fixed, so \( N\propto1/T \).
2. A negative sign means a loss of molecules.

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

Explanation:
1. Ideal behaviour needs negligible molecular volume and negligible intermolecular forces.
2. Low pressure makes the gas dilute and high temperature makes the kinetic energy large compared with the attractive energy.
3. So a real gas behaves ideally at low pressure and high temperature.

Teacher's Note:
1. Gases approach ideal behaviour as the density falls and the temperature rises.
2. High pressure and low temperature lead to liquefaction.

Answer: (a) \( \dfrac{b^{2}}{4a} \)

Explanation:
1. At equilibrium \( \dfrac{dU}{dx}=0 \): \( -\dfrac{12a}{x^{13}}+\dfrac{6b}{x^{7}}=0 \), so \( x^{6}=\dfrac{2a}{b} \).
2. \( U_{eq}=\dfrac{a}{x^{12}}-\dfrac{b}{x^{6}}=\dfrac{ab^{2}}{4a^{2}}-\dfrac{b^{2}}{2a}=-\dfrac{b^{2}}{4a} \).
3. \( D=0-U_{eq}=\dfrac{b^{2}}{4a} \).

Teacher's Note:
1. Equilibrium is where the force, \( -\dfrac{dU}{dx} \), is zero.
2. The potential at infinity is zero.

Answer: (b) \( P_0A \)

Explanation:
1. The wire keeps the volume fixed, so \( P\propto T \) and the gas pressure becomes \( 2P_0 \).
2. Each piston has \( 2P_0A \) pushing outward and \( P_0A \) from the atmosphere pushing inward.
3. The wire must supply the difference: \( T=2P_0A-P_0A=P_0A \).

Teacher's Note:
1. Draw the free-body diagram of one piston.
2. The wire holds the volume constant, so apply Gay-Lussac's law.

Answer: (a) 16 cm

Explanation:
1. Let the mercury inside rise \( h \) cm above the outer level. The air column is \( 8+46-h=54-h \) and its pressure is \( 76-h \).
2. Boyle's law: \( 76\times8=(76-h)(54-h) \), so \( h^{2}-130h+3496=0 \), giving \( h=38 \) cm.
3. Air column \( =54-38=16 \) cm.

Teacher's Note:
1. Pressure of the trapped air equals atmospheric pressure minus the mercury column above the outer level.
2. Choose the root with \( h<76 \) cm.