JEE Physics Kinetic Theory MCQs Set 08

Answer: (c) \( \frac54 \)

Explanation:
1. For an ideal gas \( PM=\rho RT \), so \( M=\frac{\rho RT}{P} \).
2. At the same T: \( \frac{M_A}{M_B}=\frac{\rho_A}{\rho_B}\cdot\frac{P_B}{P_A}=2.5\times\frac12=\frac54 \).

Teacher's Note:
1. Write the gas law in terms of density and molar mass.
2. Same temperature lets the factor RT cancel.

Answer: (d) \( \frac{Pm}{kT} \)

Explanation:
1. \( P=nkT \), where n is the number of molecules per unit volume, so \( n=\frac{P}{kT} \).
2. Density \( \rho=nm=\frac{Pm}{kT} \).

Teacher's Note:
1. The number density n and mass per molecule m give the mass density.
2. Check the units: Pa times kg over J is kg per cubic metre.

Answer: (c) 250 K

Explanation:
1. At constant volume \( \frac{\Delta P}{P}=\frac{\Delta T}{T} \).
2. \( 0.004=\frac1T\Rightarrow T=250 \) K.

Teacher's Note:
1. Pressure is proportional to absolute temperature at constant volume.
2. The answer is in kelvin; it corresponds to -23 degrees Celsius.

Answer: (a) \( \frac{2P_0T}{T+T_0} \)

Explanation:
1. Total number of moles is conserved: \( \frac{2P_0V}{RT_0}=\frac{PV}{RT_0}+\frac{PV}{RT} \).
2. \( \frac{2P_0}{T_0}=P\left(\frac1{T_0}+\frac1T\right)\Rightarrow P=\frac{2P_0T}{T+T_0} \).

Teacher's Note:
1. The pressure becomes equal in both containers, but the number of moles in each is different.
2. Check: if T = T0, the pressure stays at P0.

Answer: (d) \( 546^\circ\)C

Explanation:
1. Conservation of moles: \( \frac{2P_0V}{273R}=\frac{PV}{273R}+\frac{PV}{TR} \) with \( P=1.5P_0 \).
2. \( 2=1.5\left(1+\frac{273}{T}\right)\Rightarrow\frac{273}{T}=\frac13\Rightarrow T=819 \) K.
3. \( T=819-273=546^\circ\)C.

Teacher's Note:
1. The bulb in ice stays at 273 K.
2. Convert the answer from kelvin back to Celsius.

Answer: (a) \( \frac13\rho(\bar c)^2 \)

Explanation:
1. Pressure arises from the random motion of the molecules relative to the container.
2. A uniform motion of the whole system does not change this random motion, so \( P=\frac13\rho\bar c^2 \) is unchanged.

Teacher's Note:
1. Pressure depends on the random (thermal) velocities only.
2. A moving container does not increase the gas pressure, just as a moving bottle of soda is not at higher pressure.

Answer: (b) \( \langle v^4\rangle \)

Explanation:
1. Velocities point in all directions equally, so the odd powers of the velocity (vector) average to zero: \( \langle v\rangle=\langle v^3\rangle=\langle v^5\rangle=0 \).
2. An even power such as \( v^4 \) is always positive, so its average cannot be zero.

Teacher's Note:
1. Odd powers keep the sign (direction) and cancel; even powers are always positive.
2. This is why we use the root mean square speed instead of the average velocity.

Answer: (d) 4.24

Explanation:
1. \( v_{rms}=\sqrt{\frac{2^2+3^2+4^2+5^2+6^2}{5}} \).
2. \( =\sqrt{\frac{4+9+16+25+36}{5}}=\sqrt{18}\approx4.24 \).

Teacher's Note:
1. Square first, then average, then take the root.
2. The rms speed is larger than the mean speed of 4.

Answer: (c) \( \frac{kT}{m} \)

Explanation:
1. By equipartition, each translational degree of freedom carries \( \frac12kT \): \( \frac12m\langle v_y^2\rangle=\frac12kT \).
2. \( \langle v_y^2\rangle=\frac{kT}{m} \).

Teacher's Note:
1. Total mean-square speed is 3kT/m, shared equally among x, y and z.
2. Equipartition gives kT/2 per quadratic term in the energy.

Answer: (a) the value of \( v_{rms} \) of the molecules of gas is more than the value of the escape velocity

Explanation:
1. The moon's escape velocity (about 2.4 km/s) is small, and the thermal speeds of many gas molecules on its sunlit surface exceed it.
2. Such molecules escape over time, so the moon cannot retain an atmosphere.

Teacher's Note:
1. The earth retains heavy gases because its escape velocity (11.2 km/s) is much larger than the thermal speeds.
2. A planet keeps a gas only if v_rms is well below the escape velocity.

Answer: (b) \( P_2=P_1,\ T_2=\frac{T_1}{2} \)

Explanation:
1. Total kinetic energy \( =N\cdot\frac32kT \) is unchanged while N doubles, so \( T_2=\frac{T_1}{2} \).
2. \( P=\frac{NkT}{V} \): \( P_2=\frac{2Nk(T_1/2)}{V}=P_1 \).

Teacher's Note:
1. Pressure is proportional to the total translational kinetic energy per unit volume.
2. Same total energy in the same volume means the same pressure.

Answer: (c) \( \frac{n_1T_1+n_2T_2+n_3T_3}{n_1+n_2+n_3} \)

Explanation:
1. Total energy is conserved: \( \frac32(n_1+n_2+n_3)kT=\frac32n_1kT_1+\frac32n_2kT_2+\frac32n_3kT_3 \).
2. \( T=\frac{n_1T_1+n_2T_2+n_3T_3}{n_1+n_2+n_3} \).

Teacher's Note:
1. All three gases are monatomic, so the energy per molecule is the same function of T.
2. The mixture temperature is the number-weighted mean of the temperatures.

Answer: (d) The average of the distances travelled between two successive collisions is the mean free path.

Explanation:
1. In kinetic theory collisions are perfectly elastic and last a negligible time compared with the time between collisions, so the first statement is false.
2. Intermolecular forces are neglected except during collisions, and the molecules have a distribution of speeds.
3. The mean free path is by definition the average distance between two successive collisions.

Teacher's Note:
1. Memorise the assumptions: point masses, elastic collisions, no mutual forces, random motion.
2. Molecular speeds follow the Maxwell distribution.

Answer: (b) \( r^2 \)

Explanation:
1. \( \lambda=\frac{1}{\sqrt2\,\pi d^2n} \), where \( d=2r \) is the molecular diameter.
2. Hence \( \lambda\propto\frac1{r^2} \).

Teacher's Note:
1. The collision cross-section of a molecule is proportional to the square of its diameter.
2. Larger molecules collide more often and have a shorter mean free path.

Answer: (c) \( 2\sqrt2:1 \)

Explanation:
1. At the same pressure and temperature the number density n is the same, and the collision rate \( \propto n\bar v\propto\frac1{\sqrt M} \).
2. \( \frac{\text{He}}{\text{O}_2}=\sqrt{\frac{32}{4}}=\sqrt8=2\sqrt2 \), that is \( 2\sqrt2:1 \).

Teacher's Note:
1. Lighter molecules move faster and hit the walls more often.
2. The molecular masses are 4 and 32.

Answer: (d) it does not depend on temperature

Explanation:
1. For an ideal gas \( C_V=\frac f2R \) and \( C_P=\left(\frac f2+1\right)R \), which contain only the degrees of freedom and R.
2. They are therefore independent of temperature (within the range in which f is constant).

Teacher's Note:
1. Specific heats of real gases change with temperature only when new degrees of freedom get excited.
2. For an ideal gas of fixed f, the energy is exactly proportional to T.

Answer: (a) \( m=-c \) and \( n=-1 \)

Explanation:
1. \( P+\frac{aT^2}{V}=(RT+b)V^{-c} \), so \( P=(RT+b)V^{-c}-aT^2V^{-1} \).
2. Comparing with \( P=AV^m-BV^n \): \( m=-c \) and \( n=-1 \).

Teacher's Note:
1. Isolate P and compare the powers of V.
2. At a fixed temperature, the coefficients A = RT + b and B = aT squared are constants.

Answer: (a) \( a=16b \)

Explanation:
1. For one gram of gas, \( c_p-c_v=\frac RM \), where M is the molar mass.
2. \( a=\frac R2 \) for hydrogen (M = 2) and \( b=\frac R{32} \) for oxygen (M = 32).
3. \( \frac ab=\frac{32}{2}=16\Rightarrow a=16b \).

Teacher's Note:
1. Gram specific heats differ from molar ones by the factor 1/M.
2. The lighter gas has the larger difference of gram specific heats.

Answer: (c) diatomic

Explanation:
1. \( C_V=\frac R{0.4}=2.5R=\frac52R \).
2. \( C_V=\frac f2R\Rightarrow f=5 \), which is the value for a diatomic gas (3 translational + 2 rotational).

Teacher's Note:
1. Cv = 3R/2 for monatomic and 5R/2 for diatomic gases.
2. Gamma = 1 + R/Cv = 1.4 here.