JEE Physics Kinetic Theory MCQs Set 20

Answer: (b) 606 K

Explanation:
1. \( v_{rms} \propto \sqrt{T/M} \), so \( T_{SO_2}/64 = 303/32 \).
2. \( T_{SO_2} = 606 \) K.

Teacher's Note:
1. Equal rms speeds need \( T/M \) to be equal.
2. \( SO_2 \) is twice as heavy as \( O_2 \), so it needs twice the temperature.

Answer: (a) 1200 K

Explanation:
1. \( v_{rms} \propto \sqrt{T} \).
2. Doubling v needs 4 times T: \( 4\times300 = 1200 \) K.

Teacher's Note:
1. Speed varies as the square root of temperature.
2. To double the speed, quadruple the absolute temperature.

Answer: (c) Remain the same

Explanation:
1. The uniform motion of the lorry adds the same velocity to every molecule and does not change their random motion relative to the container.
2. Temperature depends only on the random thermal motion, so it stays the same.

Teacher's Note:
1. Temperature measures random molecular motion, not bulk motion.
2. A uniform velocity of the whole container does not heat the gas.

Answer: (c) \( 313^\circ \)C

Explanation:
1. Average KE \( \propto T \); initial T is 293 K.
2. Twice the energy needs 586 K, which is \( 313^\circ \)C.

Teacher's Note:
1. Double the kelvin temperature, not the Celsius value.
2. Convert back to Celsius at the end.

Answer: (c) \( 1527^\circ \)C

Explanation:
1. \( PV/T \) is constant: \( T_2 = 300\times2\times3 = 1800 \) K.
2. \( 1800-273 = 1527^\circ \)C.

Teacher's Note:
1. Use the combined gas law with kelvin temperatures.
2. Convert the final answer back to Celsius.

Answer: (c) 900 m\(^3\)

Explanation:
1. \( V_2 = V_1\frac{P_1T_2}{P_2T_1} = 500\times\frac{1}{0.5}\times\frac{270}{300} \).
2. \( V_2 = 900 \) m\(^3\).

Teacher's Note:
1. Apply \( P_1V_1/T_1 = P_2V_2/T_2 \).
2. Halving the pressure doubles the volume; the lower temperature then reduces it by 10 percent.

Answer: (c) 2V

Explanation:
1. \( V_2 = V\times\frac{600}{300} = 2V \).

Teacher's Note:
1. At constant pressure volume is proportional to kelvin temperature.
2. 27 and 327 degrees Celsius are 300 K and 600 K.

Answer: (a) 0.07

Explanation:
1. Volume of the balloon stays fixed at the same pressure, so \( n\propto1/T \).
2. Fraction escaping \( = 1-\frac{293}{313} = 0.064 \approx 0.07 \).

Teacher's Note:
1. The amount of gas that fits in a fixed volume at fixed pressure is inversely proportional to T.
2. Compute the fraction relative to the original amount.

Answer: (b) 0.98

Explanation:
1. Total \( PV \) is conserved: \( 1.40\times0.10 + 0.7\times0.15 = 0.245 \).
2. \( P = \frac{0.245}{0.25} = 0.98 \) MPa.

Teacher's Note:
1. At constant temperature the sum of PV terms is conserved when gases mix.
2. The final pressure lies between the two initial pressures.

Answer: (d) 5

Explanation:
1. Moles of \( O_2 = 8/32 = 0.25 \); moles of \( N_2 = 7/28 = 0.25 \).
2. Each gas contributes half of the total pressure, so the nitrogen alone exerts 5 atm.

Teacher's Note:
1. Partial pressure is proportional to the mole fraction.
2. Equal moles share the total pressure equally.

Answer: (d) \( \frac{3RT}{2E} \)

Explanation:
1. \( E = \frac32 kT = \frac32\frac{R}{N_A}T \).
2. \( N_A = \frac{3RT}{2E} \).

Teacher's Note:
1. Use \( k = R/N_A \).
2. Helium is monatomic, so only 3 translational degrees of freedom contribute.

Answer: (c) Low pressure and high temperature

Explanation:
1. A real gas behaves ideally when molecular volume and intermolecular forces are negligible.
2. This happens at low pressure (large separation) and high temperature (high kinetic energy).

Teacher's Note:
1. Ideal behaviour needs dilute and hot conditions.
2. Low temperature or high pressure brings out intermolecular forces.

Answer: (b) \( P/2 \)

Explanation:
1. \( \lambda = \frac{kT}{\sqrt2\pi d^2P} \), so \( \lambda\propto 1/P \) at constant T.
2. Doubling \( \lambda \) halves the pressure.

Teacher's Note:
1. Mean free path varies inversely with number density and hence with pressure at fixed T.
2. Higher temperature increases \( \lambda \) at fixed pressure.

Answer: (b) 3.2 Å

Explanation:
1. \( \lambda = \frac{1}{\sqrt2\pi d^2 n} \Rightarrow d^2 = \frac{1}{\sqrt2\pi\times2.7\times10^{25}\times0.8\times10^{-7}} \).
2. \( d^2 \approx 1.04\times10^{-19} \), so \( d \approx 3.2\times10^{-10} \) m = 3.2 Å.

Teacher's Note:
1. Molecular diameters are of the order of an angstrom.
2. Substitute the number density carefully; the power of ten decides the unit.

Answer: (d) 10375 J/kg K

Explanation:
1. \( C_P - C_V = 4150 \) and \( C_P = 1.4C_V \).
2. \( 0.4C_V = 4150 \), so \( C_V = 10375 \) J/kg K.

Teacher's Note:
1. Solve the two simultaneous relations.
2. The gas is hydrogen-like (\( r = R/M \approx 4157 \)).

Answer: (b) \( 1-\frac1\gamma \)

Explanation:
1. \( W/Q = \frac{R}{C_P} = \frac{C_P-C_V}{C_P} = 1-\frac{1}{\gamma} \).

Teacher's Note:
1. Work done at constant pressure is \( nR\Delta T \), heat is \( nC_P\Delta T \).
2. For a monoatomic gas the fraction is 2/5.

Answer: (b) 1400 J kg\(^{-1}\)K\(^{-1}\)

Explanation:
1. Molar mass \( M = 0.795\times22.4\times10^{-3} = 0.0178 \) kg/mol.
2. A polyatomic gas has \( C_V = 3R \) per mole, so specific heat \( = \frac{3\times8.31}{0.0178} \approx 1400 \) J kg\(^{-1}\)K\(^{-1}\).

Teacher's Note:
1. Convert the density at STP to molar mass.
2. Specific heat is molar heat capacity divided by molar mass.

Answer: (c) 19/13

Explanation:
1. \( C_V = \frac{2(5R/2)+1(3R/2)}{3} = \frac{6.5R}{3} \); \( C_P = C_V + R = \frac{9.5R}{3} \).
2. \( \gamma = \frac{9.5}{6.5} = \frac{19}{13} \).

Teacher's Note:
1. Add the heat capacities (mole weighted) first, then take the ratio.
2. The result lies between 7/5 and 5/3.

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

Explanation:
1. Same temperature gives the same average translational KE, so (1) is true.
2. Hydrogen is diatomic with extra rotational energy, so its average energy is greater: (3) is true.
3. The speeds differ because the masses differ.

Teacher's Note:
1. Translational energy depends only on T.
2. Total internal energy also counts rotational degrees of freedom.

Answer: (b) (2) and (3) only

Explanation:
1. Total moles \( = \frac{2P_0V_0}{RT_0} = \frac{PV_0}{RT_0}+\frac{PV_0}{2RT_0} = \frac{3PV_0}{2RT_0} \).
2. So \( P = \frac43P_0 \).
3. \( n = \frac{PV_0}{2RT_0} = \frac{2P_0V_0}{3RT_0} \).

Teacher's Note:
1. Moles are conserved when the gas flows between the containers.
2. Pressure equalises, but the amounts in the two vessels differ.

Answer: (c) Mean free path is directly proportional to the pressure of the gas

Explanation:
1. \( \lambda = \frac{1}{\sqrt2\pi d^2n} = \frac{kT}{\sqrt2\pi d^2P} \).
2. So \( \lambda\propto 1/P \), not P. Statement (c) is wrong.

Teacher's Note:
1. Mean free path is inversely proportional to pressure at fixed temperature.
2. It is proportional to T at fixed pressure.

Answer: (c) 2

Explanation:
1. The two gases contribute partial pressures P each in volume V at temperature T.
2. Total pressure \( = P+P = 2P \), so \( x = 2 \).

Teacher's Note:
1. Dalton's law: pressures of non-reacting gases add.
2. Equivalent to doubling the number of moles at the same V and T.

Answer: (b) 2

Explanation:
1. \( v_{rms}\propto\sqrt T \); \( T_1 = 200 \) K and \( T_2 = 800 \) K.
2. \( K = \sqrt{800/200} = 2 \).

Teacher's Note:
1. Pressure has no effect on the rms speed of an ideal gas at given T.
2. Convert Celsius to kelvin first.

Answer: (b) 2

Explanation:
1. Force \( = n\times\) change of momentum per ball \( = n(2mu) \).
2. Doubling u doubles the force, so \( K = 2 \).

Teacher's Note:
1. Force equals momentum transferred per second.
2. The change in momentum is linear in u, so the force is too (for fixed n).

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. \( \lambda = \frac{1}{\sqrt2\pi d^2n} \propto \frac1n \propto \frac1\rho \), so A is true.
2. R correctly defines the mean free path but does not explain the inverse variation.

Teacher's Note:
1. Mean free path is inversely proportional to number density.
2. A definition is not an explanation.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. \( v_{rms} = \sqrt{3kT/m} > v_{mp} = \sqrt{2kT/m} \), so A is true.
2. The distribution has a long tail of fast molecules, which raises the mean of the squares more than the peak position: R explains A.

Teacher's Note:
1. The Maxwell curve is asymmetric with a high-speed tail.
2. The tail pulls rms above the most probable speed.