JEE Physics Kinetic Theory MCQs Set 06

Answer: (b) Statement I is true but Statement II is false

Explanation:
1. Temperatures: -73 °C = 200 K and 527 °C = 800 K, so \( \frac{v_2}{v_1}=\sqrt{\frac{800}{200}}=2 \): Statement I is true.
2. For an ideal gas \( PV=\frac23E_{trans} \), not equal to E: Statement II is false.

Teacher's Note:
1. v_rms is proportional to √T with T in kelvin.
2. Total translational kinetic energy is (3/2)PV.

Answer: (c) \( \frac{25}{21} \)

Explanation:
1. Monoatomic: \( \gamma_1=\frac53 \). Rigid diatomic: \( \gamma_2=\frac75 \).
2. \( \frac{\gamma_1}{\gamma_2}=\frac53\times\frac57=\frac{25}{21} \).

Teacher's Note:
1. gamma = 1 + 2/f, where f is the number of degrees of freedom.
2. A rigid diatomic molecule has 5 degrees of freedom.

Answer: (d) \( \frac72R \)

Explanation:
1. A diatomic molecule has 3 translational and 2 rotational degrees of freedom, each giving \( \frac12R \).
2. One vibrational mode contributes two quadratic terms (kinetic and potential), i.e. \( 2\times\frac12R=R \).
3. \( C_v=\frac32R+R+R=\frac72R \).

Teacher's Note:
1. A vibrational mode counts for two degrees of freedom in the energy.
2. Each quadratic energy term gives R/2 per mole.

Answer: (b) \( \gamma\propto T^0 \)

Explanation:
1. For an ideal gas \( C_p \) and \( C_v \) are constants, depending only on the degrees of freedom.
2. So \( \gamma=1+\frac2f \) does not depend on temperature: \( \gamma\propto T^0 \).

Teacher's Note:
1. Specific heats of an ideal gas are constant in the temperature range where degrees of freedom are fixed.
2. The ratio is therefore constant.

Answer: (a) -273 °C

Explanation:
1. At constant volume \( P\propto T_{kelvin} \) for a low-density (ideal) gas.
2. The pressure extrapolates to zero at absolute zero, so every line meets the axis at \( -273^\circ\text{C} \).

Teacher's Note:
1. This extrapolation is how absolute zero was first estimated.
2. Real gases liquefy before reaching it, so the lines are extended by dotted extrapolation.

Answer: (c) Proportional to the square root of the temperature

Explanation:
1. \( v_{rms}=\sqrt{\frac{3RT}{M}}\propto\sqrt T \).

Teacher's Note:
1. The kinetic energy of the molecules is proportional to T and to v².
2. At fixed temperature lighter gases move faster.

Answer: (c) 278 kPa

Explanation:
1. Constant volume: \( \frac{P_2}{P_1}=\frac{T_2}{T_1}=\frac{309}{300} \).
2. \( P_2=270\times\frac{309}{300}=278.1\ \text{kPa}\approx278\ \text{kPa} \).

Teacher's Note:
1. Temperatures must be in kelvin: 27 °C = 300 K and 36 °C = 309 K.
2. The volume of the tyre is taken as constant.

Answer: (b) 28

Explanation:
1. \( \frac{v_{rms}}{v_{avg}}=\frac{\sqrt{3RT/M}}{\sqrt{8RT/\pi M}}=\sqrt{\frac{3\pi}{8}}=\sqrt{\frac{66}{56}}=\sqrt{\frac{33}{28}} \).
2. \( \frac{\alpha+5}{\alpha}=\frac{33}{28}\Rightarrow28\alpha+140=33\alpha\Rightarrow\alpha=28 \).

Teacher's Note:
1. Use v_rms = √(3RT/M) and v_avg = √(8RT/πM).
2. The ratio does not depend on the gas or the temperature.

Answer: (b) 1 : 1

Explanation:
1. Average kinetic energy per molecule \( =\frac32kT \) depends only on the temperature.
2. Both gases are at the same temperature, so the ratio is 1 : 1.

Teacher's Note:
1. The ratio by mass is a distractor.
2. Per mole or per molecule, translational KE is the same for all ideal gases at the same T.

Answer: (a) Proportional to absolute temperature

Explanation:
1. \( \overline{KE}=\frac32kT \), which is proportional to the absolute temperature and independent of the nature of the gas.

Teacher's Note:
1. Pressure and volume change the density of the gas but not the energy per molecule.
2. This is the kinetic interpretation of temperature.