JEE Physics Kinetic Theory MCQs Set 16

Answer: (b) \( T_A\lt T_B\lt T_C \)

Explanation:
1. \( PV=nRT\Rightarrow\dfrac1V=\dfrac{P}{nRT} \), so the slope of a 1/V against P line is \( \dfrac1{nRT} \).
2. The largest slope means the smallest temperature, so \( T_A\lt T_B\lt T_C \).

Teacher's Note:
1. Slope of this graph is inversely proportional to temperature. The steepest line is the coolest gas.

Answer: (a) 600 K

Explanation:
1. \( P=\dfrac k{T^2} \) and \( PV=nRT \) give \( V\propto T^3 \).
2. \( \dfrac{T_2}{T_1}=\left(\dfrac{8V}{V}\right)^{1/3}=2 \), so \( T_2=600 \) K.

Teacher's Note:
1. Eliminate pressure between the two relations to link V and T directly. The cube root of 8 is 2.

Answer: (d) \( \sqrt{\dfrac{15}{2}}\ \text{km/s} \)

Explanation:
1. \( v_{rms}=\sqrt{\dfrac{1^2+2^2+3^2+4^2}{4}}=\sqrt{\dfrac{30}{4}}=\sqrt{\dfrac{15}{2}}\ \text{km/s} \).

Teacher's Note:
1. Square the speeds, average them and then take the root. The mean speed 2.5 km/s is smaller than the rms value.

Answer: (b) \( \dfrac27 \)

Explanation:
1. At very high temperature a diatomic molecule has translation \( \tfrac32kT \), rotation \( kT \) and vibration \( kT \) (kinetic plus potential).
2. Total = \( \tfrac72kT \); rotational fraction = \( \dfrac{1}{7/2}=\dfrac27 \).

Teacher's Note:
1. At high temperature vibrational modes also contribute. Each vibrational mode carries kT in total.

Answer: (b) 900 m/s

Explanation:
1. \( v_{rms}=\sqrt{3RT/M} \) depends only on temperature and molar mass.
2. Both are unchanged, so the speed remains 900 m/s.

Teacher's Note:
1. Pressure does not appear in the rms speed formula for a given temperature. Changing pressure at constant T changes the density, not the speed.

Answer: (c) \( m/4 \)

Explanation:
1. In equilibrium the pressure is the same in both flasks. Mass \( =\dfrac{PVM}{RT}\propto\dfrac VT \).
2. \( \dfrac{m_2}{m_1}=\dfrac{V_2/T_2}{V_1/T_1}=\dfrac{1/200}{2/100}=\dfrac14 \).

Teacher's Note:
1. Use the ideal gas law for each flask at equal pressure. Mass is proportional to volume over temperature.

Answer: (b) \( \dfrac{3P_0V_0}{4} \)

Explanation:
1. At \( V=V_0 \): \( P=\dfrac{P_0}{2} \).
2. Translational energy of one mole = \( \tfrac32RT=\tfrac32PV=\tfrac32\cdot\dfrac{P_0}{2}V_0=\dfrac{3P_0V_0}{4} \).

Teacher's Note:
1. Translational kinetic energy is three halves of PV for any ideal gas. The diatomic nature affects only the rotational part.

Answer: (c) C

Explanation:
1. On an isotherm \( PV \) is constant. Initial \( PV=3\times4=12 \).
2. A: 35, B: 24, C: 12, D: 18. Only C matches.

Teacher's Note:
1. An isothermal process keeps the product of pressure and volume constant. Compute the product for each state.

Answer: (a) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1

Explanation:
1. Both statements are true. Ideal gas behaviour needs negligible intermolecular forces and negligible molecular volume.
2. At low pressure and high temperature the molecules are far apart and move fast, so these conditions hold. Statement-2 therefore explains Statement-1.

Teacher's Note:
1. Real gases follow the ideal gas law when the molecules rarely interact. Low density and high kinetic energy achieve that.

Answer: (a) A corresponds to He, B to H2 and C to O2

Explanation:
1. \( PV=\dfrac mMRT \), so the slope is \( \dfrac{mR}{M} \) and is inversely proportional to the molar mass for equal masses.
2. Molar masses: H2 = 2, He = 4, O2 = 32. Slope order H2 > He > O2, i.e. B = H2, A = He, C = O2.

Teacher's Note:
1. For equal masses, the lightest gas has the most moles and hence the steepest PV against T line.

Answer: (c) 93.6 nm

Explanation:
1. \( \lambda=\dfrac{kT}{\sqrt2\pi d^2P}=\dfrac{1.38\times10^{-23}\times293}{1.414\times3.142\times(3.10\times10^{-10})^2\times1.013\times10^5} \).
2. \( \lambda\approx9.4\times10^{-8}\ \text{m}\approx93.6\ \text{nm} \).

Teacher's Note:
1. Use the absolute temperature, 293 K. Mean free path decreases with pressure and with the square of the atomic diameter.