JEE Physics Kinetic Theory of Gases MCQs Set 07

Answer: (d) 0.13 kg/m\(^3\)

Explanation:
1. 4 g of H\(_2\) is 2 mol and 8 g of He is 2 mol, so the total is 4 mol and the total mass is 12 g.
2. Volume at NTP \(= 4 \times 22.4 = 89.6\) L \(= 0.0896\) m\(^3\).
3. Density \(= 0.012/0.0896 = 0.134\) kg/m\(^3\).

Teacher's Note:
1. Use moles to find the volume at NTP, since 1 mol occupies 22.4 L.

Answer: (b) decreased

Explanation:
1. The line is \(T = T_0 + kV\) with \(T_0 > 0\).
2. \(P = nRT/V = nR(T_0/V + k)\).
3. As \(V\) increases, \(T_0/V\) falls, so the pressure decreases.

Teacher's Note:
1. Only a line through the origin on a T-V graph means constant pressure. A positive intercept means that the pressure falls.

Answer: (a) A straight line through the origin

Explanation:
1. For \(n\) moles, \(PV = nRT\).
2. This is a straight line through the origin with slope \(nR\).

Teacher's Note:
1. A constant product \(PV\) is a feature of an isothermal process, not of the PV versus T relation.

Answer: (c) \(192^\circ\)

Explanation:
1. In equilibrium both gases have the same pressure and temperature, so the volumes are proportional to the number of moles.
2. \(n_1 : n_2 = (m/32) : (m/28) = 7 : 8\), so \(V_2/V_1 = 8/7\).
3. The angle for gas 2 is \(\alpha = 360^\circ \times 8/15 = 192^\circ\).

Teacher's Note:
1. Equal pressure and temperature mean volume is proportional to moles. Lighter molar mass means more moles for the same mass.

Answer: (a) The H\(_2\) line is steepest, He is in between and CO\(_2\) has the least slope.

Explanation:
1. \(PV = (m/M)RT\), so the slope is \(mR/M\).
2. For equal masses the slope is inversely proportional to the molar mass.
3. \(M\) is 2 for H\(_2\), 4 for He and 44 for CO\(_2\), so H\(_2\) has the steepest line and CO\(_2\) the least.

Teacher's Note:
1. More moles per gram gives a larger slope. Hydrogen has the most moles per gram.

Answer: (c) 2

Explanation:
1. On an isotherm \(PV = nRT\) is constant.
2. For \(T_1\), \(PV = 4 \times 10^5\) and for \(T_2\), \(PV = 1 \times 10^5\), so \(T_1/T_2 = 4\).
3. \(v_{rms} \propto \sqrt{T}\), so the speed ratio is \(\sqrt{4} = 2\).

Teacher's Note:
1. First get the temperature ratio from \(PV\) and then take the square root.

Answer: (c) the average rotational kinetic energy of a molecule is \(mv^2/4\)

Explanation:
1. \(v^2 = 3kT/m\), so \(kT = mv^2/3\).
2. A diatomic molecule has two rotational degrees of freedom, so the average rotational energy is \(2 \times \tfrac{1}{2}kT = kT = mv^2/3\), not \(mv^2/4\).
3. The total average energy is \(\tfrac{5}{2}kT = 5mv^2/6\), so statement d is true.
4. The Maxwell distribution has no upper limit, so a is true.

Teacher's Note:
1. Count degrees of freedom: 3 translational and 2 rotational for a rigid diatomic molecule.

Answer: (c) number of molecules in the gas

Explanation:
1. \(pV = NkT\) where \(N\) is the number of molecules and \(k\) is Boltzmann's constant.
2. So \(pV/kT = N\).

Teacher's Note:
1. With \(R\) in place of \(k\), \(pV/RT\) would instead be the number of moles.