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CBSE 11th Standard Physics Subject Kinetic Theory Ncert Exemplar 1 Mark Questions With Solution 2021

Published on: 26/05/2021

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Questions:

Generic 2

5 x 2 = 10

Calculate the ratio of the mean free paths of the molecules of two gases having molecular diameters \(1\overset { 0 }{ A } \) and \(2\overset { 0 }{ A } \). The gases may be considered under identical conditions of temperature, pressure and volume.

Two molecules of a gas have speeds of 9 \(\times\)10⁶ ms^-1 and 1 \(\times\) 10⁶ ms^-1, respectively. What is the root mean square speed of these molecules ?

The volume of a given mass of a gas at 27⁰C, 1 atm is 100 cc. What will be its volume at 327⁰C?

When air is pumped into a cycle tyre the volume and pressure of the air in the tyre both are increased." What about Boyle's law in this case?

Calculate the number of atoms in 39.4 g gold. Molar mass of gold is 197g mol^-1.

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Answers Key:

Generic 2

5 x 2 = 10

As, we know, mean free path,
λ ∝ 1/d²
Given, d₁= 1A^o and d₂ = 2A^o⇒ λ₁ : λ₂=4:1

\(v_{r m s}=\sqrt{\frac{v_1^2+v_2^2}{2}} \)
\(=\sqrt{\frac{\left(9 \times 10^6\right)^2+\left(1 \times 10^{\circ}\right)^2}{2}}=\sqrt{\frac{(81+1) \times 10^{12}}{2}}\)
\(=\sqrt{41} \times 10^6 \mathrm{~ms}^{-1}\)

Keeping p constant, we have
V²/2 = \(\frac{V_1T_2}{T_1}\)
= \(\frac{100×600}{300}\)
= 200cc

When air is pumped, more molecules are pumped and Boyle's law is stated for situation where number of molecules constant.

Molar mass of gold is 197 g mol^-1 , the number of atoms
= 6 x 10²³
∴ Number of atoms in 39.4 g
= \(\frac{6.0×10^{23} ×39.4}{197}\)
= 1.2 x 10²³