Question

An ideal gas is enclosed in a cylinder at pressure of 2 atm and temperature, 300 K. The mean time between two successive collisions is 6 × 10^-8 s. If the pressure is doubled and temperature is increased to 500 K, the mean time between two successive collisions will be close to
1. 4 × 10^-8 s
2. 3 × 10^-6 s
3. 2 × 10^-7 s
4. 0.5 × 10^-8 s

Answer 1

Correct Answer - Option 1 : 4 × 10^-8 s

Concept:

Mean time elapsed between two successive collisions:

Mean time elapsed between two successive, collisions is \(t = \frac{\lambda }{v}\)

where, λ = mean free path length and

v = mean speed of gas molecule

\(\therefore {\rm{\;\;t}} = \frac{{\left( {\frac{{{{\rm{k}}_{\rm{B}}}{\rm{t}}}}{{\sqrt 2 {\rm{\pi }}{{\rm{d}}^2}{\rm{p}}}}} \right)}}{{\sqrt {\frac{8}{{\rm{\pi }}}\cdot\frac{{{{\rm{k}}_{\rm{B}}}{\rm{T}}}}{{\rm{M}}}} }} \Rightarrow {\rm{t}} = \frac{{{\rm{C}}\sqrt {\rm{T}} }}{{\rm{p}}}\)

where,\({\rm{\;C}} = \frac{1}{{4{{\rm{d}}^2}}}\sqrt {\frac{{{{\rm{k}}_{\rm{B}}}{\rm{M}}}}{{\rm{\pi }}}} = a\) constant for a gas.

Calculation:

Since we know

\({\rm{t}} = \frac{{{\rm{C}}\sqrt {\rm{T}} }}{{\rm{p}}}\)

where,\({\rm{\;C}} = \frac{1}{{4{{\rm{d}}^2}}}\sqrt {\frac{{{{\rm{k}}_{\rm{B}}}{\rm{M}}}}{{\rm{\pi }}}} \)

\({\rm{So}},{\rm{\;\;}}\frac{{{{\rm{t}}_2}}}{{{{\rm{t}}_1}}} = \frac{{\sqrt {{{\rm{T}}_2}} }}{{\sqrt {{{\rm{T}}_1}} }}\cdot\left( {\frac{{{{\rm{p}}_1}}}{{{{\rm{p}}_2}}}} \right)\)      ----(2)

Here given,\({\rm{\;}}\frac{{{{\rm{p}}_1}}}{{{{\rm{p}}_2}}} = \frac{1}{2},\sqrt {\frac{{{{\rm{T}}_2}}}{{{{\rm{T}}_1}}}} = \sqrt {\frac{{500}}{{300}}} = \sqrt {\frac{5}{3}} \)

and t₁ = 6 × 10^-8 s

Substituting there values in (1), we get

\({{\rm{t}}_2} = 6 \times {10^{ - 8}} \times \sqrt {\frac{5}{3}} \times \frac{1}{2}\)

= 3.86 × 10^-8 s

= 4 × 10^-8 s