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

Answers (1)

Root mean square velocity -

V_{rms}= \sqrt{\frac{3RT}{M}}

= \sqrt{\frac{3P}{\rho }}
 

- wherein

R = Universal gas constant

M = molar mass

P = pressure due to gas

\rho = density

V_{rms}\propto \sqrt{T}

V_{rms}\propto \frac{mean free path}{time between successive collision}

and mean free path = Y = \frac{kT}{\sqrt{2\pi \sigma ^{2}P}}

V_{rms}\propto \frac{Y}{b}

V_{rms}\propto \frac{T}{\sqrt{p}\times t} ......1

but V_{rms}\propto \sqrt{T} ..........2

From 1 and 2

\sqrt{T} \propto \frac{T}{\sqrt{P}\times t}

t\propto \frac{\sqrt{T}}{p}

\frac{t_{2}}{t_{1}} = \sqrt{\left ( \frac{T_{2}}{T_{1}} \right )\times \left ( \frac{P_{1}}{P_{2}} \right )} = \sqrt{\frac{500}{300}\times \frac{P_{1}}{2P_{1}}} = \sqrt{\frac{5}{6}}

t_{2} = \sqrt{\frac{5}{6}}t_{1}

t_{2} \approx 4\times 10^{-8}s

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