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  • In ideal gas is closed in a cylinder at a pressure of  1 atm and temperature 200 K. The mean time between two consecutive collision is  <img alt="6 \times 10^{-6} \mathrm{~S}" src="https:

In ideal gas is closed in a cylinder at a pressure of  1 atm and temperature 200 K. The mean time between two consecutive collision is  6 \times 10^{-6} \mathrm{~S}. If the pressure is doubled and temperature is increased to 600k. The time between two consecutive collision will be close to


 

Option: 1

4 \times 10^{-8}


Option: 2

7.34 \times 10^{-6}


Option: 3

0.5 \times 10^{-8}


Option: 4

2 \times 10^{-7}


Answers (1)

best_answer

Root mean square velocity

V_{\text {max }}=\sqrt{\frac{3 R T}{M}}=\sqrt{\frac{3 P}{\rho}}

R is the universal gas constant
M is  the molar mass
P is the pressure due to gas
\rho is the density

V_{r m s} \propto \sqrt{T}V_{r m s} \propto mean\,\, free \,\,path/ time\,\, between \,\,successive \,\,collisions

mean free path

Y=\frac{K T}{\sqrt{2 \pi \sigma^2} \rho}, \quad V_{r m s} \propto Y / b \text { and } V_{r m s} \propto \frac{T}{\sqrt{P}} \times t\,\,(i) \\

                                                               but\,\, V_{r m s} \propto {\sqrt{T}}\,...(ii) \\

\frac{t_2}{t_1}=\sqrt{\frac{T_2}{T_1} \times \frac{P_1}{P_2}}=\sqrt{\frac{600}{200} \times \frac{P_1}{2 P_1}}=\sqrt{\frac{3}{2}} \\

t_2=\sqrt{\frac{3}{2}} \times t_1=1.22 \times 6 \times 10^{-6}=7.34 \times 10^{-6}

 

Posted by

Shailly goel

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