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  • If the successive collision time interval of an ideal gas at a pressure of 2 atm and Temperature is 300

If the successive collision time interval of an ideal gas at a pressure of 2 atm and Temperature is 300 K is 6.0×10-8. What will be the new vision time if the pressure is doubled and tempercutue is set to 500k.

Option: 1

4 \times 10^{-8} \mathrm{~s} \\


Option: 2

3 \times 10^{-6} \mathrm{~S} \\


Option: 3

0.5 \times 10^{-8} \mathrm{~S} \\


Option: 4

2 \times 10^{-7} \mathrm{~S} \\


Answers (1)

best_answer

V_{\text {rms }}=\sqrt{\frac{3 R T}{m}}=\sqrt{\frac{3 p}{\rho}}V_{\text {rms }}=\sqrt{\frac{3 R T}{m}}=\sqrt{\frac{3 p}{\rho}}

V_{ rms} \propto  mean free path / time btw collisions mean free path 
V=\frac{K T}{\sqrt{2 \pi \sigma^2 \rho}}
{ V _{rms }} \propto \frac{T}{\sqrt{p}} \times t \rightarrow (1)
{ V _{rms }}\propto \sqrt{T} \rightarrow (2)
t \propto \sqrt{\frac{T}{p}}
\frac{T_2}{T_1}=\sqrt{\frac{T_2}{T_1} \times \frac{P_1}{P_2}} \\
\frac{T_2}{T_1}=\sqrt{\frac{500}{300} \times \frac{P_1}{2 P_1}}=\sqrt{\frac{5}{6}} \\
T_2=\sqrt{\frac{5}{6}+1 }
T_2=4 \times 10^{-8} \mathrm{~S} \\

Posted by

vishal kumar

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