Questions
The average speed of molecules of oxygen gas is 450 m/s. If the mean free path is 1.1 × 10-7 m, then the average number of collisions per second is
a
9.1×107
b
4.1×108
c
4.1×109
d
4.1×1012
detailed solution
Correct option is C
vav=450 m/s;λ=1.1×10−7m.We know that λ¯=vrmscollision frequency=vav0.9213×collision frequency[Since vav=0.9213×vrms1.1×10−7=4500.9213×collision frequencyCollision frequency =4500.9213×1.1×10−7=450×1070.9213×1.1=4.44×109s−1≅4.1×109Talk to our academic expert!
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If the mean free path of atoms of an ideal gas at pressure P is doubled, then the pressure of gas will become
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