Heat and Thermodynamics 4 Question 9

10. An ideal gas is enclosed in a cylinder at pressure of 2atm and temperature, 300K. The mean time between two successive collisions is 6×108s. If the pressure is doubled and temperature is increased to 500K, the mean time between two successive collisions will be close to (2019 Main, 12 Jan II)

(a) 4×108s

(b) 3×106s

(c) 2×107s

(d) 0.5×108s

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Solution:

  1. Mean time elapsed between two successive, collisions is t=λv

where, λ= mean free path length and

v= mean speed of gas molecule 

t=(kBt2πd2p)8πkBTMt=CTp

where, C=14d2kBMπ= a constant for a gas.

So, t2t1=T2T1(p1p2)

Here given, p1p2=12,T2T1=500300=53

and

t1=6×108s

Substituting there values in (i), we get

t2=6×108×53×12=3.86×108s4×108s



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