1.

If radius of O_(2) molecule =40A. T=27^(@)C annd P=1atm. Find the time of relaxation.

Answer»

`10^(-10)sec`
`10^(-12)sec`
`10^(-14)sec`
`10^(-8)sec`

Solution :`tau=(lamda)/(V_(RMS))=(1)/(sqrt(2)PIND^(2))(sqrt(m_(o)))/(sqrt(2RT))`
Now `n=(N)/(V)=(muN_(a))/(V)`
`PV=muRT`
`(mu)/(V)=(P)/(RT)`
so `n=(P)/(RT)xxN_(a)`
`tau=(sqrt(m_(o))RT)/(sqrt(2)mu.PN_(a)d^(2)sqrt(3RT))`
`tau=(sqrt(m_(o)RT))/(sqrt(6)xx3.14xx10^(5)xx6.02xx10^(23)xx16xx10^(-18))`
`=(sqrt(32xx3xx8.3xx10^(-1)))/(sqrt(6)xx3.14xx10^(5)xx6.02xx10^(23)xx16xx10^(-18))`
`=(sqrt(96xx0.83))/(sqrt(6)xx3.14xx6.02xx16xx10^(10))`
`=(4xxsqrt(0.83))/(3.14xx6.02xx16xx10^(10))`
`=(0.9)/(3.14xx6xx4xx10^(10))`
`=0.01xx10^(-10)`
`=10^(-12)` sec


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