1.

Calculate the values of `sigma`, `l` (mean free path),`Z_(1)` and `Z_(11)` for oxygen at `300 K` at a pressure of `1 atm`. Given `b=3.183xx10^(-2) dm^(3)mol^(-1)`.

Answer» Number of molecules per unit volume
`N^(*)=(P)/(kT)`, where `k=` Boltzmann constant
`:. N^(*)=(1 atm)/((1.38xx10^(-23)JK^(-1))(300K))`
`=0.241xx10^(22)m^(-3)`
`=2.41xx10^(22)m^(-3)`
The van der Waals constant is
`b=4N_(A)((4)/(3)pir^(3))`
`:. r=((3b)/(16 pi N_(A)))^(1//3)`
`=((3xx3.183xx10^(-2)dm^(3) mol^(-1))/(16xx3.14xx6.023xx10^(23)mol^(-1)))^(1//3)`
`=1.467xx10^(-9)dm`
Therefore,
`sigma` (collision diameter)`=2r`
`=2xx1.467xx10^(-9) dm`
`=2.934xx10^(-10)m`
Average speed
`u_(av)=swrt((8RT)/(pi M))`
`={(8(8.314JK^(-1)mol^(-1))(300 K))/((3.14)(0.032 kg mol^(-1)))}^(1//2)`
`=445.6 m s^(-1)`
Mean free path
`l=(1)/(sqrt(2)pi sigma^(2)N^(*))`
`=(1)/((1.414)(3.14)(2.934xx10^(-10)m)^(2)(241xx10^(23)m^(-3)))`
`=3.18xx10^(-7)m`
`Z_(1)=sqrt(2)pi sigma^(2)u_(av)N^(*)`
`=(1.414)(3.14)(2.934xx10^(-10)m)^(2)(445.6 m s^(-1))xx(241.0xx10^(23)m^(-3))`
`=1.397xx10^(9)s^(-1)`
`Z_(11)=(1)/(2)Z_(1)N^(*)`
`=(1)/(2)(1.397xx10^(9)s^(-1))(241.0xx10^(23)m^(-3))`
`=1.68xx10^(34)m^(-3)s^(-1)`


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