Find the number of molecules in `1cm^(3)` of an ideal gas at `0^(@)C`and at a pressure of `10^(-5)mm` of mercury.

Explanation

`V= 1 cm^3, T=0^@C,`
` P= (10^-5) mm of Hg `
` n=(PV)/(RT)=(rhogh) xx (V/(RT)) `
` =(13.6 xx 980 xx 10^-6 xx 1)/ (8.31 xx 273)`
`=(5.874) xx (10^13) `
no of molecules`
`=N xx n`
`= 6.023 xx (10^23) xx 5.874 xx (10^13)`
`=(35.384 xx 10^10)`
` = (3.538 xx 10^11)`