Calculate the number of molecules is 2 c.c of the perfect gas at `27^(@)C` at a pressure of 10 cm. of mercury. Mean kinetic energy of each molecule at `27^(@)C = 7 xx 10^(-4) erg` and acceleration due to gravity `= 980 cm s^(-2)`.

Answer

`5.71 xx 10^(8)`

Explanation

K.E of the given gas at `27^(@)C`
`=1/2 MC^(2) = 3/2 PV`
`=3/2 xx (10 xx 13.6 xx 980) xx 2 ergs`
`:.` No. Of molecules in the given volume
`=(K.E. "of given gas")/(K.E."per molecule of a gas")`
`(3/2 xx 10 xx 13.6 xx 980 xx2)/(7 xx 10^(-4)`
`=5.71 xx 10^(8)`