The specific heats of a gas are 0.125 and `0.075` cal/g Calculate the atomicity and molecular weight of gas Identify the gas .

Explanation

`C_(P) = 0.125 cal//g`
`C_(V) = 0.075 cal//g`
`:' C_(P) -C_(v) = (R )/(M)`
`:. 0.125 -0.075 = (2)/(M)`
`:. M =40`
Also `(C_(p))/(C_(v)) =gamma = (0.125)/(0.075) =1.66`
Thus gas is monoatomic with mol wt `=40 `and so Argon .