`5.6` litre of an unknown gas at `NTP` requires 12.5 calorie to raise its temperature by `10^(@)C` at constant volume Calculate
(a) `C_(v)` of gas
(b) atomicity of gas .

Options

A. 7 cal , 1
B. 5 cal , 1
C. 7 cal , 2
D. 5 cal , 1

Explanation

We know , `n = (PV)/(RT) = (1 xx 5.6)/(0.0821 xx 273) = 0.25`
Also
`q = (n xx C_(v) xx ? T) = 0.25 xx C_(v) xx 10`
`therefore " " 12.5 = 0.25 xx C_(v) xx 10`
`therefore " " C_(v) = 5` calorie
Also `C_(p)= C_(v) + R = 5 + 2 = 7` calorie
So , `lambda = (C_(p))/(C_(v)) = (7)/(3) = 1.40`
Thus , gas is diatomic .