When 0.1 mol of a gas absorbs `41.75 J` of heat at constant volume, the rise in temperature occurs equal to `20^(@)C`. The gas must be

Options

A. monoatomic
B. diatomic
C. triatomic
D. polyatomic

Answer

B

Explanation

Atomicity of gas is deiced by the ratio `C_(P)//C_(V)`.
`q = nC_(V) Delta T`
Thus molar heat capacity at constant volume
`C_(V) = (q)/(n Delta T)`
`= (41.75 J)/((0.1 mol)(20 K))`
`20.88 J K^(-1) mol^(-1)`
For an ideal gas,
`C_(P) - C_(V) = R`
or `C_(P) = R + C_(V)`
`= (8.314) + (20.88)`
`= 29.19 J K^(-1) mol^(-1)`
Finally,
`(C_(P))/(C_(V)) = (29.19)/(20.88) = 1.40`
This implies than the gas is diatomic.