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. triatomic
B. diatomic
C. polyatomic
D. monoatomic

Answer

B

Explanation

`C_(V)` (heat absorbed per deg.rise per mole)
`=(41.75 J)/(0.1 mol xx20^(@))=20.875 JK^(-1)mol^(-1)`
`C_(P)=C_(V)+R=20.875+8.314JK^(-1) mol^(-1)`
`=29.189 JK^(-1)mol^(-1)`
`(C_(P))/(C_(V))=(29.189)/(20.875)=1.40`
`:.` The gas is monoatomic