When 0.1mole of a gas absorbs 41.75 J of heat, the rise in temperature occursequal to 20^(@)C. The gas must be

When 0.1mole of a gas absorbs 41.75 J of heat, the rise in temperature

1.	When 0.1mole of a gas absorbs 41.75 J of heat, the rise in temperature occursequal to 20^(@)C. The gas must be
Answer» triatomic diatomic polyatomic monoatomic Solution :`C_(V) ` ( heat absorbedper degree rise per MOLE) `=( 41.75J)/( 0.1 mol xx 20^(@)) = 20.875 JK^(-1) mol^(-1)` `C_(p) = C_(v) + R = 20.875 + 8.314 JK^(-1) mol^(-1)` `= 29.189 JK^(-1) mol^(-1)` `(C_(p))/( C_(v)) = ( 29.189)/( 20.875) = 1.40`. Hence, the GAS is diatomic.