Two moles of a monoatomic gas is mixed with six moles of a diatomic gas. The molar specific heat of the mixture at constant volume is :
- 32R
- 74R
- 52R
- 94R
Answer
(4) 94R
Solution
Monoatomic: CV1= 32R; diatomic: CV2= 52R
internal energies add: CV= n1CV1+n2CV2n1+n2
= 2 32R+6 52R2+6= 3R+15R8
CV= 94R
Answer: (4)