One mole of a monoatomic ideal gas is mixed with one mole of a diatomic ideal gas . The molar specific heat of the mixture at constant volume is

Options

A. 1.5R
B. 2R
C. 2.5R
D. 3R

Answer

B

Explanation

For a monoatomic gas, there are 3 degrees of freedom and `(C_(V))_(1)=3/2R`
For a diatomic ideal gas, there are 5 degrees of freedom and `(C_(V))_(2)=5/2R`
`therefore(C_(V))_("mixture")=(n_(1)C_(V_(1))+n_(2)C_(V_(2)))/(n_(1)+n_(2))`
`=((1xx3/2R)+(1xx5/2R))/(1+1)`
`=(4R)/2=2R`