The molar specific heat at constant volume of gas mixture is `(13R)/(6)`. The gas mixture consists of

Options

A. 2 moles of `O_2` and 4 moles of `H_2`
B. 2 moles of `O_2` and 4 moles of argon
C. 2 moles of argon and 4 moles of `O_2`
D. 2 moles of `CO_2` and 4 moles of argon

Answer

C

Explanation

`C_V=(n_1C_(V_1)+n_2C_(V_2))/(n_1+n_2)`
In option (c),
`C_V=((2)(3/2R)+(4)(5/2R))/(6)`
`=(13R)/(6)`