A gas mixture contains `n_1` moles of a monoatomic gas and `n_2` moles of gas of rigid diatomic molecules. Each molecule in monoatomic and diatomic gas has 3 and 5 degrees of freedom respectively. If the adiabatic exponent `(C_p)/(C_v)` for this gas mixture is 1.5, then the ratio `n_1/n_2` will be

Options

A. 1
B. 1.5
C. 2
D. 2.5

Answer

A

Explanation

For a gas mixture
`C_v = (n_1 C_(v_1) + n_2C_(v_2) )/(n_1 + n_2) " and " C_p = (n_1 C_(p_1) + n_2 C_(p_2) )/(n_1 + n_2)`
For monoatomic gas ,
`n = n_1 , C_(v_1) = 3/2 R, C_(p_1) = 5/2 R`
For diatomic gas ,
` n = n_2 , C_(v_2) = 5/2 " and C_(p_2) = 7/2 R`
Now given for gas mixture
`(C_p)/(C_v) = 1/5`
so , ` 1.5 (n_1/n_2 (5/2) + 7/2)/(n_1/n_2 (3/2) + 5/2) = 9/2 (n_1/n_2) + 15/2 = (n_1/n_2) 5 + 7`
` rArr n_1/n_2 = 1`