Question

The molar specific heat at constant volume of gas mixture is $\frac{13R}{6}$. The gas mixture consists of

• A. $2moles$ of $O_2$ and $4moles$ of $H_2$
• B. $2moles$ of $O_2$ and $4moles$ of argon
• C. $2moles$ of argon and $4moles$ of $O_2$
• D. $2moles$ of $C{O}_2$ and $4moles$ of argon

Answer 1

Answer: (C)

$2moles$ of argon and $4moles$ of $O_2$

Molar specific heat of a mixture is given by:
$C_v=\frac{n_1{C}_{v_1}+n_2{C}_{v_2}}{n_1+n_2}$
For a monoatomic gas:
$C_v=\frac{3}{2}R$
For a diatomic gas:
$C_v=\frac{5}{2}R$
Plugging in all values with all the options we see that option C matches.
Hence option C is the correct answer.