Question

Two moles of an ideal gas with $\frac{C_P}{C_V}=\frac{5}{3}$ are mixed with $3$ moles of another ideal gas with $\frac{C_P}{C_V}=\frac{4}{3}$. The value of $\frac{C_P}{C_V}$ for the mixture is :

• A. 1.50
• B. 1.45
• C. 1.47
• D. 1.42

Answer 1

Answer: (D)

${\gamma }_{mixture}=\frac{n_1{C}_{P_1}+n_2{C}_{P_2}}{n_1{C}_{V_1}+n_2{C}_{V_2}}=\frac{n_1\frac{{\gamma }_{1}R}{{\gamma }_1-1}+n_2\frac{{\gamma }_{2}R}{{\gamma }_2-1}}{\frac{n_{1}R}{{\gamma }_1-1}+\frac{n_{2}R}{{\gamma }_2-1}}$
on rearranging we get,
$\frac{n_1+n_2}{{\gamma }_{mix}-1}=\frac{n_1}{{\gamma }_1-1}+\frac{n_2}{{\gamma }_2-1}$
$\frac{5}{{\gamma }_{mic}-1}=\frac{3}{1/3}+\frac{3}{2/3}$
$\frac{5}{{\gamma }_{mic}-1}=9+3=12$
$\Rightarrow {\gamma }_{mixture}=\frac{17}{12}=1+\frac{5}{12}$
${\gamma }_{mix}=1.42$