Question

If one mole of a monoatomic gas ($\gamma$ = 5/ 3) is mixed with one mole of a diatomic gas ($\gamma$ = 7 / 5), the value of $\gamma$ for the mixture is

• A. 1.40
• B. 1.50
• C. 1.53
• D. 3.07

Answer 1

Answer: (B)

${\gamma }_1=\frac{5}{3}$ means gas is monoatomic or $C_v{}_1=\frac{3}{2}R$
${\gamma }_2=\frac{7}{5}$ means gas is diatomic or $C_v{}_2=\frac{5}{2}R$
C${}_v$ (of the mixture)
$=\frac{n_1{C}_v{}_1+n_2{C}_v{}_2}{n_1+n_2}=\frac{(1)\left(\frac{3}{2}R\right)+(1)\left(\frac{5}{2}R\right)}{1+1}=2R$
$C_p$ (of the mixture) $=C_v+R=3R$
$\therefore {\gamma }_{mixture}=\frac{C_p}{C_v}=\frac{3R}{2R}=1.5$
$\therefore$ Correct option is (b).