Question

Two moles of helium $\left(He\right)$ are mixed with four moles of hydrogen $\left(H_2\right)$. The value of $\gamma$ of the mixture is

• A. $1.72$
• B. $1.46$
• C. $1.67$
• D. $2$

Answer 1

The correct option is B $1.46$

Given,
No. of moles of helium $\left[He\right],\left(n_1\right)=2$
No. of moles of hydrogen $\left[H_2\right],\left(n_2\right)=4$
Since, hydrogen is diatomic and helium is monoatomic
${\left(C_V\right)}_{H_2}=\frac{5R}{2};{\left(C_V\right)}_{He}=\frac{3R}{2}$
$\therefore {\left(C_P\right)}_{H_2}=\frac{7R}{2};{\left(C_P\right)}_{He}=\frac{5R}{2}$
We know,
${\gamma }_{\text{mixture}}=\frac{n_1{C}_{P_1}+n_2{C}_{P_2}}{n_1{C}_{V_1}+n_2{C}_{V_2}}$
$=\frac{2\left(\frac{5R}{2}\right)+4\left(\frac{7R}{2}\right)}{2\left(\frac{3R}{2}\right)+4\left(\frac{5R}{2}\right)}=\frac{19}{13}=1.46$