Question

$n_1$ and $n_2$ moles of two ideal gases of the thermodynamics constant ${\gamma }_1$ and ${\gamma }_2$ respectively are mixed. $C_p/C_v$ for the mixture is

• A. $\frac{{\gamma }_1+{\gamma }_2}{2}$
• B. $\frac{n_1{\gamma }_1+n_2{\gamma }_2}{n_1+n_2}$
• C. $\frac{n_1{\gamma }_2+n_2{\gamma }_1}{n_1+n_2}$
• D. $\frac{n_1{\gamma }_1({\gamma }_2+1)+n_2{\gamma }_2({\gamma }_1-1)}{n_1({\gamma }_1-1)+n_2({\gamma }_1-1}$

Answer 1

The correct option is D $\frac{n_1{\gamma }_1({\gamma }_2+1)+n_2{\gamma }_2({\gamma }_1-1)}{n_1({\gamma }_1-1)+n_2({\gamma }_1-1}$

Given : $n_1$ and $n_2$ moles of two ideal gases of the thermodynamics constant ${\gamma }_1$ and ${\gamma }_2$ respectively are mixed. Cp/Cv for the mixture is

Solution :

We have a formula for this as $\frac{n_1+n_2}{\gamma -1}$ = $\frac{n_1}{{\gamma }_1-1}+\frac{n_2}{{\gamma }_2-1}$

One simplifying this formula we get

$\gamma =\frac{n_1{\gamma }_1({\gamma }_2+1)+n_2{\gamma }_2({\gamma }_1-1)}{n_1({\gamma }_1-1)+n_2({\gamma }_1-1)}$

The correct Opt = D