Question

Calculate the specific heat capacity $C_v$ of a gaseous mixture consisting of $v_1$ moles of a gas of adiabatic exponent ${\gamma }_1$ and $v_2$ moles of another gas of adiabatic exponent ${\gamma }_2$.

• A. $C_v=\frac{R}{\gamma -1}=\frac{R}{v_1+v_2}(\frac{v_1}{{\gamma }_1+1}+\frac{v_2}{{\gamma }_2-1})$
• B. $C_v=\frac{R}{\gamma -1}=\frac{R}{v_1-v_2}(\frac{v_1}{{\gamma }_1-1}+\frac{v_2}{{\gamma }_2-1})$
• C. $C_v=\frac{R}{\gamma -1}=\frac{R}{v_1+v_2}(\frac{v_1}{{\gamma }_1+1}+\frac{v_2}{{\gamma }_2-1})$
• D. $C_v=\frac{R}{\gamma -1}=\frac{R}{v_1+v_2}(\frac{v_1}{{\gamma }_1-1}+\frac{v_2}{{\gamma }_2-1})$

Answer 1

The correct option is D $C_v=\frac{R}{\gamma -1}=\frac{R}{v_1+v_2}(\frac{v_1}{{\gamma }_1-1}+\frac{v_2}{{\gamma }_2-1})$

The internal energy of an ideal gas of mass $m$ is given by
$U=\frac{PV}{\gamma -1}$
Internal energy is an extensive property.
$\therefore U_{mix}=U_1+U_2⟹\frac{P_{mix}}{\gamma -1}=\frac{P_1}{{\gamma }_1-1}+\frac{P_2}{{\gamma }_2-1}$
(as volume is same for all)
From the formula $PV=nRT$
$P_{mix}V=(v_1+v_2)RT$; $P_{1}V=v_{1}RT$; $P_{2}V=v_{2}RT$
$\therefore P_1=\frac{v_1}{v_1+v_2}{P}_{mix}$ and $P_2=\frac{v_2}{v_1+v_2}{P}_{mix}$
$\therefore \frac{1}{\gamma -1}=\frac{1}{v_1+v_2}(\frac{v_1}{{\gamma }_1-1}+\frac{v_2}{{\gamma }_2-1})$
$C_v=\frac{R}{\gamma -1}=\frac{R}{v_1+v_2}(\frac{v_1}{{\gamma }_1-1}+\frac{v_2}{{\gamma }_2-1})$