Question

Calculate the value of $\gamma =C_p/C_v$ for a gaseous mixture consisting of $v_1=2.0$ moles of oxygen and $v_2=3.0$ moles of carbon dioxide. The gases are assumed to be ideal.

• A. $1.33$
• B. $1.45$
• C. $2.1$
• D. $1.7$

Answer 1

Answer: (A)

$1.33$

${\nu }_1=2$ $O_2({\gamma }_1=1.4)$
${\nu }_2=3$ $C{O}_2({\gamma }_2=1.3)$
Let us assume adiabatic exponent of $O_2$ and $C{O}_2$ are ${\gamma }_1$ and ${\gamma }_2$
Internal energy of system:
$U={\nu }_1\frac{R}{{\gamma }_1-1}T+{\nu }_2\frac{R}{{\gamma }_2-1}T=({\nu }_1+{\nu }_2)\frac{RT}{r-1}$

or, $\frac{{\nu }_1}{{\gamma }_1-1}+\frac{{\nu }_2}{{\gamma }_2-1}=\frac{{\mu }_1+{\nu }_2}{r-1}$

or, $\frac{{\nu }_1({\gamma }_2-1)+{\nu }_2({\gamma }_1-1)}{({\gamma }_1-1)({\gamma }_2-1)}=\frac{{\nu }_1+{\nu }_2}{(r-1)}$

or, $r=\frac{({\gamma }_1-1)({\gamma }_2-1)({\nu }_1+{\nu }_2)}{{\nu }_1({\gamma }_2-1)+{\nu }_2({\gamma }_1-1)}+1$

or, $r=\frac{{\nu }_1{\gamma }_1({\gamma }_2-1)+{\nu }_2{\gamma }_2({\gamma }_1-1)}{{\nu }_1({\gamma }_2-1)+{\nu }_2({\gamma }_1-1)}=1.33=\gamma$