Question

2 mole of an ideal monoatomic gas is mixed with 3 mole of an ideal diatomic gas at room temperature then $\gamma$ for the mixture is____

Answer 1

Total internal energy of gaseous mixture

$U = U _{1}+ U _{2}=\left( E _{ t }\right)_{1}+\left( E _{ t }+ E _{ t }\right)_{2}$

$=2\left(3 \times \frac{1}{2} RT \right)+3\left(3 \times \frac{1}{2} RT +2 \times \frac{1}{2} RT \right)=\frac{21}{2} RT$

$Now , C _{ v }=\frac{1}{ n } \cdot \frac{ d U }{ dT }=\frac{1}{5} \cdot \frac{ d }{ dT }\left(\frac{21}{2} RT \right)=\frac{21}{10} R$

$C _{ P }= C _{ V }+ R =\frac{21}{10} R + R =\frac{31}{10} R$

$\gamma=\frac{ C _{ p }}{ C _{ v }}=\frac{31 / 10}{21 / 10}=1.476$