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Q.
If one mole of a monoatomic. gas is mixed with one mole of a triatomic gas the value of $\gamma$ for the mixture is
Thermodynamics
Solution:
For monoatomic gas $C_{V_{1}}=\frac{3 R}{2}$
For triatomic gas $C_{V_{2}}=3 R$
For the mixture of 1 mole of monoatomic gas and 1 mole of triatomic gas,
$C_{V}=\frac{1\left(\frac{3 R}{2}\right)+1(3 R)}{2}=\frac{9 R}{4}$
$C_{P}=\frac{9 R}{4}+R=\frac{13 R}{4}$
$\Rightarrow \gamma_{\text{mix}}=\frac{C_{P}}{C_{V}}=\frac{13}{9}=1.44$