Heat and Thermodynamics 4 Question 29
31. If one mole of a monoatomic gas $(\gamma=5 / 3)$ is mixed with one mole of a diatomic gas $(\gamma=7 / 5)$, the value of $\gamma$ for the mixture is
$(1988,2 M)$
(a) 1.40
(b) 1.50
(c) 1.53
(d) 3.07
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Solution:
- $\gamma _1=\frac{5}{3}$ means gas is monoatomic or $C _{V _1}=\frac{3}{2} R$ $\gamma _2=\frac{7}{5}$ means gas is diatomic or $C _{V _2}=\frac{5}{2} R$
$C _V$ (of the mixture)
$$ \begin{gathered} =\frac{n _1 C _{V _1}+n _2 C _{V _2}}{n _1+n _2}=\frac{(1)\left(\frac{3}{2} R\right)+(1)\left(\frac{5}{2}\right) R}{1+1}=2 R \\ C _p(\text { of the mixture })=C _V+R=3 R \\ \therefore \quad \gamma _{\text {mixture }}=\frac{C _p}{C _V}=\frac{3 R}{2 R}=1.5 \end{gathered} $$
