A) 1.4
B) 1.54
C) 1.59
D) 1.62
Correct Answer: D
Solution :
[d]: For 16 g of helium,\[{{\mu }_{1}}=\frac{16}{4}=4\] For 16 g of oxygen, \[{{\mu }_{2}}=\frac{16}{32}=\frac{1}{2}\] For mixture of gases, \[{{C}_{V}}=\frac{{{\mu }_{1}}{{C}_{{{V}_{1}}}}+{{\mu }_{2}}{{C}_{{{V}_{2}}}}}{{{\mu }_{1}}+{{\mu }_{2}}}\]where \[{{C}_{V}}=\frac{f}{2}R\] \[{{C}_{P}}=\frac{{{\mu }_{1}}{{C}_{{{P}_{1}}}}+{{\mu }_{2}}{{C}_{{{P}_{2}}}}}{{{\mu }_{1}}+{{\mu }_{2}}}\]where\[{{C}_{P}}=\left( \frac{f}{2}+1 \right)R\] For helium, \[f=3,{{\mu }_{1}}=4\] For oxygen,\[f=5,{{\mu }_{2}}=\frac{1}{2}\] \[\therefore \]\[\frac{{{C}_{P}}}{{{C}_{V}}}=\frac{\left( 4\times \frac{5}{2}R \right)+\left( \frac{1}{2}\times \frac{7}{2}R \right)}{\left( 4\times \frac{3}{2}R \right)+\left( \frac{1}{2}\times \frac{5}{2}R \right)}=\frac{47}{29}=1.62.\]
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