A) 24.2/15
B) 15/23
C) 27/17
D) 17/27
Correct Answer: A
Solution :
[a] \[\gamma \] for \[He=5/3;\] \[\gamma \] for \[~{{O}_{2}}=7/5\] 8 g of He is equal to 2 moles of He and 16 g of \[~{{O}_{2}}=\frac{1}{2}mole\] of \[~{{O}_{2}}=7/5\] gas. Total one has \[2{\scriptstyle{}^{1}/{}_{2}}\] moles. \[\therefore \] The weighted average is \[=\left( 2\times \frac{5}{3}+\frac{7}{5}\times \frac{1}{2} \right)\times \frac{2}{5}=\left\{ \frac{10}{3}+\frac{7}{10} \right\}\times \frac{2}{5}\] \[=\frac{121}{30}\times \frac{2}{5}=\frac{24.5}{15}\]
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