A gaseous mixture consists of 16g of helium and 16 g of oxygen. The ratio `(C_p)/(C_v)` of the mixture is

Options

A. 1.62
B. 1.55
C. 1.81
D. 1.45

Answer

A

Explanation

No. of moles of helium = `16/4` = 4
No. of moles of oxygen = `16/32 = 1/2`
For monoatomic helium, `C_(V)=3/2R`
For diatomic oxygen, `C'_(V)=5/2R`
For the mixture, `(C_(V))_("mixture")`
`=(nC_(V)+n'C_(V'))/(n+n)=(4xx3/2R+1/2xx5/2R)/(4+1/2)`
`(C_(V))_("mixture")=(6R+5/4R)/(9/2R)=29/18R`
`thereforegamma_("mixture")=C_(P)/C_(V)=(C_(V)+R)/C_(V)=1+R/C_(V)`
`thereforegamma_("mixture")=1+R/(29/18R)=1+18/29=47/29`
`thereforegamma=1.62`