If `"7 gm N"_(2)` is mixed with 20 gm Ar, there `C_(p)//C_(v)` of mixture will be :

Options

A. `(17)/(6)`
B. `(11)/(7)`
C. `(17)/(11)`
D. `(17)/(13)`

Answer

C

Explanation

For `N_(2), f=5`
For `Ar, f=3`
`"For "N_(2),C_(v)=(5)/(2)R," "C_(P)=(7)/(2)R," moles "=(1)/(4)`
`"Ar, "C_(b)=(3)/(2)R," "C_(p)=(5)/(2)R," moles "=(1)/(2)`
`C_("Pmix")=(n_(1)C_(P_(1))+n_(2)C_(P_(2)))/(n_(1)+n_(2))`
`=((1)/(4)xx(7)/(2)R+(1)/(2)xx(5)/(2)R)/((1)/(4)+(1)/(2))=((7)/(8)R+(5)/(4)R)/((3)/(4))=(17R)/(8)xx(4)/(3)=(17R)/(6)`
`C_("Vmix")=(n_(1)C_(V_(1))+n_(2)C_(V_(2)))/(n_(1)+n_(2))=((1)/(4)xx(5)/(2)R+(1)/(2)xx(3)/(2)R)/((1)/(4)+(1)/(2))`
`=(11R)/(8)xx(4)/(3)=(11R)/(6)`
`Y_("mix")=(C_("P mix"))/(C_("V mix"))=(17)/(11)`