Two moles of an ideal gas with `(C_(P))/(C_(V))= (5)/(3)` are mixed with 3 moles of another ideal gas with `(C_(P))/(C_(V))= (4)/(3)`. The value of `(C_(P))/(C_(V))` for the mixture is

Options

A. 1.38
B. 1.42
C. 1.5
D. 1.7

Answer

B

Explanation

`C_(P)- C_(V)= R`
and for first gas `(C_(P_(1)))/(C_(V_(1)))= (5)/(3) rArr C_(P_(1)) = (5)/(3) C_(V_(1))`
Now `C_(P_(1))- C_(V_(1))= R`
`:. (5)/(3) C_(V_(1))- C_(V_(1))= R`
`:. 5C_(V_(1))- 3C_(V_(1)) = 3R`
`:. 2C_(V_(1))= 3R`
`:. C_(V_(1)) = (3R)/(2)`
and `C_(P_(1)) = C_(V_(1)) + R`
`=(3R)/(2) + R`
`:. C_(P_(1)) = (5R)/(2)`
Now for second gas `(C_(P_(2)))/(C_(V_(2))) = (4)/(3)`
`:. C_(P_(2)) = (4)/(3) C_(V_(2))`
and `C_(P_(2))- C_(V_(2)) R`
`:. (4)/(3) C_(V_(2))-C_(V_(2))R`
`:. 4C_(V_(2)) -3C_(V_(2)) = 3R`
`:. C_(V_(2)) = 3R`
and `C_(P_(2)) = C_(V_(2)) + R`
`=3R + R`
`:. C_(P_(2)) = 4R`
`:.` For mixture `(C_(P))/(C_(V)) = (n_(1)C_(P_(1)) + n_(2) C_(P_(2)))/(n_(1) C_(V_(1)) + n_(2) C_(V_(2)))`
`n_(1)=2` mole, `n_(2)=3` mole
`:. (C_(P))/(C_(V)) = (2 xx (5R)/(2) + 3 xx 4R)/(2 xx (3R)/(2) + 3 xx 3R)`
`=(5R + 12R)/(3R + 9R) = (17R)/(12R)`
`=(17)/(12)= 1.4166`
`=1.42`