Mole fraction of component `A` in vapour phase is `chi_(1)` and that of component `A` in liquid mixture is `chi_2`, then (`p_(A)^@`)= vapour pressure of pure A, `p_(B)^@` = vapour pressure of pure B), the total vapour pressure of liquid mixture is

Options

A. `(P_(A)^(@)chi_(2))/chi_(1)`
B. `(P_(A)^(@)chi_(1))/chi_(2)`
C. `(P_(A)^(@)chi_(1))/chi_(2)`
D. `(P_(B)^(@)chi_(2))/chi_(1)`

Answer

A

Explanation

`chi_(1)=(P_(A)^(@)chi_(A))/(P_(Total))=(P_(A)^(@)chi_(2))/(P_(Total))`
`:. (P_(Total)=(P_(A)^(@)chi_(2))/chi_(1)`