Give relation between partial pressure and mole fraction.

Explanation

Suppose at the temperature (T), three gases, enclosed in the volume (V), exert partial pressure `p_(1), p_(2)` and `p_(3)` respectively then, Ideal gas equation pV = nRT therefore,
(i) `p_(1)=(n_(1)RT)/(V) " "` …..(Eq. -i)
(ii) `p_(2)=(n_(2)RT)/(V) " "` .......(Eq.-ii)
(iii) `p_(3)=(n_(3)RT)/(V) " "` ......(Eq. -iii)
where, `n_(1), n_(2)` and `n_(3)` are number of moles of these gases.
According to Dalton.s Law, equation of total pressure is written as,
`p_("total")=(p_(1)+p_(2)+p_(3))`
`= ((n_(1)RT)/(V))+((n_(2)RT)/(V))+((n_(3)RT)/(V))`
`p_("total")=(n_(1)+n_(2)+n_(3))(RT)/(V) " "` .....(Eq. -iv)
Mole fraction by ratio of partial pressure and total pressure `(chi)` : Partial pressure is divided by total pressure, (Eq. -i / Eq. -iv)
`(p_(1))/(p_("total"))=((n_(1)RT)/(V))(V)/((n_(1)+n_(2)+n_(3))RT)`
`=((n_(1))/(n_(1)+n_(2)+n_(3)))((RTV)/(VRT))`
`therefore (p_(1))/(p_("total"))=(n_(1))/(n_(1)+n_(2)+n_(3)) " "` ......(Eq.-v)
but, `(n_(1)+n_(2)+n_(3))=` Total mole (n)
`(p_(1))/(p_("total"))=(n_(1))/(n) " "` .....(Eq. -vi)
Mole fraction of `1^(st)` gas `(chi)=(n_(1))/(n)`
`therefore (p_(1))/(p_("total"))=chi_(1)` and `p_(1)=chi_(1)p_("total")` .....(Eq. -vii)
Similarly for gas 2 and 3,
`p_(2)=chi_(2)p_("total")` and `p_(3)=chi_(3)p_("total")`
So general equation is `p_(i)=chi_(1)p_("total")` ......(Eq. - viii)
Partial pressure = mole fraction `xx` total pressure
Where, `p_(i)=` Partial pressure of gas
`chi_(i)=` Mole fraction of gas
If total pressure of a mixture of gases is known, the (Eq. - viii) can be used to find out pressure cxerted by individual gases.