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Give relation between partial pressure and mole fraction. |
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Answer» Solution :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. |
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