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Derive the relationship between partial pressure of gas and its mole fraction. |
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Answer» <P> SOLUTION :Consider a mixture of THREE gases containing `n_(1)` mole of gas A and `n_(2)` mole of gas B and `n_(3)` mole of gas C in a vessel of volume V let individual partial pressure be `P_(A), P_(B) and p_(C)` and total pressure be P.`wkt PV= nRT or P=(nRT)/(V)` `P_(A)=(n_(1)RT)/(V) ""...(1)` `P_(B)=(n_(2)RT)/(V) ""...(2)` `P_(C)=(n_(3)RT)/(V) ""...(3)` According to Dalton.s law, `P=P_(A)+P_(B)+P_(C)` `P\=(n_(1)RT)/(V)+(n_(2)RT)/(V)+(n_(3)RT)/(V)` `P=(n_(1)+n_(2)+n_(3))(RV)/(V) ""...(4)` From eqatuin (1) and (4) `(p_(A))/(P)=(n_(1))/(n_(1)+n_(2)+n_(3)) or p_(A)=[ (n_(1))/(n_(1)+n_(2)+n_(3))]p` From and (2) and (4) `(p_(B))/(P)=(n_(2))/(n_(1)+n_(2)+n_(3)) or p_(A)=[ (n_(2))/(n_(1)+n_(2)+n_(3))]p` From and (3) and (4) `(p_(C))/(P)=(n_(3))/(n_(1)+n_(2)+n_(3)) or p_(A)=[ (n_(3))/(n_(1)+n_(2)+n_(3))]p` `:.` Partial pressure = mole fraction `xx` total pressure |
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