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Derive a relationship between K_b,T_b and DeltaH_(vap) for a liquid. |
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Answer» Solution :Let us consider a PURE liquid. At its boiling point, `T_b` , its VAPOUR pressure `P^@` WOULD be equal to the external pressure. `:.P^@=P_(ext) ` at temperature `T_b`. Now let us consider a non -volatile solute dissolved in the liquid. When the solution reaches temperature `T_b`, the vapour pressure of the SYSTEM, would be less than the external pressure. When the solution is heated further and it reaches its boiling `T_b^**`, the vapour pressure of the solution would be equal to `P_(ext)` which is equal to the `P^@` of the pure liquid at its boiling point. Since vapour pressure of a system are the `K_P`’s for the respective equilibrium, In `K_(PT_2)/K_(PT_1)=(DeltaH)/R[1/T_1-1/T_2]` In `P^@/P=(DeltaH_(vap))/R[1/T_b-1/T_b^**]=(DeltaH_(vap))/R[(DeltaT_b)/(T_bT_b^**)]` Assuming the solution to be highly dilute, `T_b^**` would be very close to `T_b`. `:. InP^@/P=(DeltaH_(vap))/R[(DeltaT_b)/(T_b^(2))]` `-InP^@/P=(DeltaH_(vap))/R[(DeltaT_b)/(T_b^(2))]` `-In[1-((P^@-P)/P^@)]=(DeltaH_(vap))/R[(DeltaT_b)/(T_b^(2))]` `-In[1-X_(solute)]=(DeltaH_(vap))/R[(DeltaT_b)/(T_b^(2))]` Since `X_(solute)` is very small, we can make the approximation that in (1 - x) = - x (when x is very small).`:.X_(solute)=(DeltaH_(vap))/R[(DeltaT_b)/(T_b^(2))]` or`n/(n+N)=(DeltaH_(vap))/R[(DeltaT_b)/(T_b^(2))]` Ignoring n in comparison to N in the denominator `n/N=(DeltaH_(vap))/R[(DeltaT_b)/(T_b^(2))]` `n/(W_(solvent)/(M_(solvent)))=(DeltaH_(vap))/R[(DeltaT_b)/(T_b^(2))]`or `n/(W_(solvent))=(DeltaH_(vap))/(M_(solvent)R)[(DeltaT_b)/(T_b^(2))]` Multiplying by 1000 on both the sides, we get `n/(W_(solvent))xx1000=(DeltaH_(vap))/(M_(solvent)R)[(DeltaT_b)/(T_b^(2))]`xx1000 `m=(DeltaH_(vap))/(M_(solvent)R)[(K_bm)/(T_b^(2))]`xx1000 `K_b=(RT_b^(2)M_(solvent))/(1000DeltaH_(vap))`Similarly, `K_"f"=(RT_b^(2)M_(solvent))/(1000DeltaH_("fus"))` |
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