Thermodynamically using a cyclic process, reversibly at a constant temperature, this law follows steps.
- Suppose the solution has n1 moles of solute and n2 moles of solvent. Using a piston and cylinder arrangement, the quantity of solvent removed contains 1 mole of solute, and the quantity removed is
So the solution is huge, and its concentration, its osmotic pressure is not changed. The volume removed contains 1 mole of solute, the work done w1 on the system\[\displaystyle {{W}_{1}}=-PV=-RT—–(1)\] - The quantity removed solvent is converted reversibly into the vapour at pressure P of a solvent. If the comparison of vapour volume of liquid is neglected, and work gained for 1 mole of vapour is PV, and moles of the work gained (w2);\[\displaystyle {{W}_{2}}=\frac{{{{n}_{2}}}}{{{{n}_{1}}}}P{{V}_{1}}\]
- Vapour expanded till a pressure falls to Ps and Ps = vapour pressure of the solution
Work done by the system per mole of the vapour is,
\[\displaystyle \left( {\frac{{P+{{P}_{s}}}}{2}} \right)\left( {{{V}_{2}}-{{V}_{1}}} \right)—–(2)\]
= Mean pressure during small expansion
= moles of the total vapour work done (w3) of the system as follow;
\[\displaystyle {{W}_{3}}=\frac{{{{n}_{2}}}}{{{{n}_{1}}}}\left( {\frac{{P+{{P}_{s}}}}{2}} \right)\left( {{{V}_{2}}-{{V}_{1}}} \right)—–(3)\] - Vapour pressure Ps in contact with a solution is equilibrium and condensed reversibly so that system regains its initial state, and work done (w4) on the system condensing gas into a liquid at pressure Ps is as follows;
\[\displaystyle {{W}_{4}}=-\frac{{{{n}_{2}}}}{{{{n}_{1}}}}{{P}_{s}}{{V}_{2}}—–(4)\]The whole cycle is at a constant temperature, with no transformation of heat into work, and the system back to its initial state. i.e. work done by the system must be equal to work done by the system.
\[\displaystyle {{W}_{1}}+{{W}_{2}}+{{W}_{3}}+{{W}_{4}}=0\]
= Mean pressure during small expansion\[\displaystyle \begin{array}{l}-RT+\frac{{{{n}_{2}}}}{{{{n}_{1}}}}P{{V}_{1}}+\frac{{{{n}_{2}}}}{{{{n}_{1}}}}\left( {\frac{{P+{{P}_{s}}}}{2}} \right)\left( {{{V}_{2}}-{{V}_{1}}} \right)-\frac{{{{n}_{2}}}}{{{{n}_{1}}}}{{P}_{s}}{{V}_{2}}=0\\\\-RT+\frac{{{{n}_{2}}}}{{{{n}_{1}}}}\left[ {P{{V}_{1}}+\left( {\frac{{P+{{P}_{s}}}}{2}} \right)\left( {{{V}_{2}}-{{V}_{1}}} \right)-{{P}_{s}}{{V}_{2}}} \right]=0\\\\-RT+\frac{{{{n}_{2}}}}{{{{n}_{1}}}}\left[ {P{{V}_{1}}+\frac{{P{{V}_{2}}}}{2}-\frac{{P{{V}_{3}}}}{2}+\frac{{{{P}_{s}}{{V}_{2}}}}{2}-\frac{{{{P}_{s}}{{V}_{1}}}}{2}-{{P}_{s}}{{V}_{2}}} \right]=0\\\\-RT+\frac{{{{n}_{2}}}}{{{{n}_{1}}}}\left[ {\frac{{P{{V}_{1}}}}{2}-\frac{{P{{V}_{2}}}}{2}-\frac{{{{P}_{s}}{{V}_{1}}}}{2}-\frac{{{{P}_{s}}{{V}_{2}}}}{2}} \right]=0\\\\-RT+\frac{{{{n}_{2}}}}{{{{n}_{1}}}}\left[ {P\left( {\frac{{{{V}_{1}}+{{V}_{2}}}}{2}} \right)-{{P}_{s}}\left( {\frac{{{{V}_{1}}+{{V}_{2}}}}{2}} \right)} \right]=0\\\\-RT+\frac{{{{n}_{2}}}}{{{{n}_{1}}}}\left( {P-{{P}_{s}}} \right)\left( {\frac{{{{V}_{1}}+{{V}_{2}}}}{2}} \right)=0\\\\-RT+\frac{{{{n}_{2}}}}{{{{n}_{1}}}}\left( {P-{{P}_{s}}} \right)V=0-----(5)\\\\V=\frac{{{{V}_{1}}+{{V}_{2}}}}{2}=\text{ volume of 1 mole of vapour =}\frac{{RT}}{P}.\\\\\text{Equation }\left( \text{5} \right)\text{ is as below}\text{.}\\\\-RT+\frac{{{{n}_{2}}}}{{{{n}_{1}}}}\left( {P-{{P}_{s}}} \right)\frac{{RT}}{P}=0\\\\Or\\-1+\frac{{{{n}_{2}}}}{{{{n}_{1}}}}\left( {P-{{P}_{s}}} \right)\frac{1}{P}=0\\\\Or\end{array}\]
\[\displaystyle \frac{{P-{{P}_{s}}}}{P}=\frac{{{{n}_{1}}}}{{{{n}_{2}}}}\]
It is Raoult’s law.
