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Derive the value of critical constants the Van der Waals constants. |
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Answer» Solution :Derivation of critical constants from the Van der Waals constant: Van der Waals equation is, `(P + (an^(2))/(V^(2))) (V-nb) = nRT` for 1mole From this equation, the values of critical constant `P_(C), V_(C)` and `T_(C)` are derived in terms of a and b the Van der Waals constants. `(P + (a)/(V^(2))) (V-b) = RT` ...(1) On expanding the equation (1) `PV + (a)/(V) -Pb-(ab)/(V^(2)) - RT=0`...(2) Multiplying equation (2) by `V^(2)/P`, `(V^(2))/(P) (PV + (a)/(V)- Pb- (ab)/(V^(2)) - RT)=0` `V^(3) + (aV)/(P) -bV^(2) - (ab)/(P) - (RTV^(2))/(P) = 0` ...(3) Equation (3) is rearranged in the powers of V. `V^(3) - [(RT)/(P) +b] V^(2) + (aV)/(P) - (ab)/(P) = 0` ...(4) The above equation (4) is an cubic equation of V, which can have three roots. At the critical POINT, all the three values of V are equal to the critical volume `V_(C)`. i.e. `V= V_(C)` `V-V_(C) = 0`...(5) `(V-V_(C))^(3) = 0`...(6) `(V^(3) - 3V_(C) V^(2)) + 3V_(C)^(2) V-V_(C)^(3) = 0` ...(7) As the equation (4) is identical with equation (7), comparing the .V. terms in (4) and (7), `-3V_(C) V^(2) = -[(RT_(C))/(P_(C)) + b] V^(2)` ...(8) `3V_(C) = b + (RT_(C))/(P_(C))` ...(9) `3V_(C)^(2) = (a)/(P_(C))` ...(10) `V_(C)^(3) = (ab)/(P_(C))` ...(11) Divide equation (11) by (10) `(V_(C)^(3))/(3V_(C)^(2)) = (ab//P_(C))/(a//P_(C))` `(V_(C))/(3) = b` `:. V_(C) = 3b`...(12) When equation (12) is substituted in (10) `3V_(C) = b+ (RT_(C))/(P_(C))` `3 xx 3b =b + (RT_(C))/(a//27b^(2))` `9b-b= (RT_(C))/(a) xx 27b^(2)` `8b= (T_(C).R 27b^(2))/(a)` `:. T_(C) = (8ab)/(27Rb^(2)) = (8a)/(27Rb)` `:. T_(C) = (8a)/(27Rb)` ...(14) Substituting the values of `V_(C)` and `P_(C)` in equation (9) Critical constants a and b can be calculated using Van der Waals constants as follows:  ..(5)
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