[diablo]

Hi,

Problem:

The enthalpy of evaporation of water (100°C) = 2250J/g
calculate (assume ideal behaviour):

a) vapour pressure at 100°C (Rault's law)
b) the boiling point under an external pressureat 1.013x10⁵Pa  of a solution of glucose (M = 180) in 1kg water (apply ClausiusClapeyorn equation)

in both cases 50g Glucose in 1kg water

My calculation would be:

a)  Rault's law: P_X = N_X x P_X
     N_X = molefraction
     P_X = Boiling point of pure Water

--> nGlucose . 0,27mol ( calculated with m/M = 50g/180gmol^-1)
     n Water ...55,5mol (calculated with m/M = 1000g/18gmol^-1)
     thus N_X = n_glc/n_glc+n_wat = 0,0048

and this put in Rault's law.... --> but how do iget the vapour pressure of pure water ( is it possible to calculate it with the enthalphy?)

b)  clausius-clapeyron- equation:

ln p =  -dH / (RxT) +C

then transform equation --> T = (-dH / (lnp - C)) / R

dH = 2250 J/g
p = 1,013x10⁵Pa
C = 50g/L (Glucose)


are these calculations ok?^^

thx ,
Diabloooooo

[maakii]

My calculation would be:

a)  Rault's law: P_X = N_X x P_X
     N_X = molefraction
     P_X = Boiling point of pure Water

a)For Raoult's Law, Px = Nx x P*x, where P*x would be the vapour pressure of pure water, and not its boiling point..

In this case, the vapour pressure at 100 degrees is P*x, which is obtained by knowing that at boiling, saturated vapour pressure is equal to                 ?

Use Raoult's law, and you can find out Px for 50g glucose in 1kg water at 100 degrees.

b)To get the vapour pressure of the solution in a), you can use The Clausius Clapeyron equation: after you integrated it you will get ln(p1/p2) = - dH_vap/R (1/T1 - 1/T2) [sorry I don't recognise your form of the equation..] ..now..you have got one set of p1 and T1 values, these values are at the boiling point. Then you have your p2 value, which is the value you calculated, sub these in, and you can find the boiling point