# How do you derive the boiling point elevation equation using Raoult’s law and Clausius-Clapeyron relation?

I know that Raoult's law is $$P_{solution}=X_{solvent}P_{solvent}$$ and Clausius-Clapeyron equation is $$ln(\frac{P_{2}}{P_{1}})=-\frac{L}{R}(\frac{1}{T_{2}}-\frac{1}{T_{1}})$$ But how do you derive the boiling point elevation which is: $$\Delta T=K_{b}m i$$

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What you have done by adding an involatile solute to the liquid is to lower the activity of the solvent. This lowers the vapor pressure of the liquid.

Now if we assume that we live in a perfect fluffy nice world where all activity functions are equal to one. Then we need to merely consider the mole fraction of the liquid which is solvent (x1).

For example we can have water which when pure is 55.4 moles per litre.

If we assume that the molar volume (volume occupied by a mole of solute) is the same no matter what the concentration is in the solvent, for arguments sake we have a sugar solution. For solution of glucose in water the glucose occupies a volume of 112 ml per mole of glucose. Thus if the mole fraction of glucose (x2) increrases then the number of moles of water in one litre will decrease which will make x1 drop.

If we assume at a constant temperture that the vapor pressure is proportional to x1, this is as we assume that activity of the water is proportional to x1. As for a dilute solution where x2 is small we can write

x1 = 1 - k[2]

Where [2] is the concentration of the solute in moles per litre and k is a constant.

Now as pressure is given by the equation

P = P* x1

Where P* is the vapor pressure of pure water then for our ideal system we can understand that there will be a linear decrease of vapor pressure as the concentration of the glucose increases.

To restore the vapor pressure back to the original value we need to increase the temperture.

Now we can go from the Clausius-Clapeyron equation to give us

P = k exp (-Delta H vapp / RT)

For the water to boil then we need the vapor pressure to be equal to the atomospheric pressure, let us assume that we have exactly 1 atmosphere pressure ansd the boiling point of water is 373 K.

We can then write the equation for the mixture

P = x1 k exp (-Delta H vapp / RT)

The gas constant R is 8.31

If Delta H vapp is 40.8 x 10^3

We can make some progress, our value of k will be 519695

I used excel to then consider the tempertures required to get a vapour pressure of one for dilute solutions of glucose and I found that the boiling point of the solutions was described by the line T = m[2] + c where m was 0.4921 and c was 373. I used the solver to obtain the boiling points for the solutions.

Thus I have shown that ΔT=Kb Mi is true at least for dilute ideal solutions.

Now this will only work for dilute solutions, when we get concentrated glucose which will be what I call "damp sugar" things are going to change. One of my professional interests is in very concentrated solutions and the way that the activity functions change. I have an interest in things like 5 moles per litre solutions of sodium chloride and mixtures of things like choline chloride and ethylene glycol.

Firstly for fun I considered "damp sugar" AKA 4 M glucose, for a range of glucose solutions where the concentration ranged upto 5 moles per litre I found that we best describe the system with a polynomial.

T = 373 + k1[2] + k2[2]^2 + k3[2]^3

Where k1 = 0.548, k2 = 0.0098 and k3 = 0.0186

For some systems things work better if we consider the amount of solute per kilo of solvent (m) which is different to the concentration per litre (M).

I recalculate for moles per kilo.

For very strong solutions concentrations expressed in moles per kilo of solvent tend towards infinity so I tend not to like this method of expressing concentration.

If we consider the polynomial which describes the boiling point of our ideal sugar solution then we get

T = 373 + 0.5101m - 0.00034(m^2)

When the concentration of sugar is low then regardless of if we express the concentration in moles per litre or moles per kilo of solvent the boiling point will increase in a linear manner. It is only when we get strong solutions that things will become more complex.