The Clausius-Clapeyron Equation

Ch 3, Lesson E, Page 11 - The Clausius-Clapeyron Equation

• We start with the Clapeyron Equation and then we make a couple of assumptions.
• As long as we are not near the critical point, the molar volume of the sat’d vapor is MUCH larger than the molar volume of the sat’d liquid.
• For water near atmospheric pressure, the molar volume of sat’d vapor is about 1000 times larger than the molar volume of the sat’d liquid.
• So, at ordinary pressures, this is a pretty good assumption.
• Next, since we are operating at modest pressures, we assume that the vapor phase is an ideal gas.
• This assumption is not always reliable, especially for polar compounds like water.
• We saw that in the previous lesson.
• Nonetheless, it is common practice to make this assumption and make corrections after the fact.
• This 2nd assumption allows us to use the ideal gas EOS to eliminate the molar volume of the sat’d vapor from the equation.
• If we boldly assume that ΔHvap is constant, we have an ordinary differential equation that is not too difficult to solve.
• But wait a minute, we just learned that ΔHvap decreases as temperature increases towards the critical point !
• That is true but FAR from the critical point ΔHvap doesn’t change by much over a modest range of temperatures.
• For example, ΔHvap of water decreases by just 1% as the temperature changes from 50oC to 60oC.
• Now, we perform some algebraic and slight of hand and do a bit of calculus to make the differential equation easier to solve.
• Since we assumed that dhvap is a constant, the whole right-hand side of the equation is a constant.
• Cool.  That makes it easy to integrate.
• The result is the Clausius-Clapeyron Equation and C is just a constant of integration.
• OK, so why is this important or useful ?