Question

Can anyone explain me properly about Clausius-Clapeyron's Equation.

Answer 1

The vapor pressure of a liquid increases as the temperature increases. This is why puddles left after a rainstorm evaporate faster on a hot day than on a cold day.

The vapor pressure of a liquid increases faster as the temperature nears the boiling point of the liquid- the data is not a straight line. However, it turns out that the plot of the log of the vapor pressure vs. 1/T is a straight line with a slope equal to that of -DHvap/R. We can use this fact to derive a simple equation that relates the vapor pressure at certain temperatures to the heat of vaporization, the Clausius-Clapeyron Equation.

The equation for a straight line is y=mx+b, where m is the slope and b is the y-intercept. For our case, the y values are ln(Pvap) and the x values 1/T and the slope m = -DHvap/R. Thus, the equation for the line is

ln(Pvap) = -DHvap/RT + b

If we discover the vapor pressure at two separate temperatures, we have two points on the same line.


ln(P2) = -DHvap/RT2 + b
ln(P1) = -DHvap/RT1 + b
If we subtract equation 1 from equation 2, the b's cancel and we're left with
ln(P2) - ln(P1) = -DHvap/R * (1/T2- 1/T1)
or
ln(P2/P1) = -DHvap/R * (1/T2- 1/T1) which is known as the Clausius-Clapeyron Equation. This equation will let us figure out the vapor pressure of a liquid at any temperature if we know the heat of vaporization, or find out the heat of vaporization if we know the vapor pressure at two different temperatures.
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Example 1: Water has a vapor pressure of 24 mmHg at 25oC and a heat of vaporization of 40.7 kJ/mol. What is the vapor pressure of water at 67oC?

Solution: Simply use the Clausius-Clapeyron Equation to figure out the vapor pressure. We have to be a bit careful about the units of R: the units we're using are kJ, so R = 8.31x10-3 kJ/mol K.

ln(P2/P1) = -DHvap/R * (1/T2- 1/T1)
ln(P2/24) = 40.7 kJ/8.31x10-3 kJ/mol K *(1/340- 1/298)
ln(P2/24) = 2.03
P2/24 = 7.62
P2 = 182 mmHg