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The Clausius-Clapeyron Equation

### Skills to develop

- Apply the Clausius-Clapeyron equation to estimate the vapor
pressure at any temperature.
- Estimate the heat of phase transition from the vapor pressures
measured at two temperatures.

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The Clausius-Clapeyron Equation

The *vaporization curve*s of most liquids have similar shape. The
vapour pressure steadily increase as the temperature increases.
A good approach is to find a mathematical model for the pressure increase
as a function of temperature. Experiments showed that the pressure *P*,
enthalpy of vaporization, D*H*_{vap},
and temperature *T* are related,
*P* = *A* exp (- D*H*_{vap} / *R T*)
where *R* (= 8.3145 J mol^{-1} K^{-1}) and *A* are
the gas constant and unknown constant. This is known as the Clausius-
Clapeyron equation. If *P*_{1} and *P*_{2}
are the pressures at two temperatures *T*_{1} and
*T*_{2}, the equation has the form:
*P*_{1} D*H*_{vap} 1 1
ln (---) = ---- (--- - ---)
*P*_{2} *R* *T*_{2} *T*_{1}

The Clausius-Clapeyron equation allows us to estimate the vapor pressure
at another temperature, if the vapor pressure is known at some temperature,
and if the enthalpy of vaporization is known.

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Example 1

**
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The vapor pressure of water is 1.0 atm at 373 K, and the enthalpy of
vaporization is 40.7 kJ mol**^{-1}. Estimate the vapor pressure at
temperature 363 and 383 K respectively.
*Solution*

Using the Clausius-Clapeyron equation, we have:

P_{363} = 1.0 exp (- (40700/8.3145)(1/363 - 1/373)

= 0.697 atm

P_{383} = 1.0 exp (- (40700/8.3145)(1/383 - 1/373)

= 1.409 atm

Note that the increase in vapor pressure from 363 K to 373 K is 0.303 atm,
but the increase from 373 to 383 K is 0.409 atm. The increase in vapor
pressure is not a linear process.
*Discussion*

We can use the Clausius-Clapeyron equation to construct the entire
vaporization curve. There is a deviation from experimental value, that
is because the enthalpy of vaporization various slightly with temperature.

The Clausius-Clapeyron equation applies to any phase transition. The following
example shows its application in estimating the heat of sublimation.
####
Example 2

**
****
The vapor pressures of ice at 268 and 273 are 2.965 and 4.560 torr respectively.
Estimate the heat of sublimation of ice.
***Solution*

The enthalpy of sublimation is
D*H*_{sub}.
Use a piece of paper and derive the Clausius-Clapeyron equation
so that you can get the form:

D*H*_{sub}
= *R* ln (*P*_{268} / *P*_{268})
(1/268 - 1/273)

= 8.3145*ln(2.965/4.560) / (1/268 - 1/273)

= 52370 J mol^{-1}
Note that the heat of sublimation is the sum of heat of melting
and the heat of vaproization.
*Discussion*

Show that the vapor pressure of ice at 274 K is higher than that of water
at the same temperature.

Note the curve of vaporization is also called the curve of evaporization.

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Confidence Building Questions

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