Chem 351 Help — The Clapeyron Equation

The Clapeyron Equation: What Determines the Slope of a Phase Boundary?

We have seen that phase equilibrium requires the chemical potential of a substance to be the same in every phase present. For two phases, \(\alpha\) and \(\beta\),

\[ \mu^\alpha = \mu^\beta \]

A phase boundary on a phase diagram represents all combinations of temperature and pressure for which this equality is satisfied. The next question is therefore:

How does the equilibrium pressure change when the temperature changes?

To answer this question, consider a small movement along a phase boundary. Since the two phases remain in equilibrium,

\[ d\mu^\alpha = d\mu^\beta \]

The differential form of the chemical potential is

\[ d\mu = \bar V\,dp - \bar S\,dT \]

where \(\bar V\) is the molar volume and \(\bar S\) is the molar entropy.

Substituting this expression for each phase gives

\[ \bar V^\alpha dp - \bar S^\alpha dT = \bar V^\beta dp - \bar S^\beta dT \]

Rearranging and collecting like terms leads to

\[ \frac{dp}{dT} = \frac{\Delta \bar S} {\Delta \bar V} \]

This relationship is known as the Clapeyron equation.

Physical Interpretation

The Clapeyron equation shows that the slope of a phase boundary depends on two quantities:

\(\Delta \bar S\), the change in molar entropy during the phase transition
\(\Delta \bar V\), the change in molar volume during the phase transition

Most phase transitions increase entropy:

\[ \Delta \bar S > 0 \]

because liquids are more disordered than solids and gases are more disordered than liquids.

Therefore, the sign of the phase-boundary slope is usually determined by the sign of \(\Delta \bar V\).

For example, when a liquid vaporizes,

\[ \Delta \bar V > 0 \]

and therefore

\[ \frac{dp}{dT} > 0 \]

meaning that vapor pressure increases with temperature.

Why Water Is Unusual

Most substances expand when they melt:

\[ \Delta \bar V_{\text{fusion}} > 0 \]

and therefore have a positive solid-liquid phase-boundary slope.

Water behaves differently because ice is less dense than liquid water. As a result,

\[ \Delta \bar V_{\text{fusion}} < 0 \]

for the melting process.

Consequently,

\[ \frac{dp}{dT} < 0 \]

and the solid-liquid phase boundary of water has a negative slope.

This means that increasing pressure favors the liquid phase and slightly lowers the melting point of ice.

Big picture: The Clapeyron equation connects the shape of a phase diagram to the entropy and volume changes associated with a phase transition. It provides a direct thermodynamic explanation for why phase boundaries slope the way they do and explains unusual behavior such as the negative solid-liquid phase boundary of water.

Worked examples

Worked Example: Pressure Dependence of the Melting Point of Water

How much pressure must be applied to lower the melting point of water from \( 273.15\ \text{K} \) to \( 273.14\ \text{K} \)

Assume that the enthalpy of fusion and volume change remain constant over this small temperature range.

Solution

For the melting process

\[ \text{ice} \rightarrow \text{liquid water} \]

the Clapeyron equation is

\[ \frac{dp}{dT} = \frac{\Delta H_{\text{fus}}} {T\Delta V} \]

Rearranging gives

\[ dp = \frac{\Delta H_{\text{fus}}} {T\Delta V} dT \]

Using values near 0 °C:

\[ \Delta H_{\text{fus}} = 6.009\times10^3\ \text{J mol}^{-1} \]

and

\[ \Delta V = -1.63\times10^{-6}\ \text{m}^3\text{mol}^{-1} \]

The desired temperature change is

\[ \Delta T = 273.14-273.15 = -0.010\ \text{K} \]

Assuming \(T\) remains approximately constant at 273.15 K,

\[ \Delta p \approx \frac{(6.009\times10^3\ \text{J mol}^{-1}) (-0.010\ \text{K})} {(273.15\ \text{K}) (-1.63\times10^{-6}\ \text{m}^3\text{mol}^{-1})} \] \[ \Delta p = 1.35\times10^5\ \text{Pa} \]

Converting to atmospheres:

\[ \Delta p = \frac{1.35\times10^5\ \text{Pa}} {1.01325\times10^5\ \text{Pa atm}^{-1}} \] \[ \Delta p = 1.33\ \text{atm} \]

Therefore,

\[ \boxed{\Delta p \approx 1.3\ \text{atm}} \]

of additional pressure is required to lower the melting point of water by

\[ 0.010\ \text{K} \]

Physical interpretation: Water is unusual because liquid water occupies less volume than ice. Increasing the pressure therefore favors the liquid phase and lowers the melting point. However, the effect is quite small—lowering the melting point by only 0.01 K requires more than one atmosphere of additional pressure.

Practice

Question 1
The Clapeyron equation relates the slope of a phase boundary to which two thermodynamic quantities? A. Enthalpy and heat capacity B. Entropy change and volume change C. Pressure and temperature D. Gibbs energy and entropy
Show Answer Answer: B The Clapeyron equation is \[ \frac{dp}{dT} = \frac{\Delta S}{\Delta V} \] The slope of a phase boundary therefore depends on the entropy change and volume change associated with the phase transition.

Question 2
Most substances have a positive slope for the solid-liquid phase boundary. Which statement best explains this behavior? A. Melting decreases entropy and decreases volume. B. Melting increases entropy and increases volume. C. Melting decreases entropy and increases volume. D. Melting increases entropy and decreases volume.
Show Answer Answer: B For most substances, melting increases both entropy and molar volume: \[ \Delta S > 0 \qquad \Delta V > 0 \] Consequently, \[ \frac{dp}{dT} = \frac{\Delta S}{\Delta V} > 0 \] and the solid-liquid phase boundary has a positive slope.

Question 3
Why does water have a negative slope for its solid-liquid phase boundary? A. Ice has a larger entropy than liquid water. B. Water contracts when it freezes. C. Liquid water occupies less volume than ice. D. The enthalpy of fusion of water is negative.
Show Answer Answer: C Water is unusual because ice is less dense than liquid water. As a result, \[ \Delta V_{\text{fusion}} < 0 \] for the melting process. Since \[ \frac{dp}{dT} = \frac{\Delta S}{\Delta V} \] the negative volume change produces a negative phase-boundary slope.

Question 4
The melting point of water decreases by approximately 0.008 K for every atmosphere of additional pressure applied. Estimate the pressure increase required to lower the melting point by 0.040 K.
Show Answer The pressure change is proportional to the temperature change: \[ \frac{0.040\ \text{K}} {0.008\ \text{K atm}^{-1}} = 5.0\ \text{atm} \] Therefore, \[ \boxed{\Delta p \approx 5\ \text{atm}} \] This calculation illustrates that pressure has only a modest effect on the melting point of water. Significant changes in melting temperature require several atmospheres of additional pressure.

Key points (one glance)

The Clapeyron equation describes the slope of a phase boundary on a pressure-temperature phase diagram: \[ \frac{dp}{dT} = \frac{\Delta S}{\Delta V} \]
The equation follows directly from the condition for phase equilibrium: \[ \mu^\alpha = \mu^\beta \]
The slope of a phase boundary depends on the entropy change and volume change associated with the phase transition.
Most phase transitions involve an increase in entropy: \[ \Delta S > 0 \]
Vaporization and sublimation involve large positive volume changes: \[ \Delta V > 0 \] and therefore have positive phase-boundary slopes.
Most substances expand upon melting: \[ \Delta V_{\text{fusion}} > 0 \] and therefore have positive solid-liquid phase-boundary slopes.
Water is unusual because liquid water occupies less volume than ice: \[ \Delta V_{\text{fusion}} < 0 \] giving the solid-liquid phase boundary a negative slope.
Increasing pressure favors the phase with the smaller molar volume.
Increasing temperature generally favors the phase with the larger molar entropy.

Big picture: The Clapeyron equation connects the thermodynamic properties of a phase transition to the geometry of a phase diagram. By relating the slope of a phase boundary to the entropy and volume changes accompanying a phase change, it explains how temperature and pressure work together to determine which phase is thermodynamically stable.