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Phase changes and the Clausius-Clapeyron relation

Phase boundaries as lines of equal chemical potential, the Clapeyron slope $dP/dT = L/(T\Delta V)$, the Clausius-Clapeyron equation for vapor pressure, and boiling at altitude and in a pressure cooker.

Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.

1. What you will learn

By the end of this lesson you will be able to derive the Clapeyron relation from equal chemical potentials, find the slope of a phase boundary, use the Clausius-Clapeyron equation for vapor pressures and boiling points, and read a phase diagram.

2. What you already have

You know latent heats from the lesson on enthalpy, that at fixed temperature and pressure a system minimizes its Gibbs free energy, and that particles flow from high chemical potential to low. This lesson applies all three to the boundaries between solid, liquid and gas.

3. Words for this lesson

4. Phases coexist where their chemical potentials are equal

At fixed temperature and pressure a substance takes whichever phase has the lowest molar Gibbs free energy — its chemical potential $\mu$. Two phases coexist only where their chemical potentials are equal, and that single equation, $\mu_1(T, P) = \mu_2(T, P)$, traces a line on the pressure-temperature plane: the phase boundary.

Move a little way along the boundary. Both chemical potentials change, and since they stay equal, their changes are equal: $d\mu_1 = d\mu_2$. For a pure substance $d\mu = -s\,dT + v\,dP$, with $s$ and $v$ the molar entropy and volume. So $-s_1\,dT + v_1\,dP = -s_2\,dT + v_2\,dP$, and rearranging,

$$\frac{dP}{dT} = \frac{s_2 - s_1}{v_2 - v_1} = \frac{L}{T\Delta V}.$$

The last step uses $\Delta s = L/T$, the entropy of a phase change from the entropy lesson. This is the Clapeyron relation. A large latent heat or a small volume change makes the boundary steep.

For a liquid and its vapor, the vapor's volume dwarfs the liquid's, and treating the vapor as an ideal gas gives $\Delta V \approx RT/P$. Then $dP/dT = LP/(RT^2)$, which separates and integrates, if $L$ is roughly constant, to the Clausius-Clapeyron equation:

$$\ln\frac{P}{P_0} = -\frac{L}{R}\left(\frac{1}{T} - \frac{1}{T_0}\right).$$

It relates the vapor pressure at one temperature to that at another, or, turned around, the boiling point at one pressure to that at another.

5. The method, step by step, and how to check it

1. Molar quantities. Use the latent heat per mole and the volume change per mole, or both per kilogram; never mix them. Water's $2257$ kJ/kg is $40.7$ kJ/mol.
2. Kelvin. Every $T$ in the Clapeyron relation and in Clausius-Clapeyron is absolute.
3. Sign of $\Delta V$. Take $\Delta V$ and $L$ in the same direction, for instance from solid to liquid. Nearly always $\Delta V > 0$; for water freezing it is not.
4. Integrated form. To find a boiling point at a new pressure, solve for $1/T$ first: $1/T = 1/T_0 - (R/L)\ln(P/P_0)$. Invert only at the end.

Checks. A liquid-vapor boundary rises with temperature, so higher pressure must give a higher boiling point, and a lower pressure a lower one. Solid-liquid boundaries are steep, tens of atmospheres per kelvin or more, because $\Delta V$ is tiny. Clausius-Clapeyron is accurate to a few percent over tens of kelvin; over wider ranges the latent heat changes — water's rises to $44$ kJ/mol at room temperature — and the estimate drifts. Near the critical point it fails entirely, since there the latent heat vanishes and the vapor is no longer ideal.

6. Why ice's melting line leans backward

Most solids are denser than their liquids, so melting expands them, $\Delta V > 0$, and the melting line leans forward: squeezing a solid raises its melting point. Water is the famous exception. Ice's open hexagonal lattice, held apart by hydrogen bonds, takes about $9$ percent more room than liquid water, so on melting $\Delta V = -1.63 \times 10^{-6}$ m³/mol. With $L = 6010$ J/mol at $273$ K,

$$\frac{dP}{dT} = \frac{6010}{273 \times (-1.63 \times 10^{-6})} = -1.35 \times 10^{7}\ \text{Pa/K},$$

about $-135$ atmospheres per kelvin. Squeezing ice lowers its melting point, but only by one degree for every $135$ atmospheres.

The same backward lean is why lakes freeze from the top down and why water pipes burst in winter. The expansion on freezing can exert pressures of hundreds of atmospheres on anything that confines it, and freeze-thaw cycles split rocks and crack the roads of every northern state each spring.

7. Boiling, cooking and altitude

Water boils when its vapor pressure equals the pressure around it. At sea level that happens at $100$ °C; wherever the air pressure is lower, it happens sooner. In Denver, where the pressure averages about $83$ kPa, Clausius-Clapeyron gives a boiling point near $95$ °C, and on the summit of Pikes Peak, at $4.3$ km, about $86$ °C. Food in boiling water cooks at the water's temperature, not the flame's, so pasta takes longer and eggs cook more slowly at altitude, and package directions carry high-altitude adjustments.

A pressure cooker does the opposite. Its sealed lid lets the pressure build to about $15$ psi above atmospheric, roughly $205$ kPa absolute, and the boiling point rises to about $121$ °C. Because cooking reactions speed up roughly twofold for every ten degrees, the food cooks about three times as fast. The same temperature, $121$ °C, is the standard for autoclaves that sterilize hospital instruments and canned food: at that temperature, held for fifteen minutes, even the heat-resistant spores of Clostridium botulinum are killed.

8. Does pressure explain why ice is slippery?

A popular explanation of ice skating is that the blade's pressure melts the ice beneath it, and the skater glides on the water. The Clapeyron slope says otherwise. A $70$ kg skater standing on one blade $20$ cm long and $3$ mm wide exerts about $70 \times 9.8/(0.20 \times 0.003) = 1.1 \times 10^{6}$ Pa, some $11$ atmospheres, which lowers the melting point by less than a tenth of a degree. Skating works perfectly well at $-20$ °C, far below anything pressure melting could reach.

Current research points instead to the surface of ice itself: even well below freezing, the outermost molecular layers are disordered and mobile, a so-called premelted layer, and friction heating adds a thin film of water as the blade moves. The Clapeyron relation cannot say what does make ice slippery, but it cleanly rules out the most famous explanation — an example of thermodynamics settling a question by a single number.

9. In the world: cooking at altitude

The USDA and package instructions warn cooks above $3000$ feet to adjust recipes, and Clausius-Clapeyron supplies the reason. At the $5280$ foot elevation of Denver the air pressure averages about $83.4$ kPa, and the boiling point of water falls to about $94.6$ °C. In Leadville, Colorado, at $10\,152$ feet and about $70$ kPa, it falls to about $90$ °C.

Cooking speed depends on temperature, not on how vigorously the water bubbles. Many cooking reactions roughly double their rate for every $10$ °C, so water boiling at $90$ °C cooks about half as fast as at sea level: a three-minute egg needs five or six minutes in Leadville, and beans and pasta take noticeably longer. Baking needs different adjustments, because water in batter evaporates faster at lower pressure and leavening gases expand more, so high-altitude recipes use less leavening and slightly higher oven temperatures.

The same arithmetic worked against mountaineers on Everest, where water boils at about $71$ °C: it is impossible to cook rice properly there without a pressure cooker, which is why expeditions carry them.

10. In the world: the autoclave

Every hospital, dental office and laboratory sterilizes instruments in an autoclave, a steel pressure chamber that heats water to a temperature boiling alone could never reach. The standard cycle runs at $121$ °C, which by Clausius-Clapeyron needs an absolute pressure of about $205$ kPa — the $15$ psi above atmospheric printed on older machines' gauges.

The temperature matters because microbes are killed at a rate that depends steeply on it. The spores of Geobacillus stearothermophilus, the organism used to test autoclaves, survive boiling water for hours; at $121$ °C their number falls tenfold roughly every one to two minutes, so a fifteen-minute cycle reduces a million spores to essentially none. Steam, rather than hot air, is used because it gives up its latent heat — $2.2$ MJ per kilogram — as it condenses on the cooler instruments, heating them far faster than dry air could.

Home canning uses the same physics. The USDA requires low-acid foods such as green beans to be pressure-canned rather than boiled, because only the higher temperature reliably destroys the spores of Clostridium botulinum.

11. Pressure melting does not make ice slippery

The idea that a skate's pressure melts the ice under it is repeated in textbooks and trivia books alike, and it is wrong by orders of magnitude. The Clapeyron slope of ice's melting curve is about $-135$ atmospheres per kelvin. A skater exerts about ten atmospheres, enough to lower the melting point by less than a tenth of a degree, yet skating works at twenty degrees below freezing. Whatever makes ice slippery — a disordered surface layer and friction heating, according to current research — it is not pressure melting.

A second misconception is that boiling means reaching $100$ °C. Water boils at whatever temperature makes its vapor pressure equal to the surrounding pressure: $95$ °C in Denver, $71$ °C on Everest, $121$ °C in a pressure cooker, and at room temperature in a vacuum chamber. The number $100$ belongs to one pressure, one standard atmosphere, and is a definition that dates from the original Celsius scale rather than a property of water.

12. The slope of ice's melting curve

1. Ice melts at $273$ K with $L = 6010$ J/mol; its molar volume falls by $1.63 \times 10^{-6}$ m³. Write the Clapeyron relation.

   $\dfrac{dP}{dT} = \dfrac{L}{T\Delta V}$

   Equal chemical potentials along the boundary.

2. Assign the sign of the volume change.

   $\Delta V = V_{\text{water}} - V_{\text{ice}} = -1.63 \times 10^{-6}\ \text{m}^3/\text{mol}$

   Water is denser than ice.

3. Substitute the values.

   $\dfrac{dP}{dT} = \dfrac{6010}{273 \times (-1.63 \times 10^{-6})}$

   Joules per mole over kelvin times cubic meters per mole.

4. Evaluate the result.

   $\dfrac{dP}{dT} = -1.35 \times 10^{7}\ \text{Pa/K} \approx -135\ \text{atm/K}$

   A negative, very steep slope.

5. Find the pressure to lower the melting point by $2$ K.

   $\Delta P = 1.35 \times 10^{7} \times 2 = 2.7 \times 10^{7}\ \text{Pa} \approx 270\ \text{atm}$

   Far more than any skate blade exerts.

13. Boiling water in Denver

1. The pressure in Denver is about $83.4$ kPa. Write the equation for the boiling point.

   $\dfrac{1}{T} = \dfrac{1}{T_0} - \dfrac{R}{L}\ln\dfrac{P}{P_0}$

   Clausius-Clapeyron solved for $1/T$.

2. Find the pressure ratio's logarithm.

   $\ln\dfrac{83.4}{101.3} = \ln 0.8233 = -0.1944$

   Lower pressure, negative logarithm.

3. Find $R/L$ for water.

   $\dfrac{R}{L} = \dfrac{8.31}{40\,700} = 2.042 \times 10^{-4}\ \text{K}^{-1}$

   The molar latent heat of vaporization.

4. Substitute the values.

   $\dfrac{1}{T} = \dfrac{1}{373} + 2.042 \times 10^{-4} \times 0.1944 = 0.0026810 + 0.0000397$

   Subtracting a negative adds.

5. Add the terms.

   $\dfrac{1}{T} = 0.0027207\ \text{K}^{-1}$

   A slightly larger reciprocal, so a lower temperature.

6. Invert and convert.

   $T = 367.6\ \text{K} = 94.6\ ^{\circ}\text{C}$

   Water in Denver boils about five degrees below sea-level water.

14. The vapor pressure of water at room temperature

1. Estimate the vapor pressure of water at $25$ °C from its boiling point. Write Clausius-Clapeyron.

   $\ln\dfrac{P}{P_0} = -\dfrac{L}{R}\left(\dfrac{1}{T} - \dfrac{1}{T_0}\right)$

   With $P_0 = 101.3$ kPa at $T_0 = 373$ K.

2. Find the ratio $L/R$.

   $\dfrac{L}{R} = \dfrac{40\,700}{8.31} = 4898\ \text{K}$

   A temperature: the scale on which vapor pressure changes.

3. Find the difference of reciprocals.

   $\dfrac{1}{298} - \dfrac{1}{373} = 0.0033557 - 0.0026810 = 0.0006747\ \text{K}^{-1}$

   Compute this before multiplying; it is a small difference of similar numbers.

4. Multiply the factors.

   $\ln\dfrac{P}{P_0} = -4898 \times 0.0006747 = -3.305$

   A large negative logarithm.

5. Exponentiate both sides.

   $\dfrac{P}{P_0} = e^{-3.305} = 0.0367$

   About a twenty-seventh of an atmosphere.

6. Find the pressure.

   $P = 101.3 \times 0.0367 = 3.72\ \text{kPa}$

   The estimate.

7. Compare with the measured value.

   $P_{\text{measured}} = 3.17\ \text{kPa}$

   Seventeen percent high, because water's latent heat is larger at $25$ °C, about $44$ kJ/mol, than the boiling-point value assumed constant.

8. Say what the vapor pressure means for humidity.

   $\text{relative humidity} = \dfrac{P_{\text{vapor}}}{P_{\text{sat}}(T)}$

   Air at $25$ °C holding water vapor at $1.6$ kPa is at about $50$ percent relative humidity.

15. Your turn: in a pressure cooker at $205$ kPa absolute, at what temperature does water boil?

1. Find the logarithm of the pressure ratio.

   $\ln\dfrac{205}{101.3} = \ln 2.024 = 0.705$

   Higher pressure, positive logarithm.

2. Substitute into the equation for $1/T$.

   $\dfrac{1}{T} = \dfrac{1}{373} - 2.042 \times 10^{-4} \times 0.705 = 0.0026810 - 0.0001440$

   Subtracting lowers the reciprocal.

3. Your turn: work this step out. Its working is at the end of the packet.

   Invert and convert.

16. Guided practice

Ice at $-1$ °C is squeezed to $600$ atmospheres. What happens to its melting point, and why?

• It falls, because ice is less dense than water
• It rises, as melting points do under pressure
• It does not change, since pressure affects only boiling
• It falls, because pressure adds heat to the ice

17. Guided practice

Complete the worked solution: a solid melts at $400$ K with molar latent heat $3000$ J/mol, and its molar volume grows by $5 \times 10^{-6}$ m³/mol. What is the slope of its melting curve?

1. Multiply the temperature by the volume change.

   $T\Delta V = 400 \times 5 \times 10^{-6} =$ _p m³ K/mol

   The denominator of the Clapeyron relation.

2. Divide the latent heat by it.

   $\dfrac{dP}{dT} = \dfrac{3000}{T\Delta V} =$ _s Pa/K

   Joules over cubic meters is pascals.

3. Convert to megapascals per kelvin.

   $\dfrac{dP}{dT} =$ _m MPa/K

   Divide by a million; a melting curve is steep.

18. Guided practice

Match each feature of water's phase diagram to its meaning ($3$ of them are points or lines where phases meet).

19. Practice

Starting from $101.3$ kPa at $373$ K, use Clausius-Clapeyron with $L/R = 4898$ K to estimate the vapor pressure of water at $90$ °C ($363$ K). Fill in $\ln(P/P_0)$ and the pressure in kPa.

20. Practice

For nitrogen boiling at 77 K and one atmosphere, the molar latent heat is $5570$ J/mol and the molar volume changes by $0.006315$ m³/mol. What is the slope of the phase boundary, $dP/dT$, in MPa/K?

Answer: MPa/K

21. Practice

The air pressure in Leadville, Colorado is typically $70.1$ kPa. At what temperature, in °C, does water boil there? Water boils at $373$ K at $101.3$ kPa, and its molar latent heat is $40.7$ kJ/mol. Use $R = 8.31$ J/(mol K).

Answer: °C

22. Somewhere new

A stovetop pressure cooker at sea level is set to $12$ psi above atmospheric, an absolute pressure of $184$ kPa. At what temperature, in °C, does the water inside boil? Water boils at $373$ K at $101.3$ kPa, with $L = 40.7$ kJ/mol and $R = 8.31$ J/(mol K).

Answer: °C

23. Lesson test

Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.

24. Test question

Starting from $101.3$ kPa at $373$ K, use Clausius-Clapeyron with $L/R = 4898$ K to estimate the vapor pressure of water at $80$ °C ($353$ K). Fill in $\ln(P/P_0)$ and the pressure in kPa.

25. What you can do now

You can find how a boiling or melting point shifts with pressure. Explain to someone why pasta takes longer to cook in Denver, and why skate blades do not melt the ice.

Working for the steps left to you

15. Your turn: in a pressure cooker at $205$ kPa absolute, at what temperature does water boil?, step 3