Questions: Phase Equilibrium and Clausius-Clapeyron Equation

Question 1

A pressure cooker operates at approximately 2 atm (twice atmospheric pressure). Food cooks significantly faster than in an open pot. Why?

• A. Higher pressure forces heat to transfer more rapidly from the steam into the food
• B. At higher pressure, the saturation temperature is higher (per the Clausius-Clapeyron equation), so water remains liquid above 100°C and cooks food at a higher temperature
• C. Compressed steam at 2 atm has twice the thermal energy of atmospheric steam at the same temperature
• D. Higher pressure increases the specific heat of water, allowing it to store and release more energy

Answer: (B)

The Clausius-Clapeyron equation dp/dT = h_fg/(T·Δv) tells us that saturation pressure increases with temperature. Equivalently, fixing higher pressure forces the saturation temperature higher — at 2 atm, water boils at about 121°C instead of 100°C. Food chemistry (enzymatic breakdown, protein denaturation, starch gelatinization) proceeds much faster at 121°C than 100°C — roughly doubling for every 10°C increase. The cooking speed advantage comes entirely from the elevated temperature, enabled by the higher pressure keeping water liquid above its atmospheric boiling point.

Question 2

The solid-liquid phase boundary for water on a P-T diagram has a negative slope — pressure increases as temperature decreases along the boundary. Which statement correctly identifies why this is unusual and what causes it?

• A. Water has an unusually high latent heat of fusion, making dp/dT very negative
• B. Water expands on freezing, so Δv = v_liquid − v_ice is negative; since dp/dT = h_fus/(T·Δv), the slope is negative
• C. The entropy change on melting is negative for water, reversing the Clausius-Clapeyron sign
• D. Ice has a higher specific heat than water, which inverts the Clausius-Clapeyron relationship

Answer: (B)

Most substances contract on freezing (solid denser than liquid), giving a positive Δv at melting and a positive dp/dT slope for the solid-liquid boundary. Water is anomalous: ice is less dense than liquid water (ice floats), so Δv = v_liquid − v_ice < 0. In the Clausius-Clapeyron equation dp/dT = h_fus/(T·Δv), both h_fus and T are positive, so the sign of dp/dT is determined by Δv — which is negative for water. This means the melting point decreases under pressure (ice melts under pressure), a property with important consequences for ice skating physics and glacial flow.

Question 3

According to the Clausius-Clapeyron equation, saturation pressure grows approximately linearly with temperature for typical liquid-vapor transitions over moderate temperature ranges.

• T. True
• F. False

Answer: False

Saturation pressure grows approximately exponentially with temperature, not linearly. The integrated approximate form is ln(P₂/P₁) ≈ (h_fg/R)(1/T₁ − 1/T₂), which means P grows exponentially with 1/T (or equivalently, exponentially with temperature in a linearized approximation). For water near 100°C, saturation pressure roughly doubles for every 20°C rise — a much faster-than-linear growth. This exponential relationship is why small changes in temperature near a saturation point produce large changes in pressure, and why pressure cookers and steam turbines operate in the regimes they do.

Question 4

The Clausius-Clapeyron equation can be derived by requiring that the Gibbs free energies of the two coexisting phases are equal, and that this equality must be maintained as pressure and temperature change together along the phase boundary.

• T. True
• F. False

Answer: True

At any point on a phase boundary, the two phases coexist in equilibrium, which requires equal Gibbs free energies: G_liquid = G_vapor (for liquid-vapor). As we move along the boundary (changing P and T together), this equality must remain satisfied: dG_liquid = dG_vapor. Using the thermodynamic identity dG = V dP − S dT and equating the differentials gives (V_g − V_f)dP = (S_g − S_f)dT, or dp/dT = ΔS/ΔV. Substituting ΔS = h_fg/T yields the Clausius-Clapeyron equation. The derivation thus follows directly from the condition of thermodynamic equilibrium along the phase boundary.

Question 5

What does the Clausius-Clapeyron equation tell us about how saturation pressure changes with temperature, and why does this make pressure cookers work more effectively at cooking food?

Model answer: The Clausius-Clapeyron equation dp/dT = h_fg/(T·Δv) gives the slope of the liquid-vapor boundary on a P-T diagram — i.e., how much the saturation pressure changes for a given change in saturation temperature. The integrated form shows vapor pressure grows approximately exponentially with temperature. Reading this in reverse: fixing a higher pressure (by sealing the cooker) forces the boiling point to a higher temperature. Water in a sealed pressure cooker at ~2 atm does not boil until about 121°C. At this temperature, chemical reactions in food — protein denaturation, starch gelatinization, collagen breakdown — proceed much faster (reaction rates roughly double per 10°C). The food therefore cooks in a fraction of the time. Without the Clausius-Clapeyron relationship, the water would simply boil away at 100°C regardless of pressure, which would not allow higher-temperature cooking.

The key conceptual point: pressure and saturation temperature are linked (not independent), so imposing higher pressure imposes higher temperature for the phase transition. The Clausius-Clapeyron equation quantifies this linkage.