Problem Statement
Using the Clausius-Clapeyron equation, find how much the boiling temperature of water changes when the atmospheric pressure changes by $\Delta p = 1.0\ ext{kPa}$. The specific heat of vaporization of water is $L = 2.26\ ext{MJ/kg}$, the density of saturated steam at $100° ext{C}$ is $
ho_ ext{steam} = 0.60\ ext{kg/m}^3$, and the density of water is $
ho_ ext{water} = 958\ ext{kg/m}^3$.
Given Information
- Pressure change: $\Delta p = 1.0\ ext{kPa} = 1000\ ext{Pa}$
- Latent heat of vaporization: $L = 2.26 imes 10^6\ ext{J/kg}$
- Density of steam at $100° ext{C}$: $
ho_v = 0.60\ ext{kg/m}^3$ - Density of water at $100° ext{C}$: $
ho_l = 958\ ext{kg/m}^3$ - Boiling point at standard pressure: $T_0 = 373\ ext{K}$
Physical Concepts & Formulas
The Clausius-Clapeyron equation describes how the coexistence pressure between two phases (liquid and vapour) changes with temperature. Along the liquid-vapour coexistence curve, the equilibrium condition gives:
6074 rac{dp}{dT} = rac{L}{T(v_v – v_l)}6074
where $v_v$ and $v_l$ are the specific volumes (volume per unit mass) of the vapour and liquid phases. Since $v_v \gg v_l$ for water vapour well below the critical point, $v_v – v_l pprox v_v = 1/
ho_v$.
- $\dfrac{dp}{dT} = \dfrac{L
ho_v}{T_0}$ — simplified Clausius-Clapeyron for vapour - $\Delta T = \Delta p \cdot \dfrac{T_0}{L
ho_v}$ — change in boiling point
Step-by-Step Solution
Step 1 — Write the Clausius-Clapeyron equation:
At the phase boundary, the chemical potentials of liquid and vapour are equal. The slope of the boiling curve is:
6074 rac{dp}{dT} = rac{L}{T(v_v – v_l)}6074
where $v_v = 1/
ho_v$ and $v_l = 1/
ho_l$ are the specific volumes.
Step 2 — Simplify using $v_v \gg v_l$:
Since $
ho_v = 0.60\ ext{kg/m}^3 \ll
ho_l = 958\ ext{kg/m}^3$, the specific volume of the liquid is negligible:
6074v_v – v_l pprox v_v = rac{1}{
ho_v} = rac{1}{0.60} = 1.667\ ext{m}^3/ ext{kg}6074
Therefore:
6074 rac{dp}{dT} pprox rac{L
ho_v}{T_0} = rac{2.26 imes 10^6 imes 0.60}{373}6074
Step 3 — Invert to get $dT/dp$:
6074 rac{dT}{dp} = rac{T_0}{L
ho_v}6074
For a finite pressure change $\Delta p$:
6074\Delta T = \Delta p \cdot rac{T_0}{L
ho_v}6074
Worked Calculation
6074 rac{dp}{dT} = rac{2.26 imes 10^6 imes 0.60}{373} = rac{1.356 imes 10^6}{373} = 3636\ ext{Pa/K}6074
Therefore:
6074\Delta T = rac{\Delta p}{dp/dT} = rac{1000\ ext{Pa}}{3636\ ext{Pa/K}} pprox 0.28\ ext{K}6074
Answer
6074oxed{\Delta T pprox 0.28\ ext{K per kPa}}6074
Physical Interpretation
The boiling point of water rises by about 0.28 K (0.28°C) for every 1 kPa increase in atmospheric pressure. This is why water boils at ~95°C at high altitudes (where pressure is lower) and at ~101°C in a pressure cooker at ~105 kPa. The result is a direct consequence of the Clausius-Clapeyron equation and the large specific volume of steam compared to liquid water.
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