Problem Statement
At $T=373\ \text{K}$, water vaporises with $L=40.7\ \text{kJ/mol}$ and $\Delta V = 30.0\ \text{L/mol}$. Find $dp/dT$ for the liquid-vapour coexistence curve.
Given Information
- See problem statement for all given quantities.
Physical Concepts & Formulas
This problem applies fundamental physics principles to the scenario described. The solution requires identifying the relevant conservation laws and equations of motion, then solving systematically with careful attention to units and sign conventions.
- See the step-by-step solution for the specific equations applied.
- All quantities are in SI units unless otherwise stated.
Step-by-Step Solution
Step 1 — Identify given quantities and set up the problem: $$\frac{dp}{dT} = \frac{L}{T\Delta V} = \frac{40700}{373\times30.0\times10^{-3}} = \frac{40700}{11.19} = 3637\ \text{Pa/K}$$
Step 2 — Apply the relevant physical law or equation: Converting: $3637\ \text{Pa/K} = 36.4\ \text{mbar/K}$. So a 1 K rise in boiling temperature requires about 36 mbar higher pressure.
Step 3 — Solve algebraically for the unknown: In high-altitude cooking, the lower pressure reduces the boiling point significantly (e.g., at Everest base camp, $p\approx0.5\ \text{atm}$, water boils at $\approx82°\text{C}$).
Step 4 — Substitute numerical values with units: Result: $dp/dT \approx 3640\ \text{Pa/K} \approx 0.036\ \text{atm/K}$.
Worked Calculation
$$\frac{dp}{dT} = \frac{L}{T\Delta V} = \frac{40700}{373\times30.0\times10^{-3}} = \frac{40700}{11.19} = 3637\ \text{Pa/K}$$
$$\text{Numerical result} = \text{given expression substituted with values}$$
$$\boxed{\frac{dp}{dT} = \frac{L}{T\Delta V} = \frac{40700}{373\times30.0\times10^{-3}} = \frac{40700}{11.19} = 3637\ \text{Pa/K}}$$
$$\frac{dp}{dT} = \frac{L}{T\Delta V} = \frac{40700}{373\times30.0\times10^{-3}} = \frac{40700}{11.19} = 3637\ \text{Pa/K}$$
Converting: $3637\ \text{Pa/K} = 36.4\ \text{mbar/K}$. So a 1 K rise in boiling temperature requires about 36 mbar higher pressure.
In high-altitude cooking, the lower pressure reduces the boiling point significantly (e.g., at Everest base camp, $p\approx0.5\ \text{atm}$, water boils at $\approx82°\text{C}$).
Result: $dp/dT \approx 3640\ \text{Pa/K} \approx 0.036\ \text{atm/K}$.
Answer
$$\boxed{\frac{dp}{dT} = \frac{L}{T\Delta V} = \frac{40700}{373\times30.0\times10^{-3}} = \frac{40700}{11.19} = 3637\ \text{Pa/K}}$$
Physical Interpretation
The numerical answer is physically reasonable — matching expected orders of magnitude and dimensional analysis. The result confirms the theoretical prediction and provides quantitative insight into the system’s behaviour.
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