Problem Statement
Describe the pressure-temperature phase diagram of a pure substance, identifying the triple point, critical point, and coexistence curves.
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: Coexistence curves:
Step 2 — Apply the relevant physical law or equation:
- Solid-Liquid (melting curve): Usually has positive slope (liquid denser than solid). For water: negative slope (ice less dense than water), which is anomalous.
- Liquid-Gas (vapour pressure curve): Positive slope, described by Clausius-Clapeyron: $dp/dT = L/(T\Delta V) \approx Lp/(RT^2)$, giving $p\propto e^{-L/RT}$.
- Solid-Gas (sublimation curve): Positive slope, lower pressures.
Step 3 — Solve algebraically for the unknown: Triple point: All three phases coexist. For water: $T_{tp}=273.16\ \text{K}$, $p_{tp}=611\ \text{Pa}$.
Step 4 — Substitute numerical values with units: Critical point: Liquid-gas distinction disappears. For water: $T_c=647\ \text{K}$, $p_c=218\ \text{atm}$. Above $T_c$, there is a supercritical fluid.
Worked Calculation
$$\text{Numerical result} = \text{given expression substituted with values}$$
$$\boxed{T_c}$$
Coexistence curves:
- Solid-Liquid (melting curve): Usually has positive slope (liquid denser than solid). For water: negative slope (ice less dense than water), which is anomalous.
- Liquid-Gas (vapour pressure curve): Positive slope, described by Clausius-Clapeyron: $dp/dT = L/(T\Delta V) \approx Lp/(RT^2)$, giving $p\propto e^{-L/RT}$.
- Solid-Gas (sublimation curve): Positive slope, lower pressures.
Triple point: All three phases coexist. For water: $T_{tp}=273.16\ \text{K}$, $p_{tp}=611\ \text{Pa}$.
Critical point: Liquid-gas distinction disappears. For water: $T_c=647\ \text{K}$, $p_c=218\ \text{atm}$. Above $T_c$, there is a supercritical fluid.
Answer
$$\boxed{T_c}$$
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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