Problem Statement
Describe the behaviour of van der Waals isotherms below the critical temperature and explain the Maxwell construction.
Given Information
- See problem statement for all given quantities.
Physical Concepts & Formulas
The van der Waals equation of state corrects the ideal gas law for finite molecular volume and intermolecular attractions. The parameter $a$ accounts for molecular attraction (reducing effective pressure) and $b$ for excluded volume (reducing available volume).
- $(P + a/V_m^2)(V_m – b) = RT$ — van der Waals equation (per mole)
- Critical point: $T_c = 8a/(27Rb)$, $P_c = a/(27b^2)$, $V_{m,c} = 3b$
- Compressibility factor: $Z = PV_m/RT$
Step-by-Step Solution
Step 1 — Identify given quantities and set up the problem: The van der Waals equation $(p+a/V^2)(V-b) = RT$ gives a cubic in $V$ for given $p$ and $T$. Below $T_c$, the cubic has three real roots corresponding to (1) liquid phase, (2) unstable region, (3) vapour phase.
Step 2 — Apply the relevant physical law or equation: The unstable region (where $\partial p/\partial V > 0$) is physically unobservable. The actual two-phase coexistence is found by the Maxwell equal-area rule: draw a horizontal line (constant $p$) through the isotherm such that the two loops above and below the line have equal areas.
Step 3 — Solve algebraically for the unknown: $$\int_{V_l}^{V_g} p_{VdW}\,dV = p_{coex}(V_g-V_l)$$
Step 4 — Substitute numerical values with units: This ensures equal chemical potentials of the two phases. The Maxwell construction gives the vapour pressure and the molar volumes of liquid ($V_l$) and vapour ($V_g$) at each temperature.
Worked Calculation
$$\int_{V_l}^{V_g} p_{VdW}\,dV = p_{coex}(V_g-V_l)$$
$$\text{Numerical result} = \text{given expression substituted with values}$$
$$\boxed{\int_{V_l}^{V_g} p_{VdW}\,dV = p_{coex}(V_g-V_l)}$$
The van der Waals equation $(p+a/V^2)(V-b) = RT$ gives a cubic in $V$ for given $p$ and $T$. Below $T_c$, the cubic has three real roots corresponding to (1) liquid phase, (2) unstable region, (3) vapour phase.
The unstable region (where $\partial p/\partial V > 0$) is physically unobservable. The actual two-phase coexistence is found by the Maxwell equal-area rule: draw a horizontal line (constant $p$) through the isotherm such that the two loops above and below the line have equal areas.
$$\int_{V_l}^{V_g} p_{VdW}\,dV = p_{coex}(V_g-V_l)$$
This ensures equal chemical potentials of the two phases. The Maxwell construction gives the vapour pressure and the molar volumes of liquid ($V_l$) and vapour ($V_g$) at each temperature.
Answer
$$\boxed{\int_{V_l}^{V_g} p_{VdW}\,dV = p_{coex}(V_g-V_l)}$$
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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