Problem Statement
Solve the work-energy problem: Solve the work-energy problem: Write the Helmholtz free energy of a van der Waals gas and use it to derive the equation of state. For a van der Waals gas, integrating from $U = \nu C_v T – a\nu^2/V$: $$F = U – TS = \nu C_v T – \frac{a\nu^2}{V} – \nu RT\ln\frac{(V-\nu b)T^{C_v/R}}{\nu\Lambda^3}$$ (wh
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 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Step 2 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Step 3 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Worked Calculation
$$F = U – TS = \nu C_v T – \frac{a\nu^2}{V} – \nu RT\ln\frac{(V-\nu b)T^{C_v/R}}{\nu\Lambda^3}$$
$$\frac{1}{2}mv_i^2 + mgh_i = \frac{1}{2}mv_f^2 + mgh_f$$
$$E_i – W_{\text{friction}} = E_f \implies \frac{1}{2}mv_i^2 + mgh_i = \frac{1}{2}mv_f^2 + mgh_f + f_k d$$
Answer
$$\boxed{v_f = \sqrt{2g h}}$$
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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