Problem Statement
Find the Boyle temperature $T_B$ for a van der Waals gas. Calculate for N₂ ($a = 0.136\ \text{J·m}^3/\text{mol}^2$, $b = 38.5\ \text{cm}^3/\text{mol}$).
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 second virial coefficient for a van der Waals gas is $B(T) = b – a/(RT)$. It vanishes at $T_B$:
Step 2 — Apply the relevant physical law or equation: $$T_B = \frac{a}{Rb} = \frac{0.136}{8.314\times38.5\times10^{-6}} = \frac{0.136}{3.20\times10^{-4}} \approx 425\ \text{K}$$
Step 3 — Solve algebraically for the unknown: Below $T_B$ attractive forces dominate; above, repulsive forces dominate.
Step 4 — Substitute numerical values with units: Result: $T_B \approx 425\ \text{K}$.
Worked Calculation
$$T_B = \frac{a}{Rb} = \frac{0.136}{8.314\times38.5\times10^{-6}} = \frac{0.136}{3.20\times10^{-4}} \approx 425\ \text{K}$$
$$\text{Numerical result} = \text{given expression substituted with values}$$
$$\boxed{T_B = \frac{a}{Rb} = \frac{0.136}{8.314\times38.5\times10^{-6}} = \frac{0.136}{3.20\times10^{-4}} \approx 425\ \text{K}}$$
The second virial coefficient for a van der Waals gas is $B(T) = b – a/(RT)$. It vanishes at $T_B$:
$$T_B = \frac{a}{Rb} = \frac{0.136}{8.314\times38.5\times10^{-6}} = \frac{0.136}{3.20\times10^{-4}} \approx 425\ \text{K}$$
Below $T_B$ attractive forces dominate; above, repulsive forces dominate.
Result: $T_B \approx 425\ \text{K}$.
Answer
$$\boxed{T_B = \frac{a}{Rb} = \frac{0.136}{8.314\times38.5\times10^{-6}} = \frac{0.136}{3.20\times10^{-4}} \approx 425\ \text{K}}$$
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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