Problem Statement
Find $T_c$, $p_c$, $V_c$ for a van der Waals gas in terms of $a$, $b$, $R$. Calculate for CO₂ ($a = 0.364$, $b = 42.9\ \text{cm}^3/\text{mol}$).
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: At the critical point $\partial p/\partial V = \partial^2 p/\partial V^2 = 0$ (per mole), giving:
Step 2 — Apply the relevant physical law or equation: $$T_c = \frac{8a}{27Rb},\quad p_c = \frac{a}{27b^2},\quad V_c = 3b$$
Step 3 — Solve algebraically for the unknown: For CO₂:
Step 4 — Substitute numerical values with units: $$T_c = \frac{8\times0.364}{27\times8.314\times4.29\times10^{-5}} = \frac{2.912}{9.62\times10^{-3}} \approx 303\ \text{K}$$
$$p_c = \frac{0.364}{27\times(4.29\times10^{-5})^2} = \frac{0.364}{4.97\times10^{-8}} \approx 7.3\times10^6\ \text{Pa} \approx 72\ \text{atm}$$
$$V_c = 3\times42.9 = 128.7\ \text{cm}^3/\text{mol}$$
Step 5 — Compute and check the result: Result: $T_c\approx 304\ \text{K}$, $p_c\approx 73\ \text{atm}$, $V_c\approx 129\ \text{cm}^3/\text{mol}$.
Worked Calculation
$$T_c = \frac{8a}{27Rb},\quad p_c = \frac{a}{27b^2},\quad V_c = 3b$$
$$T_c = \frac{8\times0.364}{27\times8.314\times4.29\times10^{-5}} = \frac{2.912}{9.62\times10^{-3}} \approx 303\ \text{K}$$
$$p_c = \frac{0.364}{27\times(4.29\times10^{-5})^2} = \frac{0.364}{4.97\times10^{-8}} \approx 7.3\times10^6\ \text{Pa} \approx 72\ \text{atm}$$
At the critical point $\partial p/\partial V = \partial^2 p/\partial V^2 = 0$ (per mole), giving:
$$T_c = \frac{8a}{27Rb},\quad p_c = \frac{a}{27b^2},\quad V_c = 3b$$
For CO₂:
$$T_c = \frac{8\times0.364}{27\times8.314\times4.29\times10^{-5}} = \frac{2.912}{9.62\times10^{-3}} \approx 303\ \text{K}$$
$$p_c = \frac{0.364}{27\times(4.29\times10^{-5})^2} = \frac{0.364}{4.97\times10^{-8}} \approx 7.3\times10^6\ \text{Pa} \approx 72\ \text{atm}$$
$$V_c = 3\times42.9 = 128.7\ \text{cm}^3/\text{mol}$$
Result: $T_c\approx 304\ \text{K}$, $p_c\approx 73\ \text{atm}$, $V_c\approx 129\ \text{cm}^3/\text{mol}$.
Answer
$$\boxed{V_c = 3\times42.9 = 128.7\ \text{cm}^3/\text{mol}}$$
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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