Problem Statement
Solve the fluid mechanics problem: Show that along the critical isotherm ($T=T_c$), the van der Waals pressure varies as $p – p_c \propto (V-V_c)^3$ near $V_c$. Expand the van der Waals pressure in reduced variables near the critical point. Setting $\phi = V/V_c = 1+x$ (small $x$) and $\tau=1$: $$\pi = \frac{8\tau/3}{\phi-1/3} – \fra
Given Information
- See problem statement for all given quantities.
Physical Concepts & Formulas
The van der Waals equation of state corrects the ideal gas law for finite molecular volume and intermolecular attractions. The parameter $a$ accounts for molecular attraction (reducing effective pressure) and $b$ for excluded volume (reducing available volume).
- $(P + a/V_m^2)(V_m – b) = RT$ — van der Waals equation (per mole)
- Critical point: $T_c = 8a/(27Rb)$, $P_c = a/(27b^2)$, $V_{m,c} = 3b$
- Compressibility factor: $Z = PV_m/RT$
Step-by-Step Solution
Step 1 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Step 2 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Step 3 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Worked Calculation
$$\pi = \frac{8\tau/3}{\phi-1/3} – \fra
Given Information
- Fluid density $\rho$, velocities, cross-sections, and heights as given
- Atmospheric pressure $P_0 = 1.013\times10^5\,\text{Pa}$
- $g = 9.8\,\text{m/s}^2$
Physical Concepts & Formulas
Fluid statics is governed by Pascal’s Law and the hydrostatic pressure formula $P = P_0 + \rho g h$. Archimedes’ principle states that the buoyant force equals the weight of fluid displaced: $F_b = \rho_f V g$. Fluid dynamics for ideal (incompressible, non-viscous, steady) flow uses two key results: the continuity equation $A_1 v_1 = A_2 v_2$ (mass conservation) and Bernoulli’s equation $P + \frac{1}{2}\rho v^2 + \rho g h = \text{const}$ (energy conservation per unit volume).
- $P = P_0 + \rho g h$ — hydrostatic pressure
- $F_b = \rho_f V g$ — Archimedes buoyancy
- $A_1 v_1 = A_2 v_2$ — continuity equation
- $P + \tfrac{1}{2}\rho v^2 + \rho g h = \text{const}$ — Bernoulli’s equation
- $v_{\text{efflux}} = \sqrt{2gh}$ — Torricelli’s theorem
Step-by-Step Solution
Step 1 — Identify the situation: Static (use $P = P_0+\rho gh$ and Archimedes) or dynamic (use continuity + Bernoulli).
Step 2 — Apply Bernoulli between two points at the same streamline:
$$
$$
Step 3 — Use continuity to relate $v_1$ and $v_2$: $v_2 = v_1 A_1/A_2$.
Step 4 — Solve for the unknown (pressure, velocity, or flow rate).
Worked Calculation
Substituting all values with units:
Torricelli: Tank depth $h = 2\,\text{m}$, hole at bottom:
$$
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Answer
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Answer
$$\boxed{v_{\text{efflux}} = \sqrt{2gh}}$$
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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