HCV Ch26 P3 – First Law: Isobaric (Constant Pressure) Process

Problem Statement

Solve the fluid mechanics problem: In an isobaric process, 2 mol of ideal monatomic gas is heated from 300 K to 400 K at pressure $P = 2 \times 10^5$ Pa. Find $\Delta U$, $W$, and $Q$. ($R = 8.314$ J/mol·K) See problem statement for all given quantities. This problem applies fundamental physics principles to the scenario described. T

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:

$$P_1 + \frac{1}{2}\rho v_1^2 + \rho g h_1 = P_2 + \frac{1}{2}\rho v_2^2 + \rho g h_2$$

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:

$$v = \sqrt{2gh} = \sqrt{2\times9.8\times2} = \sqrt{39.2} \approx 6.26\,\text{m/s}$$

Answer

$$\boxed{v_{\text{efflux}} = \sqrt{2gh}}$$

Physical Interpretation

6.26 m/s is the speed at which water exits a tank with a 2 m head. This equals the speed a ball would have after falling freely 2 m — Torricelli’s theorem is essentially Bernoulli applied to free fall. As the water level drops, the efflux speed decreases; refilling to maintain constant head is the principle behind constant-head reservoirs in ancient water clocks (clepsydras).


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