Chemical & Aerospace Engineering (Non-Linear Fluids) Solution

Problem Overview

We calculate the instantaneous rate of pressure change with respect to volume ( $\frac{dP}{dV}$ ) for a non-ideal gas governed by the Van der Waals equation of state:

(P + V^2)(V - 1) = 12

evaluated at the operational condition

V = 3\text{ L/mol}

and

P = 5\text{ atm}

Step-by-Step Mathematical Solution

Step 1: Formally Apply the Product Rule with Implicit Differentiation

The governing equation is the Product Rule:

\frac{d}{dV}[u \cdot v] = \frac{du}{dV} \cdot v + u \cdot \frac{dv}{dV}

(This rule is defined in Product Rule and Implicit Differentiation.)

Let $u = P + V^2$ and $v = V - 1$ . Treating $P$ as an implicit function $P(V)$ :

\frac{d}{dV}[P + V^2] \cdot (V - 1) + (P + V^2) \cdot \frac{d}{dV}[V - 1] = \frac{d}{dV}[12]

\left(\frac{dP}{dV} + 2V\right)(V - 1) + (P + V^2)(1) = 0

Step 2: Algebraically Isolate $\frac{dP}{dV}$

Distribute and group like terms:

(V - 1)\frac{dP}{dV} + 2V(V - 1) + P + V^2 = 0

(V - 1)\frac{dP}{dV} = -2V(V - 1) - P - V^2

\frac{dP}{dV} = \frac{-2V(V - 1) - P - V^2}{V - 1}

Step 3: Substitute Operational Metrics ( $V = 3\text{ L/mol}, P = 5\text{ atm}$ )

\left.\frac{dP}{dV}\right|_{(3,5)} = \frac{-2(3)(3 - 1) - 5 - (3)^2}{3 - 1} = \frac{-6(2) - 5 - 9}{2} = \frac{-12 - 14}{2} = \frac{-26}{2} = -13\text{ atm/(L/mol)}

Engineering Conclusion

At $V = 3\text{ L/mol}$ , the rate of pressure drop is $-13\text{ atm/(L/mol)}$ . In rocket combustion chambers and pressure vessels, $\frac{dP}{dV}$ is directly linked to the isothermal compressibility $\beta = -\frac{1}{V}\frac{\partial V}{\partial P}$ . A steep negative derivative indicates that small volume expansions cause rapid pressure drops, providing critical data to design fast-acting injector valves that prevent combustion blow-out or acoustic pressure spikes in liquid rocket engines.

Tony's Intertwingled Memex • Reference Manual