When air expands adiabatically (without gaining or losing heat), its pressure P and volume V are...


When air expands adiabatically (without gaining or losing heat), its pressure {eq}P {/eq} and volume {eq}V {/eq} are related by the equation {eq}PV1.4=C {/eq} where {eq}C {/eq} is a constant. Suppose that at a certain instant the volume is {eq}390 cm^3 {/eq}, and the pressure is {eq}75 kPa {/eq} and is decreasing at a rate of {eq}8 kPa/min {/eq}.

At what rate is the volume increasing at this instant?

Implicit Differentiation

When we differentiate a function, we can do so with respect to any variable. This is because a quantity may change with respect to a variety of other quantities. This means that we may have to differentiate a function implicitly, treating the variables in the function as functions themselves, which means that we apply the Chain Rule to each piece of the function.

Answer and Explanation:

In this problem, we're given an equation that relates the pressure and volume of a gas that expands adiabatically. As both the pressure and volume are changing with respect to time, they are both functions of time. We can thus find how these rates of change are related if we differentiate this expression implicitly. The left side of this equation consists of the product of two functions, so we need to use the Product Rule in this differentiation.

{eq}\frac{d}{dt} PV^{1.4} = \frac{d}{dt}C\\ \frac{dP}{dt} V^{1.4} + P \cdot 1.4 V^{0.4} \frac{dV}{dt} = 0\\ V^{1.4} \frac{dP}{dt} + 1.4PV^{0.4} \frac{dV}{dt} = 0 {/eq}

This equation relates four quantities: the current value of the pressure and volume and the current rates of change of both pressure and volume. As we have three of these four numbers, we can find the remaining by substituting in the relevant information and solving for the unknown.

{eq}(390)^{1.4} (-8) + 1.4(75)(390)^{0.4} \frac{dV}{dt} = 0\\ 1141.866 \frac{dV}{dt} =33929.732588\\ \frac{dV}{dt} = 29.714286 {/eq}

Therefore, the volume is increasing at a rate of 29.715 cubic centimeters per minute.

Learn more about this topic:

Implicit Differentiation Technique, Formula & Examples

from GRE Math: Study Guide & Test Prep

Chapter 6 / Lesson 5

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