Problem 8 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

A gas in a perfectly insulated container and at constant temperature satisfies the gas law \(p v^{1.4}=\) constant. When the pressure is 20 Newtons per \(\mathrm{cm}^{2}\) the volume is 3 liters. The gas is being compressed at the rate of 0.2 liters per minute. How fast is the pressure changing at the instant at which the volume is 2 liters?

Step-by-Step Solution

Answer

The pressure is changing at -2.94 N/min when the volume is 2 liters.

Step 1: Understanding the problem

We are given that the gas follows the law \(p v^{1.4} = \text{constant}\). Initially, when the pressure \(p = 20\, \text{N/cm}^2\) and the volume \(v = 3\, \text{liters}\). We need to find how fast the pressure is changing (\(\frac{dp}{dt}\)) when the volume \(v = 2\, \text{liters}\) and the volume is decreasing at \(0.2\, \text{liters/min}\).

Step 2: Find the constant using initial conditions

Using the initial conditions, substitute \(p = 20\) and \(v = 3\) into the gas law equation: \(pv^{1.4} = 20 \times 3^{1.4}\). Evaluating this gives us the constant value, which we will call \(k\).

Step 3: Express pressure in terms of volume

Given \(pv^{1.4} = k\), solve for \(p\) in terms of \(v\): \(p = \frac{k}{v^{1.4}}\). This will allow us to differentiate with respect to time, \(t\).

Step 4: Differentiate with respect to time

Differentiate both sides of the equation \(p = \frac{k}{v^{1.4}}\) with respect to \(t\) to find \(\frac{dp}{dt}\). Use the chain rule: \(\frac{dp}{dt} = \frac{d}{dt}\left(\frac{k}{v^{1.4}}\right) = -1.4k v^{-2.4} \frac{dv}{dt}\).

Step 5: Substitute known values into differentiated equation

Substitute \(v = 2\), \(\frac{dv}{dt} = -0.2\) (since the volume is decreasing, it is negative), and \(k = 20 \times 3^{1.4}\) into the derivative expression: \(\frac{dp}{dt} = -1.4 \times \left(20 \times 3^{1.4}\right) \times (2)^{-2.4} \times (-0.2)\).

Step 6: Calculate the rate of change of pressure

Calculate \(\frac{dp}{dt}\) using the values substituted. This will give the rate of change of pressure in Newtons per minute at the instant when the volume is 2 liters.

Key Concepts

Differentiation

When solving related rates problems, differentiation is a key concept we use to find how variables change with respect to time. Differentiation provides a means to calculate the rate of change of one quantity as another quantity is changing. In this problem, we need to differentiate with respect to time to find out how the pressure of gas is changing.
It is important to understand that there are different kinds of differentiation. What we use here is implicit differentiation. This means we differentiate with respect to time even though our original equation doesn’t include time. Instead, it includes variables like pressure and volume which are implicitly functions of time.
Here is a step-by-step approach:

• You start with an equation that relates the variables. For our gas law, it's the equation \( p \, v^{1.4} = k \).
• The next step is to express one variable in terms of others, allowing us to differentiate with respect to time.
• Finally, you take the derivative applying rules like the power rule and chain rule, and solve for the desired rate of change.

This is a powerful technique, especially in physics and engineering, for modeling and understanding dynamic systems.

Gas Laws

The problem mentions the gas law \( p \, v^{1.4} = \text{constant} \), which is a particular form related to adiabatic processes. Perfectly insulated containers in which gas expands or compresses without heat exchange tend to follow such processes.
In simpler terms, the product of the pressure \( p \) and the volume raised to a power (1.4 in this case) remains constant. This constant showcases that if one quantity gets affected by an external force, causing it to change, the other must adjust in such a way as to keep that product steady.
Gas laws like these are pivotal to understanding how gases behave under different conditions and allow predictions about how changes in one property of a gas, such as volume, can affect others, like pressure. This is fundamental in fields like thermodynamics and fluid dynamics, where understanding such relationships can help in designing systems and applications ranging from engines to air conditioners.

Chain Rule

The chain rule is a fundamental tool in calculus that enables us to differentiate composite functions. It is also crucial for problems involving related rates where variables depend on time. In our problem, we apply the chain rule to find how the pressure changes as the volume changes over time.
The rule essentially states that to differentiate a composite function, you differentiate the outer function and then multiply it by the derivative of the inner function. This becomes especially important when variables are not explicit functions of time but are implicitly related.
In practice, this means:

• You start by differentiating the outer function \( f(v) = \frac{k}{v^{1.4}} \) with respect to \( v \).
• Then multiply by the derivative of \( v \) with respect to \( t \) (\( \frac{dv}{dt} \)).

This kind of step is commonplace when you work with continuously changing systems in science and engineering. Understanding how to apply the chain rule correctly is critical because it simplifies the process of working with complex equations that describe real-world phenomena.