Problem 91 - Intermediate Algebra

Question

Solve each problem by writing a variation model. Gas Pressure. Under constant temperature, the volume occupied by a gas varies inversely to the pressure applied. If the gas occupies a volume of 20 cubic inches under a pressure of 6 pounds per square inch, find the volume when the gas is subjected to a pressure of 10 pounds per square inch.

Step-by-Step Solution

Answer

The volume is 12 cubic inches at 10 pounds per square inch.

Step 1: Understand the Inverse Variation

When something varies inversely, it can be expressed using the formula \( y = \frac{k}{x} \). In this context, the volume \( V \) varies inversely with the pressure \( P \). Our formula becomes \( V = \frac{k}{P} \), where \( k \) is a constant.

Step 2: Find the Constant (k)

We know the volume is 20 cubic inches when the pressure is 6 pounds per square inch. Plug these values into the formula to find \( k \): \( 20 = \frac{k}{6} \). Solving for \( k \), we multiply both sides by 6: \( k = 20 \times 6 = 120 \). So, our variation model is \( V = \frac{120}{P} \).

Step 3: Use the Model to Find the New Volume

Now, use our variation model to find the volume when the pressure is increased to 10 pounds per square inch. Substitute \( P = 10 \) into the model: \( V = \frac{120}{10} \). Calculate: \( V = 12\). So, the new volume is 12 cubic inches.

Key Concepts

Volume and Pressure Relationship

In the world of gases, understanding how volume and pressure relate is fundamental. These two properties are inversely proportional under constant temperature conditions.
This means that when one value increases, the other decreases. Think of it like a seesaw where if one side goes up, the other side must go down.

When you have a greater pressure applied to a gas, its volume will shrink given that temperature remains constant. This relationship can be mathematically described using the formula:

• \( V = \frac{k}{P} \)

Here, \( V \) represents the volume, \( P \) represents the pressure, and \( k \) is a constant that holds the product of volume and pressure as a steady value.

Understanding this relationship is essential for anyone studying gases, whether you're working with tanks, balloons, or any other applications.

Variation Models

Variation models are essential tools in mathematics because they help describe how one quantity changes with respect to another. There are different types of variations, and understanding which applies is crucial.
In our situation, we're dealing with inverse variation.

Inverse variation can be observed in real-world scenarios where as one value rises, another must fall to keep a constant product:

• Expressed as \( y = \frac{k}{x} \)
• "\( y \) varies inversely with \( x \)" means that multiplying \( x \) and \( y \) gives the constant \( k \)

This formula is powerful as it allows us to predict how one value changes when another does, by maintaining the constant \( k \).
For gases, this ensures that as pressure increases or decreases, we know exactly how volume will adjust.

Problem Solving in Algebra

Problem-solving using algebra can initially seem intimidating, but it's actually about breaking down the problem into manageable steps. Here's how you can tackle a gas-related inverse variation problem.

First, identify what you know and what you need to find. In our gas pressure problem:

• You are given initial values: volume and pressure.
• You need a new volume value under changed pressure.

Next, use the inverse variation formula to determine the constant \( k \):

• With given conditions, plug them into \( V = \frac{k}{P} \)

Now, find \( k \) through calculation, and establish the variation model. With this model:

• Substitute new conditions to find unknown value.

This structured approach converts complex problems into simple steps. With practice, you'll handle inverse variations and other algebraic issues effortlessly.