Problem 54 - College Algebra with Corequisite Support

Question

For the following exercises, use the given information to answer the questions. The volume of a gas held at constant temperature varies indirectly as the pressure of the gas. If the volume of a gas is 1200 cubic centimeters when the pressure is 200 millimeters of mercury, what is the volume when the pressure is 300 millimeters of mercury?

Step-by-Step Solution

Answer

800 cubic centimeters.

Step 1: Understand the Relationship

Since the volume of the gas varies indirectly with the pressure, we can set up the relationship as \( V = \frac{k}{P} \), where \( V \) is the volume, \( P \) is the pressure, and \( k \) is a constant.

Step 2: Determine the Constant of Variation

We are given that the volume \( V = 1200 \) cubic centimeters when the pressure \( P = 200 \) mm of mercury. Substituting these values into the equation \( V = \frac{k}{P} \), we get: \[ 1200 = \frac{k}{200} \]. Solving for \( k \), we multiply both sides by 200: \[ k = 1200 \times 200 = 240000 \].

Step 3: Find the Volume at New Pressure

Now that we know \( k = 240000 \), we can find the volume when the pressure is 300 mm of mercury. Using the formula \( V = \frac{k}{P} \), substitute \( k = 240000 \) and \( P = 300 \): \[ V = \frac{240000}{300} \].

Step 4: Calculate the Volume

Perform the division: \[ V = \frac{240000}{300} = 800 \] cubic centimeters. Therefore, the volume of the gas when the pressure is 300 mm of mercury is 800 cubic centimeters.

Key Concepts

Constant of Variation

When we're talking about indirect variation, the "constant of variation" is a crucial concept. Imagine it as a bridge or a connector between two variables, in our case, volume and pressure. In indirect variation, this constant keeps the two variables balanced, even as they change.
In mathematical terms, if volume \( V \) varies indirectly as pressure \( P \), the equation that describes this relationship is \( V = \frac{k}{P} \). Here, \( k \) is what we call the constant of variation.

• The constant \( k \) remains the same no matter what happens to \( V \) and \( P \), as long as all other conditions remain constant.
• By calculating \( k \), we can predict one variable if we know the other.
• In the example, \( k \) was found using the initial given values: 1200 cubic centimeters for volume and 200 mm of mercury for pressure.

To solve for \( k \), we multiplied the volume by the pressure: \( k = 1200 \times 200 = 240000 \). This calculation gives us a constant that helps find new values once we know the new pressure or volume.

Volume and Pressure Relationship

The relationship between volume and pressure in this exercise is a classic example of Boyle's Law. Boyle's Law states that for a given mass of gas at constant temperature, the volume of the gas is inversely proportional to the pressure exerted on it.With inverse or indirect variation, as volume increases, pressure decreases, and vice versa, provided temperature remains constant. This means more pressure compresses the gas, reducing its volume, and less pressure allows the gas to expand, increasing its volume.
Consider our given problem, where \( V \) and \( P \) are linked by the equation \( V = \frac{k}{P} \):

• Initially, the volume is 1200 cubic centimeters, and the pressure is 200 mm of mercury.
• We then used the constant of variation \( k = 240000 \) to find out the new volume when the pressure increases to 300 mm of mercury.

The equation is quite powerful because it mathematically explains how volume and pressure relate under certain conditions, making it easy to calculate the unknowns when any change occurs.

Algebraic Equation Solving

Algebra plays a vital role in solving indirect variation problems like this one. You need to know how to manipulate equations to find unknown values. The steps are logical and follow basic arithmetic rules.
Start by setting up your equation \( V = \frac{k}{P} \). From here, solve for any missing value using these steps:

• Plug in known values to find the constant \( k \). For our exercise, this was done by using 1200 for \( V \) and 200 for \( P \), resulting in \( k = 240000 \).
• Rearrange the equation to find the unknown. If the pressure changes, use the known \( k \) to calculate the new volume.
• Substitute the new pressure into the equation with \( k \) to solve for \( V \). So, with a pressure of 300 mm, the calculation was \( V = \frac{240000}{300} \), which simplified to 800 cubic centimeters.

By following these algebraic steps, indirect variation problems become straightforward. It's all about understanding the relationship and applying equation-solving techniques systematically.