Problem 63 - College Algebra

Question

In chemistry the volume for a certain gas is given by \(V=20 T\) , where \(V\) is measured in \(c \mathrm{c}\) and \(T\) is temperature in \(^{\circ} \mathrm{C}\) . If the temperature varies between \(80^{\circ} \mathrm{C}\) and \(120^{\circ} \mathrm{C}\) , find the set of volume values.

Step-by-Step Solution

Answer

The volume values range from 1600 cc to 2400 cc.

Step 1: Identify the Given Equation

We are given the formula for volume as \( V = 20T \), where \( V \) is the volume in cubic centimeters \( (cc) \) and \( T \) is the temperature in degrees Celsius \( (^{\circ}C) \).

Step 2: Determine Temperature Range

The temperature \( T \) varies from \( 80^{\circ}C \) to \( 120^{\circ}C \). We need to find the corresponding volumes for these temperature values.

Step 3: Calculate Volume at Lower Temperature

Substitute the lower temperature \( T = 80^{\circ}C \) into the volume equation: \[ V = 20 \times 80 = 1600 \] The volume for \( T = 80^{\circ}C \) is \( 1600 \ cc \).

Step 4: Calculate Volume at Upper Temperature

Substitute the upper temperature \( T = 120^{\circ}C \) into the volume equation: \[ V = 20 \times 120 = 2400 \] The volume for \( T = 120^{\circ}C \) is \( 2400 \ cc \).

Step 5: Determine Set of Volume Values

The volume values range continuously from \( 1600 \ cc \) to \( 2400 \ cc \). Therefore, the set of volume values is \( \{ V \mid 1600 \leq V \leq 2400 \} \).

Key Concepts

Linear Equations

Linear equations form the backbone of many mathematical models, making them an essential concept in problem-solving. They are equations that plot a straight line when graphed and involve variables raised only to the power of one. This simplicity allows for straightforward calculations and predictions, which is why they are so prevalent in both mathematics and real-world applications.
In the given exercise, the relationship between temperature and volume is expressed as a linear equation: \( V = 20T \).
Here, \( V \) is the dependent variable, representing the volume, while \( T \) is the independent variable, representing temperature. The formula depicts a direct proportionality, implying that as temperature increases, the volume also increases linearly.

• The coefficient 20 in the equation signifies the rate of change, indicating that for each one-unit increase in temperature, the volume increases by 20 units.
• This linear relationship simplifies calculations and predictions about the volume of gas as the temperature changes.

Understanding linear equations equips students with the ability to model, solve, and predict scenarios in various scientific fields, including physics, chemistry, and economics.

Temperature-Volume Relationship

The temperature-volume relationship for gases is a fundamental principle in chemistry and physics, commonly known as Charles's Law. It states that, at constant pressure, the volume of a gas is directly proportional to its absolute temperature.
In our exercise, this concept is represented by the equation \( V = 20T \), demonstrating a straightforward relationship between temperature and volume.
Here's how it works:

• When the temperature of a gas increases, the molecules move faster and further apart, resulting in an increase in volume.
• Conversely, a decrease in temperature causes the molecules to slow down and occupy less space, leading to a decrease in volume.

This principle is vital in various applications, such as predicting how changes in temperature can affect the physical state or behavior of gases. It's also crucial in understanding natural phenomena and technological processes involving gases, like weather patterns and engine designs.

Problem-Solving Steps

Approaching any mathematical problem effectively requires a structured problem-solving methodology. This method begins with identifying the given information and understanding what the problem requires.

• First, we recognize the linear equation \( V = 20T \), where volume depends on temperature.
• Next, we determine the temperature range, given as 80°C to 120°C, which helps in predicting the corresponding volumes.
• Subsequently, we plug the lower temperature (80°C) into the equation to find the initial volume, resulting in 1600 cc.
• Following this, we substitute the upper temperature (120°C) into the same equation to find the maximum volume, calculated as 2400 cc.
• Finally, we identify the complete set of possible volume values, ranging from 1600 cc to 2400 cc.

This step-by-step approach ensures that each part of the problem is addressed systematically, allowing for clean and reliable solutions. Such strategies are not only useful in solving textbook exercises but also in tackling real-life mathematical modeling challenges.