Problem 32
Question
Charles’ Law states that when a sample of gas is kept at a constant pressure, its volume V will increase directly as the temperature \(t\). Write an equation for Charles’ Law that expresses volume as a function
Step-by-Step Solution
Verified Answer
The equation for Charles' Law is \(V(t) = k \cdot t\), where \(k\) is a constant.
1Step 1: Understand the Relationship
Charles' Law describes a direct proportional relationship between the volume (\(V\)) of a gas and its temperature (\(t\)) when pressure is constant. This implies that as temperature increases, the volume increases proportionally.
2Step 2: Define the Proportionality
Since volume and temperature are directly proportional, we can express this relationship as \(V \propto t\). This means there is a constant \(k\) such that \(V = k \cdot t\), where \(k\) is the proportionality constant.
3Step 3: Write the Equation
Based on the direct relationship and the proportionality constant, the equation expressing Charles' Law in terms of volume as a function of temperature is given by \(V(t) = k \cdot t\), where \(V(t)\) denotes the volume as a function of temperature \(t\).
4Step 4: Interpret the Equation
In the equation \(V(t) = k \cdot t\), \(k\) represents how much the volume changes per unit change in temperature. This constant \(k\) is determined experimentally based on the specific gas and conditions.
Key Concepts
Understanding Direct ProportionalityExploring Volume-Temperature RelationshipThe Role of the Proportionality Constant
Understanding Direct Proportionality
In math and science, direct proportionality is a simple yet powerful concept. It tells us how two values change together in a consistent manner. For Charles' Law, we explore the direct proportionality between volume and temperature of a gas.
This means that when you keep the pressure constant, if you increase the temperature, the volume goes up too. It's like they move at the same rhythm or pace.
This means that when you keep the pressure constant, if you increase the temperature, the volume goes up too. It's like they move at the same rhythm or pace.
- If temperature doubles, volume doubles.
- If temperature triples, the volume triples too.
Exploring Volume-Temperature Relationship
The volume-temperature relationship is central to Charles' Law. This relationship indicates how sensitive the volume of a gas is to changes in temperature. Imagine a balloon expanding as it gets warmer. That expansion showcases this core idea.
In a mathematical sense, this relationship is described with the formula:\[ V = k \times t \]Here, \(V\) is the volume and \(t\) is the temperature. This formula tells us that the volume of a gas is calculated by multiplying the temperature by a constant factor, known as the proportionality constant.
In a mathematical sense, this relationship is described with the formula:\[ V = k \times t \]Here, \(V\) is the volume and \(t\) is the temperature. This formula tells us that the volume of a gas is calculated by multiplying the temperature by a constant factor, known as the proportionality constant.
- This helps predict how much a gas would expand if it got hotter.
- It is always assuming the pressure around the gas stays the same.
The Role of the Proportionality Constant
In the equation used for Charles' Law, the proportionality constant \(k\) plays an essential role. This constant helps quantify the exact relationship between volume and temperature. Quite simply, it measures how much the volume of a gas changes with each degree of temperature change.
Finding \(k\) is crucial because this value doesn't stay the same for all gases. It's determined by experimental data and can vary based on:
This constant allows scientists and engineers to design systems that can handle changes in temperature, like air conditioning units and hot air balloons.
Finding \(k\) is crucial because this value doesn't stay the same for all gases. It's determined by experimental data and can vary based on:
- The type of gas you're examining.
- The conditions under which you're studying the gas.
This constant allows scientists and engineers to design systems that can handle changes in temperature, like air conditioning units and hot air balloons.
Other exercises in this chapter
Problem 31
Simplify each expression. \(\frac{\frac{p^{3}}{2 q}}{-\frac{p^{2}}{4 q}}\)
View solution Problem 32
In 2006, the cost to mail a first-class letter was 39¢ for any weight up to and including 1 ounce. Each additional ounce or part of an ounce added 24¢ to the co
View solution Problem 32
Simplify each expression. $$ \frac{\frac{1}{b+2}+\frac{1}{b-5}}{\frac{2 b^{2}-b-3}{b^{2}-3 b-10}} $$
View solution Problem 32
Simplify each expression. \(\frac{\frac{m+n}{5}}{\frac{m^{2}+n^{2}}{5}}\)
View solution