Problem 29
Question
The volume \((V)\) of a gas varies directly as the temperature \((T)\) and inversely as the pressure \((P)\). If \(V=48\) when \(T=320\) and \(P=20\), find \(V\) when \(T=280\) and \(P=30 . \quad V=28\)
Step-by-Step Solution
Verified Answer
When \( T = 280 \) and \( P = 30 \), the volume \( V = 28 \).
1Step 1: Understanding Direct and Inverse Variation
In this problem, the volume \( V \) of a gas varies directly with temperature \( T \) and inversely with pressure \( P \). This can be expressed as the equation \( V = k \cdot \frac{T}{P} \), where \( k \) is a constant of proportionality.
2Step 2: Finding the Constant of Proportionality
We know \( V = 48 \) when \( T = 320 \) and \( P = 20 \). Substitute these values into the equation: \[ 48 = k \cdot \frac{320}{20} \]. Simplify to find \( k \): \[ 48 = k \cdot 16 \], thus \( k = \frac{48}{16} = 3 \).
3Step 3: Using the Constant to Find New Volume
Given the values \( T = 280 \) and \( P = 30 \), and using the constant \( k = 3 \) from the previous step, substitute into the formula: \[ V = 3 \cdot \frac{280}{30} \]. Simplify to find \( V \): \[ V = 3 \cdot \frac{28}{3} = 3 \cdot 9.3333 = 28 \].
Key Concepts
Proportionality ConstantGas LawsCollege Algebra
Proportionality Constant
When dealing with variations between variables, we often encounter the term 'proportionality constant.' This is denoted by the letter \( k \) and plays a pivotal role in maintaining the relationship between quantities.
In scenarios of direct variation, two quantities increase or decrease together in the same ratio. This means if one value doubles, so does the other. In the given problem, the volume \( V \) of gas directly varies with temperature \( T \). On the other hand, inverse variation implies that one quantity increases while the other decreases. Here, volume \( V \) inversely varies with pressure \( P \).
The equation representing the relationship is \( V = k \cdot \frac{T}{P} \). The constant \( k \) allows us to encapsulate the initial conditions or the specific relationship between these variables once we have one data point. After determining \( k = 3 \), it enables us to predict \( V \) for different values of \( T \) and \( P \), ensuring the mathematical relationship is preserved.
In scenarios of direct variation, two quantities increase or decrease together in the same ratio. This means if one value doubles, so does the other. In the given problem, the volume \( V \) of gas directly varies with temperature \( T \). On the other hand, inverse variation implies that one quantity increases while the other decreases. Here, volume \( V \) inversely varies with pressure \( P \).
The equation representing the relationship is \( V = k \cdot \frac{T}{P} \). The constant \( k \) allows us to encapsulate the initial conditions or the specific relationship between these variables once we have one data point. After determining \( k = 3 \), it enables us to predict \( V \) for different values of \( T \) and \( P \), ensuring the mathematical relationship is preserved.
Gas Laws
Gas laws help us understand how different conditions affect the behavior of gases. The direct and inverse variation present in the problem reflect important principles of gas laws, particularly the Combined Gas Law. This law integrates the relationships of Boyle's and Charles's laws, considering pressure, volume, and temperature.
According to Charles's Law, volume increases with temperature when pressure is constant, showing a direct relationship. Conversely, Boyle's Law illustrates that volume decreases as pressure increases (if the temperature is constant), which is an inverse relationship.
In practical applications such as this exercise, the understanding of these laws determines how a change in temperature or pressure might affect the gas volume. Using the equation: \( V = k \cdot \frac{T}{P} \), we combine both these relationships. Hence, any adjustments to \( T \) or \( P \) result in predictable changes in \( V \), emphasized by the role of the proportionality constant.
According to Charles's Law, volume increases with temperature when pressure is constant, showing a direct relationship. Conversely, Boyle's Law illustrates that volume decreases as pressure increases (if the temperature is constant), which is an inverse relationship.
In practical applications such as this exercise, the understanding of these laws determines how a change in temperature or pressure might affect the gas volume. Using the equation: \( V = k \cdot \frac{T}{P} \), we combine both these relationships. Hence, any adjustments to \( T \) or \( P \) result in predictable changes in \( V \), emphasized by the role of the proportionality constant.
College Algebra
College Algebra gives us the tools to solve complex real-world problems using mathematical principles. In this exercise, algebraic manipulation is essential for finding unknowns and validating relationships between variables.
The process entails identifying variables, setting up an equation, and solving for an unknown, often involving substitution and simplification. In our example, we use algebra to first find the constant of proportionality, initializing our equation based on the given conditions.
Once \( k \) is found, algebra streamlines predicting new scenarios by plugging in different values for known variables. This systematic approach fosters critical thinking and problem-solving, foundational skills in algebra. Ensuring procedural precision, such as simplifying fractions or verifying calculations, is crucial, especially when teaching students to effectively use algebra in varied contexts, from theoretical exercises to practical applications like analyzing gas behaviors.
The process entails identifying variables, setting up an equation, and solving for an unknown, often involving substitution and simplification. In our example, we use algebra to first find the constant of proportionality, initializing our equation based on the given conditions.
Once \( k \) is found, algebra streamlines predicting new scenarios by plugging in different values for known variables. This systematic approach fosters critical thinking and problem-solving, foundational skills in algebra. Ensuring procedural precision, such as simplifying fractions or verifying calculations, is crucial, especially when teaching students to effectively use algebra in varied contexts, from theoretical exercises to practical applications like analyzing gas behaviors.
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