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Boyle's Law states that when a sample of gas is compressed at a constant temperature, the pressure \(P\) of the gas is inversely proportional to the volume \(V\) of the gas. (a) Suppose that the pressure of a sample of air that occupies \(0.106 \mathrm{m}^{3}\) at \(25^{\circ} \mathrm{C}\) is \(50 \mathrm{kPa}\). Write \(V\) as a function of \(P\). (b) Calculate \(d V / d P\) when \(P=50 \mathrm{kPa}\). What is the meaning of the derivative? What are its units?

Short Answer

Expert verified
\(V = \frac{5.3}{P}\); \(\frac{dV}{dP} = -0.00212\, \text{m}^3/\text{kPa}\). It indicates change in volume per unit pressure change.

Step by step solution

01

Understanding Boyle's Law

According to Boyle's Law, pressure \(P\) and volume \(V\) are inversely proportional when temperature is held constant. Mathematically, this relationship can be expressed as \(P \cdot V = k\), where \(k\) is a constant.
02

Solving for k

Given that \(P = 50\, \text{kPa}\) and \(V = 0.106\, \text{m}^3\), substitute these values into the equation \(P \cdot V = k\) to find \(k\). \[50 \times 0.106 = k \]\(k = 5.3\, \text{kPa} \cdot \text{m}^3\).
03

Expressing V as a function of P

Rearrange the equation \(P \cdot V = k\) to express \(V\) as a function of \(P\): \[V = \frac{k}{P} = \frac{5.3}{P}\].
04

Calculating the derivative dV/dP

To find the derivative \(\frac{dV}{dP}\), differentiate the expression \(V = \frac{5.3}{P}\) with respect to \(P\). \[\frac{dV}{dP} = -\frac{5.3}{P^2}\].
05

Evaluate dV/dP at P = 50 kPa

Substitute \(P = 50\, \text{kPa}\) into the derivative found in Step 4 to evaluate it:\[\frac{dV}{dP} = -\frac{5.3}{(50)^2} = -\frac{5.3}{2500} = -0.00212\].
06

Interpretation and Units

The derivative \(\frac{dV}{dP} = -0.00212\, \text{m}^3/\text{kPa}\) represents how the volume changes with respect to pressure changes. The negative sign indicates that as pressure increases, volume decreases.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Inverse Proportionality
Inverse proportionality is a core concept in understanding Boyle's Law. When two variables are inversely proportional, it means that as one variable increases, the other decreases. In the context of Boyle's Law, this involves the pressure (\(P\)) and volume (\(V\)) of a gas. These variables are linked by the equation \(P \cdot V = k\), where \(k\) is a constant, assuming temperature remains unchanged.

To dive deeper, think of inverse proportionality as relationships on a see-saw. As one side goes up, the other comes down. Thus, if pressure rises while the gas is compressed, the volume must decrease in response.

For Boyle's Law: - Pressure (\(P\)) and volume (\(V\)) have an inverse association. - The relationship is expressed by \(P \cdot V = k\). - At any point, reducing the volume of a gas by, let's say half, would theoretically double the pressure, given constant temperature.
Derivative Interpretation
The derivative is a fantastically helpful tool in understanding how one variable changes in relation to another. In this context, we explore the derivative \(\frac{dV}{dP}\). This derivative tells us how the volume \(V\) changes as the pressure \(P\) changes. A derivative, therefore, is just a fancy way of saying "rate of change."

When we compute \(\frac{dV}{dP} = -0.00212\), this value shows how much volume will decrease if pressure is increased by a small increment at a given moment. In the interpretation, the negative sign is key! It simply means that with increase in pressure, the volume will decrease.

The units of \(\frac{dV}{dP}\) are \(\text{m}^3/\text{kPa}\) (cubic meters per kilopascal), meaning: - For each kilopascal increase in pressure, the volume drops by \(0.00212\) cubic meters, given our starting condition. - It's a mathematical manifestation of the inverse relationship showcased by Boyle's Law.
Differential Calculus
Differential calculus is the branch of mathematics that concerns itself with the concepts of rates of change and slopes of curves. Within the framework of Boyle's Law, differential calculus allows us to calculate derivatives such as \(\frac{dV}{dP}\). This derivative lets us measure how swiftly volume contracts with an increase in pressure.

Derivatives emerge from the fundamental concept of functions. A function describes how one quantity depends on another. For example, in \(V = \frac{k}{P}\), volume is described as a function of pressure.

This branch of calculus allows us to: - Understand relationships between variables. - Differentiate a function to find its rate of change, crucial in interpreting how systems behave under various conditions. - Apply such knowledge to real-world phenomena, like how gases react to pressure changes.

So, when you hear or calculate a derivative like \(\frac{dV}{dP} = -\frac{5.3}{P^2}\), realize that differential calculus is offering you a lens to observe change. It's like hitting pause on a dynamic process to analyze what happens at any precise point, a powerful tool in science and engineering.

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