Question

According to Boyle's law (a) \((\mathrm{dP} / \mathrm{dV})_{\mathrm{T}}=-\mathrm{K} / \mathrm{P}^{2}\) (b) \((\mathrm{dP} / \mathrm{dV})_{\mathrm{T}}=\mathrm{K} / \mathrm{V}\) (c) \((\mathrm{d} \mathrm{V} / \mathrm{dP})_{\mathrm{T}}=-\mathrm{K} / \mathrm{P}^{2}\) (d) \((\mathrm{dV} / \mathrm{dP})_{\mathrm{T}}=-\mathrm{K} / \mathrm{P}\)

Short answer

None of the options given match the derived equation from Boyle's Law.

Step-by-step solution

Recall Boyle's Law

Boyle's Law states that at constant temperature, the pressure of a gas is inversely proportional to its volume. This is expressed as \( P \times V = K \), where \( K \) is a constant value.

Derive Relationship between Pressure and Volume

Since \( P \times V = K \), differentiating both sides with respect to \( V \) at constant \( T \), we get \( P \times \frac{dV}{dV} + V \times \frac{dP}{dV} = 0 \). This simplifies to \( P + V \left(\frac{dP}{dV}\right)_{T} = 0 \).

Solve for the Derivative of Pressure with Respect to Volume

Rearrange the equation: \( V \left(\frac{dP}{dV}\right)_{T} = -P \). Therefore, \( \left(\frac{dP}{dV}\right)_{T} = -\frac{P}{V} \).

Examine the Possible Answers

Recognize that from Step 3 the derived equation \( \left(\frac{dP}{dV}\right)_{T} = -\frac{P}{V} \) matches none of the options given. Compare this with each option to deduce the correct answer.

Key concepts

Understanding Gas Laws

Gas laws are fundamental principles in chemistry and physics that describe the behavior of gases. They provide a mathematical framework to predict how gases will respond under different conditions of temperature, pressure, and volume. Gas laws are crucial because they intersect with many areas of science and real life. The most common gas laws students encounter include Boyle's Law, Charles's Law, and Avogadro's Law. These laws are derived from the ideal gas law, which combines them into one equation. By understanding these laws, we can better predict how gases change when exposed to different environments. They are particularly useful in fields like meteorology, engineering, and medicine. In our study of Boyle's Law, we specifically focus on the interplay between pressure and volume under a constant temperature scenario.

Exploring the Pressure-Volume Relationship

The pressure-volume relationship is a key aspect of Boyle's Law. It tells us that pressure and volume are inversely related when the temperature is kept constant. This means if you increase the pressure on a gas, its volume will decrease, and vice versa. Mathematically, it is expressed as - Inverse relationship: - As pressure increases, volume decreases. - As volume increases, pressure decreases.
In equation form, this relationship is shown as \( P \times V = K \), where \( P \) is the pressure, \( V \) is the volume, and \( K \) is a constant for a given amount of gas at constant temperature. This principle can be visualized by imagining a balloon. When you squeeze the balloon (increasing pressure), its size (volume) goes down. Understanding this relationship helps us solve real-world problems, like designing better hydraulic systems or improving gas storage methods.

The Role of Differential Calculus in Chemistry

Differential calculus is a powerful mathematical tool that finds extensive use in chemistry, particularly when dealing with the behaviors of gases. It allows chemists to determine how small changes in one quantity (like volume) can impact another (such as pressure). The idea is to work with rates of change, or derivatives, to gain deeper insight into physical processes.

In the context of Boyle's Law, differential calculus lets us explore how pressure changes with volume under constant temperature conditions. By taking the derivative of the equation \( P \times V = K \), we can derive relationships between pressure and volume directly, such as \( \left(\frac{dP}{dV}\right)_{T} = -\frac{P}{V} \). This derivative tells us the rate at which pressure changes with respect to volume when temperature is unchanged, highlighting the elegance and functionality of using calculus in solving chemistry problems. With calculus, complex systems become more understandable and manageable, enabling advancements in scientific research and application.