Problem 31 - Calculus Early Transcendentals: Pearson New International Edition

Question

According to the ideal gas law, the pressure, temperature, and volume of a gas are related by \(P V=k T\), where \(k\) is a constant. Find the rate of change of pressure (pounds per square inch) with respect to temperature when the temperature is \(300^{\circ} \mathrm{K}\) if the volume is kept fixed at 100 cubic inches.

Step-by-Step Solution

Answer

The rate of change of pressure with respect to temperature is \(\frac{P}{100}\) psi/°K.

Step 1: Identify Known Values

According to the ideal gas law equation, we have \(PV = kT\). We know that \(V = 100\) cubic inches and \(T = 300^{\circ} \text{K}\). The volume \(V\) is held constant.

Step 2: Differentiate the Ideal Gas Law

Since the volume \(V\) is constant, differentiate the equation \(PV = kT\) with respect to temperature \(T\): \(\frac{d}{dT}(PV) = \frac{d}{dT}(kT)\).

Step 3: Product Rule Application

Apply the product rule to the left side: \( \frac{d}{dT}(PV) = V \frac{dP}{dT} + P \frac{dV}{dT}\). Because the volume is constant, \(\frac{dV}{dT} = 0\). Therefore, \(V \frac{dP}{dT} = k\).

Step 4: Express \(\frac{dP}{dT}\)

Since \(V \frac{dP}{dT} = k\), solve for \(\frac{dP}{dT}\): \(\frac{dP}{dT} = \frac{k}{V}\).

Step 5: Find \(k\) from Known Values

Substitute the known values at \(T = 300\): From \(PV = kT\), \(k = \frac{PV}{T}\). To find \(k\), rearrange to get \(k = \frac{300P}{300}\), thus \(k = P\).

Step 6: Calculate \(\frac{dP}{dT}\) at the Given Conditions

Substituting back \(P\) for \(k\) in \(\frac{dP}{dT} = \frac{k}{100}\), we get \(\frac{dP}{dT} = \frac{P}{100}\).

Key Concepts

Ideal Gas Law

The ideal gas law is a fundamental concept in thermodynamics that describes the relationship between the pressure, volume, and temperature of an ideal gas. This is expressed with the equation \(PV = kT\), where \(P\) represents pressure, \(V\) represents volume, \(T\) represents temperature, and \(k\) is a constant. This formula captures how these three variables interact in a system.
One important aspect of the ideal gas law is understanding that each variable affects the others. For instance, if the volume of the gas is held constant, an increase in temperature will result in an increase in pressure. The deal gas law is widely used in various scientific and engineering fields because it provides a simple way to understand how gases behave under different conditions.

Differentiation

Differentiation is a mathematical process used to determine how a function changes as its input changes. In this context, we differentiate the equation \(PV = kT\) with respect to temperature \(T\). This helps us find the rate of change of pressure as temperature changes, which is what the problem requires.
By using differentiation, we can see how a small change in temperature affects the pressure of the gas when volume is kept constant. It converts an implicit understanding of the relationship into a specific rate of change, which is essential in fields ranging from physics to engineering.
This concept is particularly important in optimizing processes, tracking changes in conditions, and forecasting outcomes based on mathematical models.

Product Rule

The product rule is a key differentiation rule used when dealing with functions that are multiplied together. In this problem, the product rule is applied to the equation \(PV = kT\), specifically when differentiating \(PV\) with respect to \(T\).
Apply the product rule with:

• Function 1: \(P\)
• Function 2: \(V\)

The product rule states: \( \frac{d}{dT}(PV) = V \frac{dP}{dT} + P \frac{dV}{dT} \). Given that \(V\) is constant, \( \frac{dV}{dT} = 0\), simplifying our equation.
This simplification allows us to solve for the rate of change of the pressure \( \frac{dP}{dT} \), making the product rule an invaluable tool in calculus for finding derivatives of products of functions.

Temperature and Pressure Relationship

The relationship between temperature and pressure in a gas, as described by the ideal gas law, is crucial for understanding the behavior of gases. When examining this relationship with the gas law equation \(PV = kT\), and knowing that volume is constant, changing temperature means changing pressure.

When temperature increases, the kinetic energy of gas molecules increases, generally causing pressure to rise if the volume doesn't change. In this exercise, this concept is further expressed mathematically by finding \(\frac{dP}{dT}\), the rate of change of pressure with respect to temperature.
These relationships help predict and control reactions in closed systems efficiently, from basic science applications to complex industrial processes. Thus, understanding how temperature affects pressure under constant volume helps in designing and managing these processes effectively.