Problem 32 - Precalculus Mathematics for Calculus

Question

The pressure \(P\) of a sample of gas is directly proportional to the temperature \(T\) and inversely proportional to the volume \(V\). (a) Write an equation that expresses this variation. (b) Find the constant of proportionality if \(100 \mathrm{L}\) of gas exerts a pressure of \(33.2 \mathrm{kPa}\) at a temperature of \(400 \mathrm{K}\) (absolute temperature measured on the Kelvin scale). (c) If the temperature is increased to \(500 \mathrm{K}\) and the volume is decreased to \(80 \mathrm{L}\), what is the pressure of the gas?

Step-by-Step Solution

Answer

(a) \( P = k \cdot \frac{T}{V} \); (b) \( k = 8.3 \); (c) \( P = 51.875 \) kPa.

Step 1: Understanding Direct and Inverse Proportion

The pressure \( P \) is directly proportional to the temperature \( T \), meaning when \( T \) increases, so does \( P \). This can be expressed as \( P \propto T \). Pressure is also inversely proportional to the volume \( V \), meaning when \( V \) increases, \( P \) decreases, expressed as \( P \propto \frac{1}{V} \). Thus, \( P \) can be expressed in terms of \( T \) and \( V \) as \( P = k \cdot \frac{T}{V} \), where \( k \) is the constant of proportionality.

Step 2: Writing the Equation

Combine the expressions for direct and inverse proportion: \( P = k \cdot \frac{T}{V} \). This is the equation that represents how pressure varies with temperature and volume.

Step 3: Solving for the Constant of Proportionality

Using the given conditions, \( V = 100 \) L, \( P = 33.2 \) kPa, \( T = 400 \) K, substitute these values into the equation: \( 33.2 = k \cdot \frac{400}{100} \). Simplify to find \( 33.2 = 4k \), then solve for \( k \): \( k = \frac{33.2}{4} = 8.3 \). Therefore, the constant of proportionality \( k = 8.3 \).

Step 4: Calculating the New Pressure

With \( k = 8.3 \), temperature changed to \( 500 \) K, and volume changed to \( 80 \) L, substitute the values into the equation: \( P = 8.3 \cdot \frac{500}{80} \). Simplify to find \( P = 8.3 \cdot 6.25 \). Calculate to get \( P = 51.875 \) kPa.

Key Concepts

Direct Variation

Direct variation is a relationship between two variables where, as one variable increases, the other variable also increases. This can be expressed using the simple relation:

• If pressure \( P \) is directly proportional to temperature \( T \), then \( P \propto T \).

Direct variation is described by an equation of the form \( y = kx \), where \( k \) is the constant that doesn't change. In the context of gas laws, if you increase the temperature, the kinetic energy increases, triggering a rise in pressure, assuming volume is held constant.
In the given problem, it is straightforward to see that if you were to increase the absolute temperature \( T \), while the volume remained unchanged, the pressure \( P \) would increase proportionally according to the formula \( P = k \cdot \frac{T}{V} \). This concept helps us understand how gas behaves under varying thermal conditions.

Inverse Variation

Inverse variation describes a situation where an increase in one variable leads to a decrease in another. For gas laws, if the volume increases while maintaining the temperature constant, the pressure decreases. This is captured by the formula:

• If pressure \( P \) is inversely proportional to volume \( V \), then \( P \propto \frac{1}{V} \).

The inverse variation model is borne out of the principle that the product of the two variables remains a constant. As an example, air in a balloon gets compressed into a smaller volume, which results in greater pressure—the familiar air pressure law.
In the problem here, you have pressure inversely proportional to volume so as \( V \) decreases, \( P \) increases assuming temperature remains constant. The equation given by \( P = k \cdot \frac{T}{V} \) embodies this inverse relationship, combining both the effects of temperature and volume on pressure.

Constant of Proportionality

The constant of proportionality is the unchanging value that relates the two variables in direct and inverse variation scenarios. Think of it as a multiplier or factor that ensures the relationship between variables is maintained across different scenarios. For gases, it's often denoted by \( k \).

• If \( P = k \cdot \frac{T}{V} \), then \( k \) is determined based on known values of \( P \), \( T \), and \( V \).

In the problem, when given that 100 L of gas exerts a pressure of 33.2 kPa at a temperature of 400 K, you calculated \( k \) as \( 8.3 \). This constant, \( k \), allows you to predict future pressures if you know the temperature and volume of the gas.
It provides a baseline to compute unknown values and enables predictions about gas behavior in different conditions. Calculating \( k \) is decisive as it anchors the equation \( P = k \cdot \frac{T}{V} \) making it a reliable tool for problem-solving.