Problem 25 - University Physics with Modern Physics

Question

A tire has a gauge pressure of \(300 . \mathrm{kPa}\) at \(15.0^{\circ} \mathrm{C}\). What is the gauge pressure at \(45.0^{\circ} \mathrm{C}\) ? Assume that the change in volume of the tire is negligible.

Step-by-Step Solution

Answer

Answer: The gauge pressure at 45.0°C is approximately 330.75 kPa.

Step 1: Convert temperatures to Kelvin

First, we need to convert the given temperatures from Celsius to Kelvin. To do so, we add 273.15 to the temperatures in Celsius: T1 = 15.0 + 273.15 = 288.15 K T2 = 45.0 + 273.15 = 318.15 K

Step 2: Rearrange the formula for P2

We rearrange the equation P1/T1 = P2/T2 to find P2: P2 = P1 * (T2/T1)

Step 3: Substitute the given values and solve for P2

Now, we substitute the given values into the equation and solve for P2: P2 = 300,000 * (318.15 / 288.15) P2 ≈ 330,750 Pa

Step 4: Convert P2 to kPa

Since the initial pressure was given in kPa, we should convert the final pressure to kPa as well: P2 = 330,750 Pa / 1000 = 330.75 kPa

Step 5: State the final answer

The gauge pressure of the tire at 45.0°C is approximately 330.75 kPa.

Key Concepts

Temperature Conversion

Temperature conversion is an essential concept in science, especially when working with the Ideal Gas Law. In this exercise, you need to convert temperatures from Celsius to Kelvin to use them in calculations. The Kelvin scale is often used in scientific calculations because it starts at absolute zero, making it an absolute temperature measurement.To convert Celsius to Kelvin:

• Add 273.15 to the Celsius temperature.
• For example, a temperature of \(15.0^{\circ} \text{C}\) becomes \(288.15 \text{K}\) after conversion.
• Similarly, \(45.0^{\circ} \text{C}\) converts to \(318.15 \text{K}\).

This straightforward conversion allows you to work efficiently within thermodynamic equations without additional correction factors. Remember, Kelvin does not use the degree sign (°) like Celsius.

Pressure Calculations

Pressure calculations are critical when applying the Ideal Gas Law. In this problem, you needed to find a new gauge pressure after a change in temperature, assuming constant volume.The pressure-temperature relationship for gases can be shown by using:\[ \frac{P_1}{T_1} = \frac{P_2}{T_2} \]where:

• \( P_1 \) and \( T_1 \) are the initial pressure and temperature.
• \( P_2 \) and \( T_2 \) are the final pressure and temperature.

This formula derives from the Ideal Gas Law simplified under constant volume assumptions. Solving for \( P_2 \), you can determine the new pressure using known temperatures and initial pressure. This relationship is linear, making it predictable and reliable in closed systems.

Kelvin Scale

The Kelvin scale is fundamental in scientific calculations related to thermodynamics. Unlike other temperature scales, it starts at absolute zero, which is approximately -273.15°C. Here are some characteristics of the Kelvin scale:

• Absolute zero is the point at which molecular motion theoretically stops.
• There is no negative number in the Kelvin scale, simplifying many physical equations.
• This scale is directly proportional to the average kinetic energy of particles in a substance.

When working with gases and their properties, using the Kelvin scale avoids errors associated with negative temperatures and provides accurate relative temperature measurements. Always ensure to convert Celsius to Kelvin in these scenarios.

Gauge Pressure

Gauge pressure is the pressure relative to the ambient atmospheric pressure. It is widely used in engineering applications where the differences between internal pressure and atmospheric pressure are measured. A few key points about gauge pressure:

• It can be positive or negative, depending on whether the measured pressure is above or below atmospheric pressure.
• In this exercise, the tire's pressure given in kilopascals (kPa) is gauge pressure.
• Gauge pressure readings need to be converted to absolute pressure (by adding atmospheric pressure) when used in thermodynamic equations.

In this scenario, you are directly calculating the change in gauge pressure with temperature changes. Understanding gauge pressure helps ensure accurate assessments in practical applications like tire inflation where only relative pressure difference matters.