Chapter 19: Problem 25
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.
Short Answer
Expert verified
Answer: The gauge pressure at 45.0°C is approximately 330.75 kPa.
Step by step solution
01
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
02
Rearrange the formula for P2
We rearrange the equation P1/T1 = P2/T2 to find P2:
P2 = P1 * (T2/T1)
03
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
04
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
05
State the final answer
The gauge pressure of the tire at 45.0°C is approximately 330.75 kPa.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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 \(318.15 \text{K}\).
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.
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.
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.