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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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Most popular questions from this chapter

a) What is the root-mean-square speed for a collection of helium- 4 atoms at \(300 . \mathrm{K} ?\) b) What is the root-mean-square speed for a collection of helium- 3 atoms at 300 . K?

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