Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 17: Problem 52

A tire is pressurized to \(100 \mathrm{kPa}\) gauge in Michigan where \(T=0^{\circ} \mathrm{C}\). In Arizona the tire is at \(70^{\circ} \mathrm{C}\). Assuming a rigid tire, estimate the pressure in \(\mathrm{kPa}\) gauge. a) 120 c) 140 b) 130 d) 150

Short Answer

Expert verified
Answer: b) 130 kPa (gauge)

Step by step solution

01

Identify the known and unknown variables

We are given the initial pressure and temperature in Michigan: \(P_1 = 100\,kPa\,\mathrm{(gauge)}\) and \(T_1 = 0^\circ\mathrm{C}\). We are also given the final temperature in Arizona: \(T_2 = 70^\circ\mathrm{C}\). We need to find the final pressure \(P_2\) in \(\mathrm{kPa}\,\mathrm{ (gauge)}\).
02

Convert Celsius to Kelvin

In order to use the combined gas law, we need to convert temperatures from Celsius to Kelvin. \(T_1 = 0^\circ\mathrm{C} + 273.15 = 273.15\,\mathrm{K}\) \(T_2 = 70^\circ\mathrm{C} + 273.15 = 343.15\,\mathrm{K}\)
03

Use the combined gas law

Since the tire is rigid, the volume remains constant and we can use the following relation from the combined gas law, known as the pressure-temperature relation: \(\frac{P_1}{T_1} = \frac{P_2}{T_2}\).
04

Solve for the unknown pressure \(P_2\)

We can now solve for \(P_2\) using the values obtained in the previous steps: \(P_2 = \frac{P_1 \cdot T_2}{T_1} = \frac{100\,\mathrm{kPa} \cdot 343.15\,\mathrm{K}}{273.15\,\mathrm{K}} = 125.6\,\mathrm{kPa}\)
05

Round the answer to the nearest option

Now we round the answer we found to the nearest option: \(P_2 \approx 130\,\mathrm{kPa}\) (gauge) The correct answer is choice b) 130 kPa (gauge).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A wheel accelerates from rest with \(\alpha=6 \mathrm{rad} / \mathrm{s}^{2}\). How many revolutions are experienced in 4 s? a) \(7.64\) c) \(12.36\) b) \(9.82\) d) \(25.6\)

Estimate the average \(c_{p}\) value, in \(\mathrm{kJ} / \mathrm{kg} \cdot \mathrm{K}\), of a gas if \(522 \mathrm{~kJ} / \mathrm{kg}\) of heat are necessary to raise the temperature from \(300 \mathrm{~K}\) to \(800 \mathrm{~K}\) holding the pressure constant. a) \(1.000\) c) \(1.038\) b) \(1.026\) d) \(1.044\)

An investment pays \(\$ 6000\) at the end of the first year, \(\$ 4000\) at the end of the second year, and \(\$ 2000\) at the end of the third year. Compute the present value of the investment if a 10 percent rate of return is required. a) \(\$ 8333\) c) \(\$ 10,300\) b) \(\$ 9667\) d) \(\$ 12,000\)

The maintenance costs associated with a machine are \(\$ 2,000\) per year for the first 10 years and \(\$ 1,000\) per year thereafter. The machine has an infinite life. If interest is 10 percent, what is the present worth of the annual disbursements? a) \(\$ 16,145\) c) \(\$ 21,300\) b) \(\$ 19,678\) d) \(\$ 92,136\)

An object is moving with an initial velocity of \(20 \mathrm{~m} / \mathrm{s}\). If it is decelerating at \(5 \mathrm{~m} / \mathrm{s}^{2}\) how far does it travel before it stops? a) \(10 \mathrm{~m}\) c) \(30 \mathrm{~m}\) b) \(20 \mathrm{~m}\) d) \(40 \mathrm{~m}\)
See all solutions

Recommended explanations on Physics Textbooks

Engineering Physics

Read Explanation

Astrophysics

Read Explanation

Geometrical and Physical Optics

Read Explanation

Magnetism

Read Explanation

Energy Physics

Read Explanation

Physics of Motion

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free