/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 62 A bicycle tire is filled with ai... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 62

A bicycle tire is filled with air to a pressure of 75 psi at a temperature of \(19^{\circ} \mathrm{C}\) . Riding the bike on asphalt on a hot day increases the temperature of the tire to \(58^{\circ} \mathrm{C}\) . The volume of the tire increases by 4.0\(\%\) . What is the new pressure in the bicycle tire?

Short Answer

Expert verified
The new pressure in the bicycle tire is approximately \(79.64 \,\mathrm{psi}\).

Step by step solution

01

Convert Given Temperatures to Kelvin Scale

The Combined Gas Law uses the Kelvin temperature scale. So, we need to convert the given temperatures from Celsius to Kelvin: \(T_1 = 19^{\circ}\mathrm{C} + 273.15 = 292.15 \mathrm{K}\) \(T_2 = 58^{\circ}\mathrm{C} + 273.15 = 331.15 \mathrm{K}\)
02

Calculate the Volume Increase

The volume of the tire increases by 4%, so we need to find the relationship between initial and final volumes: \(V_2 = V_1 + 0.04V_1 = 1.04V_1\)
03

Apply the Combined Gas Law

While applying the Combined Gas Law, the terms related to volume and initial pressure will cancel out to get the final pressure. Substitute the known values and solve for \(P_2\): \(\frac{P_1V_1}{T_1}=\frac{P_2(1.04V_1)}{T_2}\) \(\frac{75(1)}{292.15}=\frac{P_2(1.04)}{331.15}\)
04

Solve for the Final Pressure

Next, we will isolate the final pressure, \(P_2\), by solving the equation: \(\frac{75(1)(331.15)}{292.15(1.04)} = P_2\) \(P_2 \approx 79.64 \,\mathrm{psi}\) Thus, the new pressure in the bicycle tire is approximately 79.64 psi.

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.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Temperature Conversion
When working with gases, especially when applying the Combined Gas Law, it is critical to convert temperatures from Celsius to the Kelvin scale.
The Kelvin scale is an absolute temperature scale, starting at absolute zero, which provides a consistent basis for calculations in thermodynamics.
To convert Celsius to Kelvin, simply add 273.15 to the Celsius temperature:
  • The formula for conversion is: \[ T(K) = T(^{\circ}C) + 273.15 \]
  • For example, converting \(19^{\circ}\mathrm{C}\) to Kelvin gives: \[ 19 + 273.15 = 292.15 \, \mathrm{K} \]
  • Similarly, \(58^{\circ}\mathrm{C}\) converts to: \[ 58 + 273.15 = 331.15 \, \mathrm{K} \]
This conversion ensures accuracy and consistency when dealing with gas law equations and calculations.
Pressure Calculation
Calculating the pressure of a gas involves understanding how variables like temperature and volume affect it, especially when these factors change.
In this exercise, the initial pressure (\( P_1 \)) is given as 75 psi (pounds per square inch) and the task is to find the new pressure after changes in temperature and volume.
Using the Combined Gas Law, which is \[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \], we can relate the initial and final states of pressure, volume, and temperature. Here are the steps:
  • Substitute known values into the equation.
  • Rearrange the equation to solve for the final pressure \( P_2 \).
  • Ensure that the volume increase and temperature changes are accurately represented.
Finally, solving the equation gives us the new pressure: approximately 79.64 psi.
Volume Change
In gas law problems, changes in volume significantly impact pressure and temperature calculations.
In this scenario, the volume of the tire increases by 4%. This means that if the initial volume is denoted as \( V_1 \), the increased volume \( V_2 \) can be calculated as:
  • \[ V_2 = 1.04 \times V_1 \]
  • This indicates that the final volume is 104% of the initial volume.
This increase must be factored into the Combined Gas Law to accurately assess how it affects the pressure.Volume changes, especially increases, generally lead to pressure adjustments when considering a gas under constant temperature conditions, due to the inverse relationship between volume and pressure.
Kelvin Scale
The Kelvin scale is the foundation for thermodynamic calculations as it provides an absolute temperature reference, unlike Fahrenheit or Celsius.
This scale will always start at absolute zero (0 K), which is the lowest possible temperature, where molecular motion stops.
Using the Kelvin scale is imperative in gas law equations because it avoids negative temperatures that can distort calculations.
  • All temperatures should ideally be converted to Kelvin when using equations like the Combined Gas Law.
  • If you face a problem with temperature in Celsius, always remember to convert it to Kelvin first.
This ensures that all thermodynamic relationships uphold their proportional differences, maintaining consistency and accuracy in physical sciences.

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

One of the chemical controversies of the nineteenth century concerned the element beryllium (Be). Berzelius originally claimed that beryllium was a trivalent element (forming \(\mathrm{Be}^{3+}\) ions) and that it gave an oxide with the formula \(\mathrm{Be}_{2} \mathrm{O}_{3}\) . This resulted in a calculated atomic mass of 13.5 for beryllium. In formulating his periodic table, Mendeleev proposed that beryllium was divalent (forming \(\mathrm{Be}^{2+}\) ions) and that it gave an oxide with the formula BeO. This assumption gives an atomic mass of \(9.0 .\) In \(1894,\) A. Combes (Comptes Rendus \(1894,\) p. 1221 ) reacted beryllium with the anion \(C_{5} \mathrm{H}_{7} \mathrm{O}_{2}^{-}\) and measured the density of the gaseous product. Combes's data for two different experiments are as follows: $$\begin{array}{lll}{\text { Mass }} & {0.2022 \mathrm{g}} & {0.2224 \mathrm{g}} \\ {\text { Volume }} & {22.6 \mathrm{cm}^{3}} & {26.0 \mathrm{cm}^{3}} \\ {\text { Temperature }} & {13^{\circ} \mathrm{C}} & {17^{\circ} \mathrm{C}} \\ {\text { Pressure }} & {765.2 \mathrm{mm} \mathrm{Hg}} & {764.6 \mathrm{mm}}\end{array}$$ If beryllium is a divalent metal, the molecular formula of the product will be \(\mathrm{Be}\left(\mathrm{C}_{5} \mathrm{H}_{7} \mathrm{O}_{2}\right)_{2} ;\) if it is trivalent, the formula will be \(\mathrm{Be}\left(\mathrm{C}_{5} \mathrm{H}_{7} \mathrm{O}_{2}\right)_{3} .\) Show how Combes's data help to confirm that beryllium is a divalent metal.

Methane \(\left(\mathrm{CH}_{4}\right)\) gas flows into a combustion chamber at a rate of \(200 . \mathrm{L} / \mathrm{min}\) at 1.50 \(\mathrm{atm}\) and ambient temperature. Air is added to the chamber at 1.00 \(\mathrm{atm}\) and the same temperature, and the gases are ignited. a. To ensure complete combustion of \(\mathrm{CH}_{4}\) to \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(g),\) three times as much oxygen as is necessary is reacted. Assuming air is 21 mole percent \(\mathrm{O}_{2}\) and 79 \(\mathrm{mole}\) percent \(\mathrm{N}_{2},\) calculate the flow rate of air necessary to deliver the required amount of oxygen. b. Under the conditions in part a, combustion of methane was not complete as a mixture of \(\mathrm{CO}_{2}(g)\) and \(\mathrm{CO}(g)\) was produced. It was determined that 95.0\(\%\) of the carbon in the exhaust gas was present in \(\mathrm{CO}_{2}\) . The remainder was present as carbon in CO. Calculate the composition of the exhaust gas in terms of mole fraction of \(\mathrm{CO}, \mathrm{CO}_{2}, \mathrm{O}_{2}, \mathrm{N}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) . Assume \(\mathrm{CH}_{4}\) is completely reacted and \(\mathrm{N}_{2}\) is unreacted.

Metallic molybdenum can be produced from the mineral molybdenite, \(\mathrm{MoS}_{2}\). The mineral is first oxidized in air to molybdenum trioxide and sulfur dioxide. Molybdenum trioxide is then reduced to metallic molybdenum using hydrogen gas. The balanced equations are $$\operatorname{MoS}_{2}(s)+\frac{7}{2} \mathrm{O}_{2}(g) \longrightarrow \operatorname{MoO}_{3}(s)+2 \mathrm{SO}_{2}(g)$$ $$\mathrm{MoO}_{3}(s)+3 \mathrm{H}_{2}(g) \longrightarrow \mathrm{Mo}(s)+3 \mathrm{H}_{2} \mathrm{O}(l)$$ Calculate the volumes of air and hydrogen gas at \(17^{\circ} \mathrm{C}\) and 1.00 atm that are necessary to produce \(1.00 \times 10^{3} \mathrm{kg}\) pure molybdenum from \(\mathrm{MoS}_{2}\) . Assume air contains 21\(\%\) oxygen by volume, and assume 100\(\%\) yield for each reaction.

A 2.747 -g sample of manganese metal is reacted with excess \(\mathrm{HCl}\) gas to produce 3.22 \(\mathrm{L} \mathrm{H}_{2}(g)\) at 373 \(\mathrm{K}\) and 0.951 atm and a manganese chloride compound \(\left(\mathrm{Mn} \mathrm{Cl}_{x}\right)\). What is the formula of the manganese chloride compound produced in the reaction?

A mixture of chromium and zinc weighing 0.362 g was reacted with an excess of hydrochloric acid. After all the metals in the mixture reacted, 225 mL dry of hydrogen gas was collected at \(27^{\circ} \mathrm{C}\) and \(750 .\) torr. Determine the mass percent of \(\mathrm{Zn}\) in the metal sample [Zinc reacts with hydrochloric acid to produce zinc chloride and hydrogen gas; chromium reacts with hydrochloric acid to produce chromium (III) chloride and hydrogen gas. \(]\)
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Analysis

Read Explanation

Making Measurements

Read Explanation

The Earths Atmosphere

Read Explanation

Kinetics

Read Explanation

Nuclear Chemistry

Read Explanation

Inorganic Chemistry

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.