/*! 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 58 An ideal gas at \(7^{\circ} \mat... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 58

An ideal gas at \(7^{\circ} \mathrm{C}\) is in a spherical flexible container having a radius of \(1.00 \mathrm{~cm}\). The gas is heated at constant pressure to \(88^{\circ} \mathrm{C}\). Determine the radius of the spherical container after the gas is heated. [Volume of a sphere \(\left.=(4 / 3) \pi r^{3} .\right]\)

Short Answer

Expert verified
The radius of the spherical container after the gas is heated is approximately \(1.23 cm\).

Step by step solution

01

Write down the Ideal Gas Law equation

The Ideal Gas Law equation is given by \(PV = nRT\), where P is the pressure, V is the volume, n is the number of moles, R is the universal gas constant, and T is the temperature in Kelvin.
02

Convert the temperatures to Kelvin

We need to convert the given temperatures from degrees Celsius to Kelvin. The formula for the conversion is \(T(K) = T(掳C) + 273.15\). Initial temperature, \(T_1 = 7掳C + 273.15 = 280.15 K\) Final temperature, \(T_2 = 88掳C + 273.15 = 361.15 K\)
03

Calculate the initial volume

The volume of a sphere is given by the formula: \(V = \frac{4}{3} \pi r^3\). Using the initial radius, \(r_1 = 1.00 cm\), we can calculate the initial volume, \(V_1\). \(V_1 = \frac{4}{3} \pi (1.00)^3 = \frac{4}{3} \pi cm^3\)
04

Use the constant pressure property

Since the pressure remains constant during the heating process, we can write the Ideal Gas Law equation for both initial and final states as follows: \(\frac{V_1}{T_1} = \frac{V_2}{T_2}\)
05

Solve for the final volume

Substituting the values of \(V_1, T_1\), and \(T_2\) into the above equation, we can solve for the final volume, \(V_2\): \(\frac{\frac{4}{3} \pi cm^3}{280.15 K} = \frac{V_2}{361.15 K}\) After rearranging the equation, we find that: \(V_2 = \frac{4}{3} \pi cm^3 \times \frac{361.15}{280.15} \approx 1.89 \pi cm^3\)
06

Find the new radius of the container

We can find the final radius \(r_2\) using the formula for the volume of a sphere. Setting \(V_2 = 1.89 \pi cm^3\), we can solve for \(r_2\): \(\frac{4}{3} \pi r_2^3 = 1.89 \pi cm^3\) Dividing both sides by \(\frac{4}{3} \pi\), we get: \(r_2^3 = 1.89 cm^3\) Take the cube root of both sides: \(r_2 \approx 1.23 cm\) The radius of the spherical container after the gas is heated is approximately 1.23 cm.

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!

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

Hydrogen cyanide is prepared commercially by the reaction of methane, \(\mathrm{CH}_{4}(g)\), ammonia, \(\mathrm{NH}_{3}\left(\mathrm{~g}\right.\) ), and oxygen, \(\mathrm{O}_{2}(\mathrm{~g})\), at high temperature. The other product is gaseous water. a. Write a chemical equation for the reaction. b. What volume of \(\mathrm{HCN}(\mathrm{g})\) can be obtained from the reaction of \(20.0 \mathrm{~L} \mathrm{CH}_{4}(g), 20.0 \mathrm{~L} \mathrm{NH}_{3}(g)\), and \(20.0 \mathrm{~L} \mathrm{O}_{2}(g) ?\) The volumes of all gases are measured at the same temperature and pressure.

We-state that the ideal gas law tends to hold best at low pressures and high temperatures. Show how the van der Waals equation simplifies to the ideal gas law under these conditions.

Consider the following reaction: $$ 4 \mathrm{Al}(s)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{Al}_{2} \mathrm{O}_{3}(s) $$ It takes \(2.00 \mathrm{~L}\) pure oxygen gas at STP to react completely with a certain sample of aluminum. What is the mass of aluminum reacted?

Small quantities of hydrogen gas can be prepared in the laboratory by the addition of aqueous hydrochloric acid to metallic zinc. $$ \mathrm{Zn}(s)+2 \mathrm{HCl}(a q) \longrightarrow \mathrm{ZnCl}_{2}(a q)+\mathrm{H}_{2}(g) $$ Typically, the hydrogen gas is bubbled through water for collection and becomes saturated with water vapor. Suppose \(240 . \mathrm{mL}\) of hydrogen gas is collected at \(30 .{ }^{\circ} \mathrm{C}\) and has a total pressure of \(1.032\) atm by this process. What is the partial pressure of hydrogen gas in the sample? How many grams of zinc must have reacted to produce this quantity of hydrogen? (The vapor pressure of water is 32 torr at \(30^{\circ} \mathrm{C}\).)

Calculate the average kinetic energies of \(\mathrm{CH}_{4}(g)\) and \(\mathrm{N}_{2}(g)\) molecules at \(273 \mathrm{~K}\) and \(546 \mathrm{~K}\).
See all solutions

Recommended explanations on Chemistry Textbooks

Ionic and Molecular Compounds

Read Explanation

Physical Chemistry

Read Explanation

Chemical Reactions

Read Explanation

Chemistry Branches

Read Explanation

Nuclear Chemistry

Read Explanation

The Earths Atmosphere

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.