/*! 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 deep-sea diver uses a gas cyli... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 62

A deep-sea diver uses a gas cylinder with a volume of \(10.0 \mathrm{~L}\) and a content of \(51.2 \mathrm{~g}\) of \(\mathrm{O}_{2}\) and \(32.6 \mathrm{~g}\) of He. Calculate the partial pressure of each gas and the total pressure if the temperature of the gas is \(19^{\circ} \mathrm{C}\).

Short Answer

Expert verified
The partial pressure of \(\mathrm{O}_{2}\) in the gas cylinder is 3.87 atm, the partial pressure of \(\mathrm{He}\) is 19.58 atm, and the total pressure is 23.45 atm.

Step by step solution

01

Convert the given information into usable units

First, we need to convert the given information into usable units. The volume of the gas cylinder is already given in liters, and we remember that 1 L equals 1,000 mL. The masses of \(\mathrm{O}_{2}\) and \(\mathrm{He}\) are given in grams. The temperature is given as a Celsius, so it must be converted to Kelvin by adding 273 to it: T(K) = 19 + 273 = 292 K.
02

Find the number of moles of O2 and He

Now, we need to find the number of moles (\(n\)) of each gas separately. To do this, we use the molar mass (MM) of each element in grams per mole. For \(\mathrm{O}_{2}\), MM = 32 g/mol, and for \(\mathrm{He}\), MM = 4 g/mol. We have: n_O2 = (51.2 g) / (32 g/mol) = 1.6 mol n_He = (32.6 g) / (4 g/mol) = 8.15 mol
03

Calculate the partial pressure of each gas

Now that we have the number of moles of each gas, we can apply the Ideal Gas Law to find the partial pressures of each gas separately. Ideal Gas Law states, PV = nRT We are given the volume (V = 10 L) and temperature (T = 292 K), and the ideal gas constant (R = 0.0821 L路atm/mol路K). Thus, the Ideal Gas Law for each gas becomes: For \(\mathrm{O}_{2}\): P_O2 = (n_O2 脳 R 脳 T) / V P_O2 = (1.6 mol 脳 0.0821 L路atm/mol路K 脳 292 K) / (10 L) = 3.87 atm For \(\mathrm{He}\): P_He = (n_He 脳 R 脳 T) / V P_He = (8.15 mol 脳 0.0821 L路atm/mol路K 脳 292 K) / (10 L) = 19.58 atm
04

Calculate the total pressure

The total pressure can be found by adding the partial pressures of both gases: Total Pressure = P_O2 + P_He = 3.87 atm + 19.58 atm = 23.45 atm So, the partial pressure of \(\mathrm{O}_{2}\) in the gas cylinder is 3.87 atm, the partial pressure of \(\mathrm{He}\) is 19.58 atm, and the total pressure is 23.45 atm.

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

Which of the following statements best explains why a closed balloon filled with helium gas rises in air? (a) Helium is a monatomic gas, whereas nearly all the molecules that make up air, such as nitrogen and oxygen, are diatomic. (b) The average speed of helium atoms is higher than the average speed of air molecules, and the higher speed of collisions with the balloon walls propels the balloon upward. (c) Because the helium atoms are of lower mass than the average air molecule, the helium gas is less dense than air. The balloon thus weighs less than the air displaced by its volume. (d) Because helium has a lower molar mass than the average air molecule, the helium atoms are in faster motion. This means that the temperature of the helium is higher than the air temperature. Hot gases tend to rise.

Arsenic(III) sulfide sublimes readily, even below its melting point of \(320^{\circ} \mathrm{C}\). The molecules of the vapor phase are found to effuse through a tiny hole at \(0.28\) times the rate of effusion of Ar atoms under the same conditions of temperature and pressure. What is the molecular formula of arsenic(III) sulfide in the gas phase?

In the Dumas-bulb technique for determining the molar mass of an unknown liquid, you vaporize the sample of a liquid that boils below \(100^{\circ} \mathrm{C}\) in a boiling-water bath and determine the mass of vapor required to fill the bulb (see drawing). From the following data, calculate the molar mass of the unknown liquid: mass of unknown vapor, \(1.012 \mathrm{~g}\); volume of bulb, \(354 \mathrm{~cm}^{3}\); pressure, 742 torr; temperature, \(99^{\circ} \mathrm{C}\).

Does the effect of intermolecular attraction on the properties of a gas become more significant or less significant if (a) the gas is compressed to a smaller volume at constant temperature; (b) the temperature of the gas is increased at constant volume?

As discussed in the "Chemistry Put to Work" box in Section \(10.8\), enriched uranium can be produced by gaseous diffusion of \(\mathrm{UF}_{6}\). Suppose a process were developed to allow diffusion of gaseous uranium atoms, \(\mathrm{U}(g)\). Calculate the ratio of diffusion rates for ${ }^{235} \mathrm{U}\( and \){ }^{238} \mathrm{U}$, and compare it to the ratio for UF \(_{6}\) given in the essay.
See all solutions

Recommended explanations on Chemistry Textbooks

Physical Chemistry

Read Explanation

Kinetics

Read Explanation

Chemical Reactions

Read Explanation

Inorganic Chemistry

Read Explanation

Chemical Analysis

Read Explanation

Chemistry Branches

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.