Introductory Chemistry: A Foundation
Introductory Chemistry: A Foundation
9th Edition
ISBN: 9781337399425
Author: Steven S. Zumdahl, Donald J. DeCoste
Publisher: Cengage Learning
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Textbook Question
Chapter 13, Problem 118AP

2 . 5 0 L container at 1 .00 atm and 48 ° C is filled with 5 . 41 g of a monatomic gas.

Determine the identity of the gas.

Assuming the 2 . 5 0 L container is a large elastic balloon, predict what will happen when 1 0.0  g of oxygen gas is added to the balloon (which already contains 5 . 41 g of the monatomic gas).

ovide values for each of the following variables. In addition, explain what is happening for each variable, incorporating the kinetic molecular theory into your explanation.

m>Temperature of gas mixture = ?K

m>Total moles of gas mixture = ?mol

m>Total pressure of gas mixture = ?atm

m>Volume of balloon = ?L

Now assuming the 2 . 5 0 L container is rigid (like a steel container), predict what will happen when 1 0.0  g of oxygen gas is added to the container (which again already contains 5 . 41 g of the monatomic gas).

ovide values for each of the following variables. In addition, explain what is happening for each variable, incorporating the kinetic molecular theory into your explanation.

m>Temperature of gas mixture = ?K

m>Total moles of gas mixture = ?mol

m>Total pressure of gas mixture = ?atm

m>Volume of rigid container = ? L

Expert Solution
Check Mark
Interpretation Introduction

(a)

Interpretation:

To determine the identity of the gas based on the pressure, volume and temperature given.

Concept Introduction:

The ideal gas equation is:

PV=nRT

Where,

P = Pressure of the gas

V = Volume of the gas

n = moles of the gas

R = Universal gas constant

T = Temperature of the gas.

Answer to Problem 118AP

The monatomic gas is Argon.

Explanation of Solution

The ideal gas equation is

PV=nRT

Where,

P = Pressure of the gas = 1.00 atm

V = Volume of the gas = 2.50 L

n = moles of the gas = ?

R = Universal gas constant = 0.0821 L.atm/mol.K

T = Temperature of the gas = -48 ° C = 225 K

Substituting the values in the given equation, we get,

PV=nRT1.00atm×2.50L=n×0.0821L.atm/mol.K×225Kn=0.135mol

Thus, the moles of the gas = 0.135 mol

From the moles of the gas, one can find the molar mass of the gas thereby identity of the gas.

Moles=MassMolar mass0.135mol=5.41gMM=40.074g/mol

The monatomic gas with this molecular weight is Argon.

Expert Solution
Check Mark
Interpretation Introduction

(b)

Interpretation:

To determine the values of different variables when another gas is added to the elastic balloon which already has a monatomic gas in it.

Concept Introduction:

The ideal gas equation is:

PV=nRT

Where,

P = Pressure of the gas

V = Volume of the gas

n = moles of the gas

R = Universal gas constant

T = Temperature of the gas.

Answer to Problem 118AP

Temperature of gas mixture = 225 K

Total moles of gas mixture = 0.447 mol

Total pressure of gas mixture = 1 atm

Volume of balloon = 8.26 L.

Explanation of Solution

Given, 10.0 g of oxygen is added.

Moles of oxygen are to be found.

Moles=MassMolar massMoles=10.0g32g/mol=0.3125 mol

Moles of oxygen = 0.3125 mol

Moles of monatomic gas = 0.135 mol

Total number of moles = 0.3125 mol + 0.135 mol = 0.447 mol

Air inside the balloon and atmospheric air pressure has very small pressure difference.

Therefore, one can consider it same and assume here that pressure of air inside balloon is equal to atmospheric pressure that is 1 atm.

Since, there is no change in temperature so, the temperature of the mixture is 225 K.

Total volume of gas mixture is found using ideal gas equation.

PV=nRT1atm×V=0.447mol×0.0821L.atm/mol.K×225KV=8.26L

Thus,

Temperature of gas mixture = 225 K

Total moles of gas mixture = 0.447 mol

Total pressure of gas mixture = 1 atm

Volume of balloon = 8.26 L.

Expert Solution
Check Mark
Interpretation Introduction

(c)

Interpretation:

To determine the values of different variables when another gas is added to the rigid steel container this already has a monatomic gas in it.

Concept Introduction:

The ideal gas equation is:

PV=nRT

Where,

P = Pressure of the gas

V = Volume of the gas

n = moles of the gas

R = Universal gas constant

T = Temperature of the gas.

Answer to Problem 118AP

Temperature of gas mixture = 225 K

Total moles of gas mixture = 0.447 mol

Total pressure of gas mixture = 3.303 atm

Volume of rigid container = 2.5 L.

Explanation of Solution

Given, 10.0 g of oxygen is added.

Moles of oxygen are to be found.

Moles=MassMolar massMoles=10.0g32g/mol=0.3125 mol

Moles of oxygen = 0.3125 mol

Moles of monatomic gas = 0.135 mol

Total number of moles = 0.3125 mol + 0.135 mol = 0.447 mol

Since, there is no change in temperature so, the temperature of the mixture is 225 K.

Since, the given container is rigid so, the volume of the mixture is 2.50 L.

Total pressure of gas mixture is found using ideal gas equation.

PV=nRTP×2.50L=0.447×0.0821L.atm/mol.K×225KP=3.303atm

Thus,

Temperature of gas mixture = 225 K

Total moles of gas mixture = 0.447 mol

Total pressure of gas mixture = 3.303 atm

Volume of rigid container = 2.5 L.

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Chapter 13 Solutions

Introductory Chemistry: A Foundation

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A sample of helium gas with a volume of...Ch. 13 - Prob. 24QAPCh. 13 - Prob. 25QAPCh. 13 - Prob. 26QAPCh. 13 - Prob. 27QAPCh. 13 - Prob. 28QAPCh. 13 - A sample of gas in a balloon has an initial...Ch. 13 - Suppose a 375mLsample of neon gas at 78Cis cooled...Ch. 13 - For each of the following sets of...Ch. 13 - For each of the following sets of...Ch. 13 - Prob. 33QAPCh. 13 - Prob. 34QAPCh. 13 - Suppose 1.25Lof argon is cooled from 291Kto 78K....Ch. 13 - Suppose a 125mLsample of argon is cooled from...Ch. 13 - Prob. 37QAPCh. 13 - Prob. 38QAPCh. 13 - Prob. 39QAPCh. 13 - Prob. 40QAPCh. 13 - Prob. 41QAPCh. 13 - If :math>1.04gof chlorine gas occupies a volume of...Ch. 13 - If 3.25moles of argon gas occupies a volume of...Ch. 13 - Prob. 44QAPCh. 13 - Prob. 45QAPCh. 13 - Prob. 46QAPCh. 13 - Prob. 47QAPCh. 13 - Prob. 48QAPCh. 13 - Prob. 49QAPCh. 13 - Prob. 50QAPCh. 13 - Prob. 51QAPCh. 13 - Determine the pressure in a 125Ltank containing...Ch. 13 - Prob. 53QAPCh. 13 - Prob. 54QAPCh. 13 - Prob. 55QAPCh. 13 - Suppose that a 1.25gsample of neon gas is confined...Ch. 13 - At what temperature will a 1.0gsample of neon gas...Ch. 13 - Prob. 58QAPCh. 13 - What pressure exists in a 200Ltank containing...Ch. 13 - Prob. 60QAPCh. 13 - Suppose a 24.3mLsample of helium gas at 25Cand...Ch. 13 - Prob. 62QAPCh. 13 - Prob. 63QAPCh. 13 - Prob. 64QAPCh. 13 - Prob. 65QAPCh. 13 - Prob. 66QAPCh. 13 - Prob. 67QAPCh. 13 - Suppose than 1.28gof neon gas and 2.49gof argon...Ch. 13 - A tank contains a mixture of 52.5gof oxygen gas...Ch. 13 - What mass of new gas would but required to fill a...Ch. 13 - Prob. 71QAPCh. 13 - Prob. 72QAPCh. 13 - A 500mLsample of O2gas at 24Cwas prepared by...Ch. 13 - Prob. 74QAPCh. 13 - Prob. 75QAPCh. 13 - Prob. 76QAPCh. 13 - Prob. 77QAPCh. 13 - Prob. 78QAPCh. 13 - Prob. 79QAPCh. 13 - Prob. 80QAPCh. 13 - Prob. 81QAPCh. 13 - Prob. 82QAPCh. 13 - Prob. 83QAPCh. 13 - Prob. 84QAPCh. 13 - Calcium oxide can be used to “scrub" carbon...Ch. 13 - Consider the following reaction:...Ch. 13 - Consider the following reaction for the combustion...Ch. 13 - Although we: generally think of combustion...Ch. 13 - m>89. Ammonia and gaseous hydrogen chloride...Ch. 13 - Calcium carbide, CaC2, reacts with water to...Ch. 13 - Prob. 91QAPCh. 13 - Prob. 92QAPCh. 13 - What volume does a mixture of 14.2gof He and...Ch. 13 - Prob. 94QAPCh. 13 - Prob. 95QAPCh. 13 - Consider the following chemical equation:...Ch. 13 - Prob. 97QAPCh. 13 - Dinitrogen monoxide, N2O, reacts with propane,...Ch. 13 - Consider the following unbalanced chemical...Ch. 13 - Prob. 100QAPCh. 13 - Prob. 101QAPCh. 13 - Prob. 102QAPCh. 13 - Prob. 103APCh. 13 - Prob. 104APCh. 13 - Prob. 105APCh. 13 - onsider the flasks in the following diagrams. mg...Ch. 13 - Prob. 107APCh. 13 - helium tank contains 25.2Lof helium m 8.40atm...Ch. 13 - Prob. 109APCh. 13 - Prob. 110APCh. 13 - Prob. 111APCh. 13 - Prob. 112APCh. 13 - Prob. 113APCh. 13 - Prob. 114APCh. 13 - Prob. 115APCh. 13 - Prob. 116APCh. 13 - Prob. 117APCh. 13 - 2.50Lcontainer at 1.00atm and 48Cis filled with...Ch. 13 - Prob. 119APCh. 13 - Prob. 120APCh. 13 - Prob. 121APCh. 13 - Prob. 122APCh. 13 - Prob. 123APCh. 13 - f a gaseous mixture is made of 3.50gof He and...Ch. 13 - Prob. 125APCh. 13 - Prob. 126APCh. 13 - f 5.l2gof oxygen gas occupies a volume of 6.21Lat...Ch. 13 - Prob. 128APCh. 13 - Prob. 129APCh. 13 - Prob. 130APCh. 13 - Prob. 131APCh. 13 - Prob. 132APCh. 13 - t what temperature does 4.00gof helium gas have a...Ch. 13 - Prob. 134APCh. 13 - f 3.20gof nitrogen gas occupies a volume of...Ch. 13 - Prob. 136APCh. 13 - mixture at 33Ccontains H2at 325torr, N2at 475torr,...Ch. 13 - Prob. 138APCh. 13 - Prob. 139APCh. 13 - he following demonstration takes place in a...Ch. 13 - onsider the following unbalanced chemical...Ch. 13 - Prob. 142APCh. 13 - Prob. 143APCh. 13 - Prob. 144APCh. 13 - Prob. 145APCh. 13 - Prob. 146APCh. 13 - Prob. 147APCh. 13 - Prob. 148APCh. 13 - Prob. 149APCh. 13 - omplete the following table for an ideal gas. mg...Ch. 13 - Prob. 151CPCh. 13 - Prob. 152CPCh. 13 - certain flexible weather balloon contains helium...Ch. 13 - Prob. 154CPCh. 13 - Prob. 155CPCh. 13 - Prob. 156CPCh. 13 - Prob. 157CP
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