Question
Chapter 13, Problem 118AP
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.

Expert Solution

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.

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.

Expert Solution

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.

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.

Expert Solution

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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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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Chapter 13 Solutions

Introductory Chemistry: A Foundation

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