General, Organic, and Biological Chemistry
General, Organic, and Biological Chemistry
7th Edition
ISBN: 9781285853918
Author: H. Stephen Stoker
Publisher: Cengage Learning
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Question
Chapter 7, Problem 7.53EP

(a)

Interpretation Introduction

Interpretation:

The pressure in atmospheres for oxygen gas at given set of conditions has to be determined.

Concept Introduction:

Ideal gas law provides the relationship between the four gas variables temperature, pressure, volume, and molar amount of a gaseous substance at specified set of conditions.  The mathematical expression for ideal gas law can be shown below,

PV=nRTWhere,P=PressureofgasV=VolumeofgasT=Temperatureofgasn=NumberofmolesofgasR=Idealgasconstant

(a)

Expert Solution
Check Mark

Explanation of Solution

Record the given data,

P = ?V = 4.00Ln = 0.72 moleT = 40°C converted to 313KR=0.821 atm.L/mole.K

Now, substitute these values in ideal gas law and do some mathematical calculation as shown in below,

PV=nRT

P=nRTV

P=(0.72mole)(0.0821atm.L/mole.K)(313K)(4.00L)=4.6 atm

Therefore, the new pressure is 4.6 atm.

(b)

Interpretation Introduction

Interpretation:

The pressure in atmospheres for oxygen gas at given set of conditions has to be determined.

Concept Introduction:

Ideal gas law provides the relationship between the four gas variables temperature, pressure, volume, and molar amount of a gaseous substance at specified set of conditions.  The mathematical expression for ideal gas law can be shown below,

PV=nRTWhere,P=PressureofgasV=VolumeofgasT=Temperatureofgasn=NumberofmolesofgasR=Idealgasconstant

(b)

Expert Solution
Check Mark

Explanation of Solution

Record the given data,

P = ?V = 4.00Ln = 4.5 moleT = 40°C converted to 313KR=0.821 atm.L/mole.K

Now, substitute these values in ideal gas law and do some mathematical calculation as shown in below,

PV=nRT

P=nRTV

P=(4.5mole)(0.0821atm.L/mole.K)(313K)(4.00L)=29 atm

Therefore, the new pressure is 29 atm.

(c)

Interpretation Introduction

Interpretation:

The pressure in atmospheres for oxygen gas at given set of conditions has to be determined.

Concept Introduction:

Ideal gas law provides the relationship between the four gas variables temperature, pressure, volume, and molar amount of a gaseous substance at specified set of conditions.  The mathematical expression for ideal gas law can be shown below,

PV=nRTWhere,P=PressureofgasV=VolumeofgasT=Temperatureofgasn=NumberofmolesofgasR=Idealgasconstant

(c)

Expert Solution
Check Mark

Explanation of Solution

Record the given data,

P = ?V = 4.00Ln = 0.72 gconvertedto(0.72 g O2×1.00 mole O232.0 g O2)=0.031molesT = 40°C converted to 313KR=0.821 atm.L/mole.K

Now, substitute these values in ideal gas law and do some mathematical calculation as shown in below,

PV=nRT

P=nRTV

P=(0.0225mole)(0.0821atm.L/mole.K)(313K)(4.00L)=0.14 atm

Therefore, the new pressure is 0.14 atm.

(d)

Interpretation Introduction

Interpretation:

The pressure in atmospheres for oxygen gas at given set of conditions has to be determined.

Concept Introduction:

Ideal gas law provides the relationship between the four gas variables temperature, pressure, volume, and molar amount of a gaseous substance at specified set of conditions.  The mathematical expression for ideal gas law can be shown below,

PV=nRTWhere,P=PressureofgasV=VolumeofgasT=Temperatureofgasn=NumberofmolesofgasR=Idealgasconstant

(d)

Expert Solution
Check Mark

Explanation of Solution

Record the given data,

P = ?V = 4.00Ln = 4.5 gconvertedto(4.5 g O2×1.00 mole O232.0 g O2)=0.14molesT = 40°C converted to 313KR=0.821 atm.L/mole.K

Now, substitute these values in ideal gas law and do some mathematical calculation as shown in below,

PV=nRT

P=nRTV

P=(0.14mole)(0.0821atm.L/mole.K)(313K)(4.00L)=0.90 atm

Therefore, the new pressure is 0.90 atm.

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Chapter 7 Solutions

General, Organic, and Biological Chemistry

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