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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Chapter 7, Problem 7.35EP

(a)

Interpretation Introduction

Interpretation:

The volume in liters for a sample of carbon dioxide gas at given set of conditions by using combined gas law has to be determined.

Concept Introduction:

Combined gas law tells that the Kelvin temperature of gas is directly proportional to the pressure and volume of a fixed amount of gas.  The mathematical expression for combined gas law can be represented as follows,

P1V1T1=P2V2T2

(a)

Expert Solution
Check Mark

Explanation of Solution

Record the given data,

P1=1.35atmV1=15.2LT1=33°Cconvertedto306KP2=3.50atmT2=35°Cconvertedto308KV2=?

Now, substitute these values in rearranged combined gas law and do some simple mathematical calculation to gent final answer as follows,

V2=V1×P1P2×T2T1

V2=15.2L×(1.35atm3.50atm)×(308K306K)=5.90L

The final volume in liters for a sample of carbon dioxide gas at given set of conditions is 5.90L.

(b)

Interpretation Introduction

Interpretation:

The pressure in atmospheres for a sample of carbon dioxide gas at given set of conditions by using combined gas law has to be determined.

Concept Introduction:

Combined gas law tells that the Kelvin temperature of gas is directly proportional to the pressure and volume of a fixed amount of gas.  The mathematical expression for combined gas law can be represented as follows,

P1V1T1=P2V2T2

(b)

Expert Solution
Check Mark

Explanation of Solution

Record the given data,

P1=1.35atmV1=15.2LT1=33°Cconvertedto306KP2=?T2=42°Cconvertedto315KV2=10.0L

Now, substitute these values in rearranged combined gas law and do some simple mathematical calculation to gent final answer as follows,

P2=P1×V1V2×T2T1

P2=1.35 atm×(15.2 L10.0 L)×(315 K306 K)=2.11atm

The final pressure in atmosphere for a sample of carbon dioxide gas at given set of conditions is 2.11atm.

(c)

Interpretation Introduction

Interpretation:

The temperature in degree Celsius for a sample of carbon dioxide gas at given set of conditions by using combined gas law has to be determined.

Concept Introduction:

Combined gas law tells that the Kelvin temperature of gas is directly proportional to the pressure and volume of a fixed amount of gas.  The mathematical expression for combined gas law can be represented as follows,

P1V1T1=P2V2T2

(c)

Expert Solution
Check Mark

Explanation of Solution

Record the given data,

P1=1.35atmV1=15.2LT1=33°Cconvertedto306KP2=7.00atmT2=?V2=0.973L

Now, substitute these values in rearranged combined gas law and do some simple mathematical calculation to gent final answer as follows,

T2=T1×V2V1×P2P1

T2=306K×(7.00atm1.35atm)×(0.973L15.2L)=102K

The obtained pressure is in Kelvin units now we have to convert this value into degree Celsius as follows,

T273

102273=171°C

The final temperature in degree Celsius for a sample of carbon dioxide gas at given set of conditions is 171°C.

(a)

Interpretation Introduction

Interpretation:

The volume in milliliters for a sample of carbon dioxide gas at given set of conditions by using combined gas law has to be determined.

Concept Introduction:

Combined gas law tells that the Kelvin temperature of gas is directly proportional to the pressure and volume of a fixed amount of gas.  The mathematical expression for combined gas law can be represented as follows,

P1V1T1=P2V2T2

(a)

Expert Solution
Check Mark

Explanation of Solution

Record the given data,

P1=1.35atmV1=15.2Lconvertedto15200mLT1=33°Cconvertedto306KP2=6.70atmT2=97°Cconvertedto370KV2=?

Now, substitute these values in rearranged combined gas law and do some simple mathematical calculation to gent final answer as follows,

V2=V1×P1P2×T2T1

V2=15200mL×(1.35atm6.70atm)×(370K306K)=3.70×103mL

The final volume in liters for a sample of carbon dioxide gas at given set of conditions is 3.70×103mL.

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Chapter 7 Solutions

General, Organic, and Biological Chemistry

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