GENERAL,ORGANIC,+BIOCHEMISTRY
GENERAL,ORGANIC,+BIOCHEMISTRY
10th Edition
ISBN: 9781260148954
Author: Denniston
Publisher: RENT MCG
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Question
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Chapter 5.1, Problem 5.2PP

(a)

Interpretation Introduction

Interpretation:

Volume that will be occupied by nitrogen gas at 100°C has to be calculated if the sample of nitrogen at 25°C has a volume of 3.00L.

Concept Introduction:

Charles’s law states that the volume of gas is directly proportional to the absolute temperature if the pressure and number of moles of the gas are kept as constant.  This can be expressed as,

    VT=kc

During temperature change, the initial and final can be related as,

    ViTi=VfTf

(a)

Expert Solution
Check Mark

Explanation of Solution

Given initial volume (Vi) of nitrogen gas is 3.00L and initial temperature (Ti) is 25°C.  Final temperature (Tf) is given as 100°C.

Initial and final temperature conversion into Kelvin scale:

    Ti=(25+273.15)K=298.15KTf=(100+273.15)K=373.15K

Charles’s law relating the volume and temperature can be given as,

    ViTi=VfTf

Rearranging the above equation to calculate the final volume,

    Vf=ViTfTi        (1)

Substituting the values in equation (1),

    Vf=(3.00L)(373.15K)(298.15K)=3.754L

Therefore, the volume occupied by nitrogen gas at 100°C is 3.754L.

(b)

Interpretation Introduction

Interpretation:

Volume that will be occupied by nitrogen gas at 150°F has to be calculated if the sample of nitrogen at 25°C has a volume of 3.00L.

Concept Introduction:

Refer part (a).

(b)

Expert Solution
Check Mark

Explanation of Solution

Given initial volume (Vi) of nitrogen gas is 3.00L and initial temperature (Ti) is 25°C.  Final temperature (Tf) is given as 150°F.

Initial temperature conversion into Kelvin scale:

    Ti=(25+273.15)K=298.15K

Final temperature conversion to Kelvin scale:

    Tf=(15032)×59+273.15=338.70K

Charles’s law relating the volume and temperature can be given as,

    ViTi=VfTf

Rearranging the above equation to calculate the final volume,

    Vf=ViTfTi        (1)

Substituting the values in equation (1),

    Vf=(3.00L)(338.70K)(298.15K)=3.408L

Therefore, the volume occupied by nitrogen gas at 150°F is 3.408L.

(c)

Interpretation Introduction

Interpretation:

Volume that will be occupied by nitrogen gas at 273K has to be calculated if the sample of nitrogen at 25°C has a volume of 3.00L.

Concept Introduction:

Refer part (a).

(c)

Expert Solution
Check Mark

Explanation of Solution

Given initial volume (Vi) of nitrogen gas is 3.00L and initial temperature (Ti) is 25°C.  Final temperature (Tf) is given as 273K.

Initial temperature conversion into Kelvin scale:

    Ti=(25+273.15)K=298.15K

Charles’s law relating the volume and temperature can be given as,

    ViTi=VfTf

Rearranging the above equation to calculate the final volume,

    Vf=ViTfTi        (1)

Substituting the values in equation (1),

    Vf=(3.00L)(273K)(298.15K)=2.746L

Therefore, the volume occupied by nitrogen gas at 273K is 2.746L.

(d)

Interpretation Introduction

Interpretation:

Volume that will be occupied by nitrogen gas at 546K has to be calculated if the sample of nitrogen at 25°C has a volume of 3.00L.

Concept Introduction:

Refer part (a).

(d)

Expert Solution
Check Mark

Explanation of Solution

Given initial volume (Vi) of nitrogen gas is 3.00L and initial temperature (Ti) is 25°C.  Final temperature (Tf) is given as 546K.

Initial temperature conversion into Kelvin scale:

    Ti=(25+273.15)K=298.15K

Charles’s law relating the volume and temperature can be given as,

    ViTi=VfTf

Rearranging the above equation to calculate the final volume,

    Vf=ViTfTi        (1)

Substituting the values in equation (1),

    Vf=(3.00L)(546K)(298.15K)=5.493L

Therefore, the volume occupied by nitrogen gas at 546K is 5.493L.

(e)

Interpretation Introduction

Interpretation:

Volume that will be occupied by nitrogen gas at 0°C has to be calculated if the sample of nitrogen at 25°C has a volume of 3.00L.

Concept Introduction:

Refer part (a).

(e)

Expert Solution
Check Mark

Explanation of Solution

Given initial volume (Vi) of nitrogen gas is 3.00L and initial temperature (Ti) is 25°C.  Final temperature (Tf) is given as 100°C.

Initial and final temperature conversion into Kelvin scale:

    Ti=(25+273.15)K=298.15KTf=(0+273.15)K=273.15K

Charles’s law relating the volume and temperature can be given as,

    ViTi=VfTf

Rearranging the above equation to calculate the final volume,

    Vf=ViTfTi        (1)

Substituting the values in equation (1),

    Vf=(3.00L)(273.15K)(298.15K)=2.748L

Therefore, the volume occupied by nitrogen gas at 0°C is 2.748L.

(f)

Interpretation Introduction

Interpretation:

Volume that will be occupied by nitrogen gas at 373K has to be calculated if the sample of nitrogen at 25°C has a volume of 3.00L.

Concept Introduction:

Refer part (a).

(f)

Expert Solution
Check Mark

Explanation of Solution

Given initial volume (Vi) of nitrogen gas is 3.00L and initial temperature (Ti) is 25°C.  Final temperature (Tf) is given as 373K.

Initial temperature conversion into Kelvin scale:

    Ti=(25+273.15)K=298.15K

Charles’s law relating the volume and temperature can be given as,

    ViTi=VfTf

Rearranging the above equation to calculate the final volume,

    Vf=ViTfTi        (1)

Substituting the values in equation (1),

    Vf=(3.00L)(373K)(298.15K)=3.753L

Therefore, the volume occupied by nitrogen gas at 373K is 3.753L.

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Chapter 5 Solutions

GENERAL,ORGANIC,+BIOCHEMISTRY

Ch. 5.1 - Prob. 5.3QCh. 5.1 - Prob. 5.4QCh. 5.1 - Prob. 5.5QCh. 5.1 - Prob. 5.6QCh. 5.2 - Prob. 5.7QCh. 5.2 - Prob. 5.8QCh. 5.2 - Prob. 5.9QCh. 5.2 - Prob. 5.10QCh. 5.2 - Prob. 5.11QCh. 5.2 - Prob. 5.12QCh. 5.3 - Prob. 5.13QCh. 5.3 - Prob. 5.14QCh. 5 - Prob. 5.15QPCh. 5 - Prob. 5.16QPCh. 5 - Prob. 5.17QPCh. 5 - Prob. 5.18QPCh. 5 - Prob. 5.19QPCh. 5 - Prob. 5.20QPCh. 5 - Prob. 5.21QPCh. 5 - Prob. 5.22QPCh. 5 - Prob. 5.23QPCh. 5 - Prob. 5.24QPCh. 5 - Prob. 5.25QPCh. 5 - Prob. 5.26QPCh. 5 - Prob. 5.27QPCh. 5 - Prob. 5.28QPCh. 5 - Prob. 5.29QPCh. 5 - Prob. 5.30QPCh. 5 - Prob. 5.31QPCh. 5 - Prob. 5.32QPCh. 5 - Prob. 5.33QPCh. 5 - Prob. 5.34QPCh. 5 - Prob. 5.35QPCh. 5 - Prob. 5.36QPCh. 5 - Prob. 5.37QPCh. 5 - Prob. 5.38QPCh. 5 - Calculate the pressure, in atm, required to...Ch. 5 - A balloon filled with helium gas at 1.00 atm...Ch. 5 - Prob. 5.41QPCh. 5 - Prob. 5.42QPCh. 5 - Prob. 5.43QPCh. 5 - The temperature on a summer day may be 90°F....Ch. 5 - Prob. 5.45QPCh. 5 - Prob. 5.46QPCh. 5 - Prob. 5.47QPCh. 5 - Prob. 5.48QPCh. 5 - A balloon containing a sample of helium gas is...Ch. 5 - The balloon described in Question 5.49 was then...Ch. 5 - Prob. 5.51QPCh. 5 - A balloon, filled with an ideal gas, has a volume...Ch. 5 - Prob. 5.53QPCh. 5 - Prob. 5.54QPCh. 5 - Prob. 5.55QPCh. 5 - Prob. 5.56QPCh. 5 - Prob. 5.57QPCh. 5 - A sealed balloon filled with helium gas occupies...Ch. 5 - A 5.00-L balloon exerts a pressure of 2.00 atm at...Ch. 5 - If we double the pressure and temperature of the...Ch. 5 - State Avogadro’s law in words. Ch. 5 - Prob. 5.62QPCh. 5 - Prob. 5.63QPCh. 5 - Prob. 5.64QPCh. 5 - Prob. 5.65QPCh. 5 - Prob. 5.66QPCh. 5 - Prob. 5.67QPCh. 5 - Prob. 5.68QPCh. 5 - Prob. 5.69QPCh. 5 - Prob. 5.70QPCh. 5 - Prob. 5.71QPCh. 5 - Prob. 5.72QPCh. 5 - Prob. 5.73QPCh. 5 - Prob. 5.74QPCh. 5 - Prob. 5.75QPCh. 5 - Calculate the pressure (atm) exerted by 1.00 mol...Ch. 5 - A sample of argon (Ar) gas occupies 65.0 mL at...Ch. 5 - A sample of O2 gas occupies 257 mL at 20°C and...Ch. 5 - Prob. 5.79QPCh. 5 - Prob. 5.80QPCh. 5 - Prob. 5.81QPCh. 5 - Calculate the volume of 6.00 mol O2 gas at 30 cm...Ch. 5 - State Dalton’s law in words. Ch. 5 - Prob. 5.84QPCh. 5 - Prob. 5.85QPCh. 5 - Prob. 5.86QPCh. 5 - Prob. 5.87QPCh. 5 - Prob. 5.88QPCh. 5 - Prob. 5.89QPCh. 5 - Prob. 5.90QPCh. 5 - Prob. 5.91QPCh. 5 - Prob. 5.92QPCh. 5 - Prob. 5.93QPCh. 5 - Prob. 5.94QPCh. 5 - Prob. 5.95QPCh. 5 - Prob. 5.96QPCh. 5 - Prob. 5.97QPCh. 5 - Prob. 5.98QPCh. 5 - Prob. 5.99QPCh. 5 - Prob. 5.100QPCh. 5 - Prob. 5.101QPCh. 5 - Prob. 5.102QPCh. 5 - Prob. 5.103QPCh. 5 - Prob. 5.104QPCh. 5 - Prob. 5.105QPCh. 5 - Prob. 5.106QPCh. 5 - Prob. 5.107QPCh. 5 - Prob. 5.108QPCh. 5 - Prob. 5.109QPCh. 5 - Prob. 5.110QPCh. 5 - Prob. 5.111QPCh. 5 - Prob. 5.112QPCh. 5 - Prob. 2MCPCh. 5 - Prob. 5MCPCh. 5 - Prob. 6MCPCh. 5 - Prob. 8MCPCh. 5 - Prob. 9MCPCh. 5 - Prob. 10MCP
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