EBK GENERAL CHEMISTRY: THE ESSENTIAL CO
EBK GENERAL CHEMISTRY: THE ESSENTIAL CO
7th Edition
ISBN: 8220106637203
Author: Chang
Publisher: YUZU
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Chapter 6, Problem 6.77QP

(a)

Interpretation Introduction

Interpretation:

The final temperature of the given sample has to be calculated.

Concept Introduction:

The ideal gas Law equation is,

PV=nRTPPressureVVolumenNumberofmolesRIdealgasconstantTTemperature

The first law of thermodynamics states that energy can be either destroyed or created but instead it can be converted from one form to other.

The equation for the first law of thermodynamics can be given as,

  ΔU=q+w

W=PdV ; Where P is the pressure, dV is the change in volume.

ΔU=CvdTq=CPdTR=Cp-Cv

Here CpandCv are the molar heat capacity at constant pressure and molar heat capacity at constant volume respectively.

(a)

Expert Solution
Check Mark

Explanation of Solution

Given,

A 1.00-mole sample of ammonia at 14.0atm and 250C in a cylinder fitted with a movable piston expands against a constant external pressure of 1.00atm. At equilibrium, the pressure and volume of the gas are 1.00atmand23.5L , respectively.

The final temperature of the sample can be calculated as follows,

P=14atmV=?T=25oC

The volume of the substance will change in accordance to the piston’s movement.

Thus,

According to the Ideal gas equation,

V=nRTP=1×0.0821×29814=1.75L

At equilibrium the volume is 23.5L thus, dV=23.51.75=21.75L ,

W=-Pdv=-(1×21.75)L=-21.75atm.L

 atm.L=101.33J

Therefore,

W=-21.75atm.L×101.33J=-2204J

First law of thermodynamics can be written as,

ΔU=q+WCvdT=CpdT+w(Cp-Cv)dT=-WdT=-WR

Substituting the known values in the above equation,

dT=2204J8.314J/K=265K

dT=TFinal-TInitial265K=TFinal-298KTFinal=265K+298KTFinal=563K

Therefore, final temperature (TFinal) of the given sample is 563K

(b)

Interpretation Introduction

Interpretation:

q,wandΔU for the given process have to be determined.

Concept Introduction:

Specific heat:

Specific heat can be defined as quantity of heat required to raise the temperature of 1g substance by 1°C.  The relationship between heat and change in temperature can be expressed by the equation given below.

    q=cmΔT

Where   q= Heat added

       c= Specific heat

    m= Mass

  ΔT= Change in temperature.

The first law of thermodynamics states that energy can be either destroyed or created but instead it can be converted from one form to other.

The equation for the first law of thermodynamics can be given as,

  ΔU=q+w

W=PdV ; Where P is the pressure, dV is the change in volume.

ΔU=CvdTq=CPdTR=Cp-Cv

Here CpandCv are the molar heat capacity at constant pressure and molar heat capacity at constant volume respectively.

(b)

Expert Solution
Check Mark

Explanation of Solution

A 1.00-mole sample of ammonia at 14.0atm and 250C in a cylinder fitted with a movable piston expands against a constant external pressure of 1.00atm. At equilibrium, the pressure and volume of the gas are 1.00atmand23.5L , respectively.

At equilibrium the volume is 23.5L thus,dV=23.51.75=21.75L,

W=-Pdv=-(1×21.75)L=-21.75atm.L

 atm.L=101.33J

Therefore,

W=-21.75atm.L×101.33J=-2204J

First law of thermodynamics can be written as,

ΔU=q+WCvdT=CpdT+w(Cp-Cv)dT=-WdT=-WR

Substituting the known values in the above equation,

dT=2204J8.314J/K=265K

q=cmΔTmolarmassofammonia=17.03mol/gc=0.0258J/g.0CΔT=265K=-8.050Cq=17.03mol/g×0.0258J/g.0C×(-8.15oC)=-3.54JoC

The equation for the first law of thermodynamics is,

  ΔU=q+w

ΔU can be determined as follows,

q=3.54JoCW=2204J

ΔU=(-3.54)+(-2204)=-2207.5J

For the given process,

W=-2204J

q=-3.54JoC

ΔU=-2207.5J

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Chapter 6 Solutions

EBK GENERAL CHEMISTRY: THE ESSENTIAL CO

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