EBK PHYSICS FOR SCIENTISTS AND ENGINEER
EBK PHYSICS FOR SCIENTISTS AND ENGINEER
9th Edition
ISBN: 8220100654428
Author: Jewett
Publisher: Cengage Learning US
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Chapter 22, Problem 22.32P

At point A in a Carnot cycle, 2.34 mol of a monatomic ideal gas has a pressure of 1 4000 kPa, a volume of 10.0 L, and a temperature of 720 K. The gas expands isothermally to point B and then expands adiabatically to point C, where its volume is 24.0 L. An isothermal compression brings it to point D, where its volume is 15.0 L. An adiabatic process returns the gas to point A. (a) Determine all the unknown pressures, volumes, and temperatures as you f ill in the following table:

Chapter 22, Problem 22.32P, At point A in a Carnot cycle, 2.34 mol of a monatomic ideal gas has a pressure of 1 4000 kPa, a

(b) Find the energy added by heat, the work done by the engine, and the change in internal energy for each of the steps A → B, BC, CD, and D → A (c) Calculate the efficiency Wnet/|Qk|. (d) Show that the efficiency is equal to 1 - TC/TA, the Carnot efficiency.

(a)

Expert Solution
Check Mark
To determine

The unknown pressures, volumes and the temperature in the table.

Answer to Problem 22.32P

The values of unknown pressures, volumes and the temperature in the table are,

  P V T
A 1400kPa 10.0L 720K
B 875kPa 16.0L 720K
C 445kPa 24.0L 549K
D 712kPa 15.0L 549K

Explanation of Solution

Given: The number of moles of a mono atomic ideal gas is 2.34mol , the pressure of an ideal gas is 1400kPa , the volume of an ideal gas is 10.0L , the temperature of an ideal gas is 720K , the volume of gas when it expands adiabatically to point C is 24.0L and the volume of gas when it is isothermally compressed is 15.0L .

Write the equation of adiabatic process DA .

PDVDγ=PAVAγPD=PA(VAVD)γ

Here,

PD is the pressure of the gas at point D .

VD is the volume of the gas at point D .

PA is the pressure of the gas at point A .

VA is the volume of the gas at point A .

γ is the ratio of specific heats of the gas.

The value of γ is 5/3 for mono atomic gas.

Substitute 1400kPa for PA , 10.0L for VA , 15.0L for VD and 5/3 for γ in above equation to find PD .

PD=(1400kPa)((10.0L)(15.0L))(5/3)712kPa

Thus, the pressure of the gas at point D is 712kPa .

Write the ideal gas equation.

PDVD=nRTDTD=PDVDnR

Here,

n is the number of moles of the gas.

R is the gas constant.

TD is the temperature of the gas at point D .

The value of gas constant is 8.314J/molK .

Substitute 712kPa for PD , 15.0L for VD , 2.34mol for n and 8.314J/molK for R in above equation to find TD .

TD={712kPa(1000Pa1kPa)}{15.0L(1m31000L)}(2.34mol)(8.314J/molK)=548.96K549K

Thus, the temperature of the gas at point D is 549K .

In isothermal process, the temperature is constant.

For isothermal process CD , the temperature at point C is equal to the temperature at point D .

The temperature of the gas at point C is,

TC=549K

Thus, the temperature of the gas at point C is 549K .

Write the ideal gas equation.

PCVC=nRTCPC=nRTCVC

Here,

PC is the pressure of the gas at point C .

VC is the volume of the gas at point C .

Substitute 2.34mol for n , 8.314J/molK for R , 549K for TC and 24.0L for VC in above equation to find PC .

PC=(2.34mol)(8.314J/molK)(549K){24.0L(1m31000L)}=445×103Pa(1kPa1000Pa)=445kPa

Thus, the pressure of the gas at point C is 445kPa .

Write the equation of adiabatic process BC .

TBVBγ1=TCVCγ1VB=VC(TCTB)1/(γ1)

Here,

TB is the temperature of the gas at point B .

VB is the volume of the gas at point B .

Substitute 24.0L for VC , 549K for TC , 720K for TB and 5/3 for γ in above equation to find VB .

VB=(24.0L)((549K)(720K))1/(5/31)=15.979L16.0L

Thus, the volume of the gas at point B is 16.0L .

In isothermal process, the temperature is constant.

For isothermal process AB , the temperature at point A is equal to the temperature at point B .

The temperature of the gas at point B is,

TB=720K

Thus, the temperature of the gas at point B is 720K .

Write the ideal gas equation.

PBVB=nRTBPB=nRTBVB

Here,

PB is the pressure of the gas at point B .

Substitute 2.34mol for n , 8.314J/molK for R , 720K for TB and 16.0L for VB in above equation to find PB .

PB=(2.34mol)(8.314J/molK)(720K){16.0L(1m31000L)}=875464.2Pa(1kPa1000Pa)875kPa

Thus, the pressure of the gas at point B is 875kPa .

Form a table and show the unknown value of pressures, volumes and temperatures.

  P V T
A 1400kPa 10.0L 720K
B 875kPa 16.0L 720K
C 445kPa 24.0L 549K
D 712kPa 15.0L 549K

Conclusion:

Therefore, the values of unknown pressures, volumes and the temperature in the table are,

  P V T
A 1400kPa 10.0L 720K
B 875kPa 16.0L 720K
C 445kPa 24.0L 549K
D 712kPa 15.0L 549K

(b)

Expert Solution
Check Mark
To determine

The energy added by heat, work done by the engine and the change in internal energy for each of the steps AB , BC , CD and DA .

Answer to Problem 22.32P

The values of energy added by heat, work done by the engine and the change in internal energy for each of the steps in the table are,

  Q W ΔEint
A +6.58kJ 6.58kJ 0
B 0 4.99kJ 4.99kJ
C 5.02kJ +5.02kJ 0
D 0 +4.99kJ +4.99kJ

Explanation of Solution

Given: The number of moles of a mono atomic ideal gas is 2.34mol , the pressure of an ideal gas is 1400kPa , the volume of an ideal gas is 10.0L , the temperature of an ideal gas is 720K , the volume of gas when it expands adiabatically to point C is 24.0L and the volume of gas when it is isothermally compressed is 15.0L .

The process AB is an isothermal process in which the temperature is constant due to which the change in temperature ΔT is 0 .

Write the equation of change in temperature for process AB .

ΔEint=nCvΔT

Here,

Cv is the molar heat capacity of the gas at constant volume.

Substitute 0 for ΔT in above equation to find ΔEint .

ΔEint=nCv(0)=0

Thus, the change in internal energy for process AB is 0 .

Write the equation of work done by the engine for process AB .

W=nRTAln(VBVA)

Substitute 2.34mol for n , 8.314J/molK for R , 720K for TA , 16.0L for VB and 10.0L for VA in above equation to find W .

W=(2.34mol)(8.314J/molK)(720K)ln((16.0L)(10.0L))=6583.54J(1kJ1000J)6.58kJ

Thus, the work done by the engine for process AB is 6.58kJ .

Write the equation of isothermal process AB .

Q=W

Substitute 6.58kJ for W in above equation to find Q .

Q=(6.58kJ)=+6.58kJ

Thus, the energy added by heat for process AB is +6.58kJ .

Write the equation of change in temperature for process BC .

ΔEint=nCv(TCTB)

Here,

Cv is the molar heat capacity of the gas at constant volume.

The value of Cv for mono atomic gas is 3R/2 .

Substitute 3R/2 for Cv in above equation.

ΔEint=n(32R)(TCTB)

Substitute 2.34mol for n , 8.314J/molK for R , 549K for TC and 720K for TB in above equation to find ΔEint .

ΔEint=(2.34mol){32(8.314J/molK)}{(549K)(720K)}=4990.14J(1kJ1000J)4.99kJ

Thus, the change in internal energy for process BC is 4.99kJ .

The process BC is an adiabatic process. In adiabatic process, the transfer of heat is zero due to which Q is zero.

Thus, the energy added by heat for process BC is 0 .

Write the equation of change in internal energy for process BC .

ΔEint=Q+WW=ΔEintQ

Substitute 4.99kJ for ΔEint and 0 for Q in above equation to find W .

W=(4.99kJ)0=4.99kJ

Thus, the work done by the engine for process BC is 4.99kJ .

The process CD is an isothermal process in which the temperature is constant due to which the change in temperature ΔT is 0 .

Write the equation of change in temperature for process CD .

ΔEint=nCvΔT

Substitute 0 for ΔT in above equation to find ΔEint .

ΔEint=nCv(0)=0

Thus, the change in internal energy for process CD is 0 .

Write the equation of work done by the engine for process CD .

W=nRTCln(VDVC)

Substitute 2.34mol for n , 8.314J/molK for R , 549K for TC , 15.0L for VD and 24.0L for VC in above equation to find W .

W=(2.34mol)(8.314J/molK)(549K)ln((15.0L)(24.0L))={5019.95J(1kJ1000J)}+5.02kJ

Thus, the work done by the engine for process CD is +5.02kJ .

Write the equation of isothermal process CD .

Q=W

Substitute +5.02kJ for W in above equation to find Q .

Q=(+5.02kJ)=5.02kJ

Thus, the energy added by heat for process CD is 5.02kJ .

Write the equation of change in temperature for process DA .

ΔEint=nCv(TATD)

Substitute 3R/2 for Cv in above equation.

ΔEint=n(32R)(TATD)

Substitute 2.34mol for n , 8.314J/molK for R , 720K for TA and 549K for TD in above equation to find ΔEint .

ΔEint=(2.34mol){32(8.314J/molK)}{(720K)(549K)}=+4990.14J(1kJ1000J)+4.99kJ

Thus, the change in internal energy for process DA is +4.99kJ .

The process DA is an adiabatic process. In adiabatic process, the transfer of heat is zero due to which Q is zero.

Thus, the energy added by heat for process DA is 0 .

Write the equation of change in internal energy for process DA .

ΔEint=Q+WW=ΔEintQ

Substitute +4.99kJ for ΔEint and 0 for Q in above equation to find W .

W=(+4.99kJ)0=+4.99kJ

Thus, the work done by the engine for process DA is +4.99kJ .

Form a table and show the value of energy added by heat, work done by the engine and the change in internal energy.

  Q W ΔEint
A +6.58kJ 6.58kJ 0
B 0 4.99kJ 4.99kJ
C 5.02kJ +5.02kJ 0
D 0 +4.99kJ +4.99kJ

Conclusion:

Therefore, the values of energy added by heat, work done by the engine and the change in internal energy for each of the steps in the table are,

  Q W ΔEint
A +6.58kJ 6.58kJ 0
B 0 4.99kJ 4.99kJ
C 5.02kJ +5.02kJ 0
D 0 +4.99kJ +4.99kJ

(c)

Expert Solution
Check Mark
To determine

The value of efficiency Wnet/|Qh| .

Answer to Problem 22.32P

The value of efficiency Wnet/|Qh| is 0.237 .

Explanation of Solution

Given: The number of moles of a mono atomic ideal gas is 2.34mol , the pressure of an ideal gas is 1400kPa , the volume of an ideal gas is 10.0L , the temperature of an ideal gas is 720K , the volume of gas when it expands adiabatically to point C is 24.0L and the volume of gas when it is isothermally compressed is 15.0L .

Calculate the net work done from the table is,

Wnet={(6.58kJ)+(4.99kJ)+(5.02kJ)+(+4.99kJ)}=1.56kJ

Write the equation for efficiency.

e=Wnet|Qh|

Here,

Qh is the energy from the hot reservoir which the energy in process AB .

Substitute 1.56kJ for Wnet and 6.58kJ for Qh in above equation.

e=(1.56kJ)|6.58kJ|=1.56kJ6.58kJ=0.237

The value of efficiency Wnet/|Qh| is 0.237 .

Conclusion:

Therefore, the value of efficiency Wnet/|Qh| is 0.237 .

(d)

Expert Solution
Check Mark
To determine

To show: The efficiency is equal to the Carnot efficiency 1TC/TA .

Answer to Problem 22.32P

The efficiency is equal to the Carnot efficiency 1TC/TA .

Explanation of Solution

Given: The number of moles of a mono atomic ideal gas is 2.34mol , the pressure of an ideal gas is 1400kPa , the volume of an ideal gas is 10.0L , the temperature of an ideal gas is 720K , the volume of gas when it expands adiabatically to point C is 24.0L and the volume of gas when it is isothermally compressed is 15.0L .

Write the equation for Carnot efficiency.

eC=1TcTh

Here,

Tc is the temperature for cold reservoir.

Th is the temperature for hot reservoir.

The value of Tc is 549K and Th is 720K .

Substitute 549K for Tc and 720K for Th in above equation to find eC .

eC=1(549K)(720K)=0.23750.237

Thus, the Carnot efficiency is 0.237 .

Write the equation for efficiency.

e=1TCTA

Substitute 549K for TC and 720K for TA in above equation to find e .

e=1(549K)(720K)=0.23750.237

The value of efficiency is 0.237 which is equal to the Carnot efficiency.

Conclusion:

Therefore, the efficiency is equal to the Carnot efficiency 1TC/TA .

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Chapter 22 Solutions

EBK PHYSICS FOR SCIENTISTS AND ENGINEER

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