EBK PHYSICS FOR SCIENTISTS AND ENGINEER
EBK PHYSICS FOR SCIENTISTS AND ENGINEER
6th Edition
ISBN: 9781319321710
Author: Mosca
Publisher: VST
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Chapter 19, Problem 80P

(a)

To determine

The volume of gas at A.

(a)

Expert Solution
Check Mark

Answer to Problem 80P

The volume at A is 19.69L .

Explanation of Solution

Given:

The pressure at A is PA=5.00atm .

The temperature at A is TA=600K .

Formula used:

The expression for volume at A is given by,

  VA=nRTAPA

Calculation:

The volume at A is calculated as,

  VA=nRTAPA=( 2.00mol)( 8.314J/ molK )( 600K)( ( 5.00atm )( 101.325kPa 1atm ))=(( 1.969× 10 2 m 3 )( 10 3 L 1 m 3 ))=19.69L

Conclusion:

Therefore, the volume at A is 19.69L .

(b)

To determine

The volume and temperature of gas at B.

(b)

Expert Solution
Check Mark

Answer to Problem 80P

The volume and temperature at B are 39.38L and 1200K respectively.

Explanation of Solution

Formula used:

The expression for volume at B is given by,

  VB=2VA

The expression for temperature at B is given by,

  TB=TAVBVA

Calculation:

The volume at B is calculated as,

  VB=2VA=2(19.69L)=39.38L

The temperature at B is calculated as,

  TB=TAVBVA=(600K)( 39.38L 19.69L)=1200K

Conclusion:

Therefore, the volume and temperature at B are 39.38L and 1200K respectively.

(c)

To determine

The temperature of gas at C.

(c)

Expert Solution
Check Mark

Answer to Problem 80P

The temperature at C is 600K .

Explanation of Solution

Formula used:

The expression for temperature at C as process is isothermal given by,

  TC=TA

Calculation:

The temperature at C is calculated as,

  TC=TA=600K

Conclusion:

Therefore, the temperature at C is 600K .

(d)

To determine

The volume of gas at C.

(d)

Expert Solution
Check Mark

Answer to Problem 80P

The volume at C is 111.4L .

Explanation of Solution

Formula used:

The expression for volume at C as process is given by,

  VC=VB( T B T C )1γ1

Calculation:

The volume at C is calculated as,

  VC=VB( T B T C )1 γ1=(39.38L)( 1200K 600K)1 .0.671=111.4L

Conclusion:

Therefore, the volume at C is 111.4L .

(e)

To determine

The work done by gas in each segment of cycle.

(e)

Expert Solution
Check Mark

Answer to Problem 80P

The work done in process AB, BC and CA are 9.98kJ , 14.96kJ and 17.29kJ respectively.

Explanation of Solution

Formula used:

The work done in process AB is given by,

  WAB=PAVA

The expression for work done in process BC is given by,

  WBC=32nRΔTBC

The expression for work done in process CA is given by,

  WCA=nRTCln(VAVC)

Calculation:

The work done in process AB is calculated as,

  WAB=PAVA=(5.00atm)(19.69L)=(( 98.45atmL)( 0.101325kJ 1atmL ))=9.98kJ

The work done in process BC is calculated as,

  WBC=32nRΔTBC=32(2.00mol)(8.314J/molK)(600K1200K)=(( 14.96× 10 3 J)( 10 3 kJ 1J ))=14.96kJ

The work done in process CA is calculated as,

  WCA=nRTCln( V A V C )=(2.00mol)(8.314J/molK)(600K)ln( 16.69L 111.4L)=(( 17.29× 10 3 J)( 10 3 kJ 1J ))=17.29kJ

Conclusion:

Therefore, the work done in process AB, BC and CA are 9.98kJ , 14.96kJ and 17.29kJ respectively.

(f)

To determine

The heat absorbed in each segment of cycle.

(f)

Expert Solution
Check Mark

Answer to Problem 80P

The heat absorbed in process AB, BC and CA are 24.9kJ , 0 and 17.29kJ respectively.

Explanation of Solution

Formula used:

The expression for heat absorbed in process AB is given by,

  QAB=ncPΔTAB

The expression for heat absorbed in process CA is given by,

  QCA=WCA

Calculation:

The heat absorbed in process AB is calculated as,

  QAB=ncPΔTAB=52(2.00mol)(8.314J/molK)(1200K600K)=(( 24.9× 10 3 J)( 10 3 kJ 1J ))=24.9kJ

The heat absorbed in process BC is 0 because the process is adiabatic.

The heat absorbed in process CA is calculated as,

  QCA=WCA=17.29kJ

Conclusion:

Therefore, the heat absorbed in process AB, BC and CA are 34.92kJ , 0 and 17.29kJ respectively.

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Chapter 19 Solutions

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