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 18, Problem 69P

(a)

To determine

The final temperature, volume, work done and heat absorbed if the expansion is isothermal.

(a)

Expert Solution
Check Mark

Answer to Problem 69P

The final temperature, volume, work done and heat absorbed are 300K , 7.80L , 1.14kJ and 1.14kJ respectively.

Explanation of Solution

Given:

The initial pressure is 400kPa .

The final pressure is 160kPa

The initial temperature is 300K .

Formula used:

The expression for initial volume is given by,

  Vi=nRTP

The expression for final volume is given by,

  Vf=ViPiPf

The expression for work done is given by,

  Wbygas=nRTlnVfVi

The expression for heat absorbed is given by,

  Qin=ΔEint+Wbygas

Calculation:

The temperature remains same for an isothermal expansion.

The initial volume is calculated as,

  Vi=nRTP=( 0.5mol)( 8.314J/ mol K)( 300K)400kPa=(( 3.12× 10 3 m 3 )( 10 3 L 1 m 3 ))=3.12L

The final volume is calculated as,

  Vf=ViPiPf=(3.12L)( 400kPa 160kPa)=7.80L

The work done by gas is calculated as,

  Wbygas=nRTlnVfVi=(0.5mol)(8.314J/molK)(300K)ln( 7.80L 3.12L)=(( 1.14× 10 3 J)( 10 3 kJ 1J ))=1.14kJ

The heat absorbed is calculated as,

  Qin=ΔEint+Wbygas=0+(1.14kJ)=1.14kJ

Conclusion:

Therefore, the final temperature, volume, work done and heat absorbed are 300K , 7.80L , 1.14kJ and 1.14kJ respectively.

(b)

To determine

The final temperature, volume, work done and heat absorbed if the expansion is adiabatic.

(b)

Expert Solution
Check Mark

Answer to Problem 69P

The final temperature, volume, work done and heat absorbed are 208K , 5.41L , 574J and 0 respectively.

Explanation of Solution

Formula used:

The expression for final temperature is given by,

  Tf=PfVfnR

The expression for final volume is given by,

  Vf=Vi( P i P f )r

The expression for work done is given by,

  W=32nRΔT

Calculation:

The final volume is calculated as,

  Vf=Vi( P i P f )r=(3.12L)( 400kPa 160kPa)35=5.41L

The final temperature is calculated as,

  Tf=PfVfnR=( 160kPa)( ( 5.41L )( 10 3 m 3 1L ))( 0.5mol)( 8.314J/ mol K)=208K

The work done by gas is calculated as,

  W=32nRΔT=32(0.5mol)(8.314J/molK)(300K280K)=574J

The heat absorbed is zero in case of adiabatic process.

Conclusion:

Therefore, the final temperature, volume, work done and heat absorbed are 208K , 5.41L , 574J and 0 respectively.

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Chapter 18 Solutions

EBK PHYSICS FOR SCIENTISTS AND ENGINEER

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