Physics for Scientists and Engineers with Modern Physics
Physics for Scientists and Engineers with Modern Physics
10th Edition
ISBN: 9781337553292
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 19, Problem 22P

In Figure P19.22, the change in internal energy of a gas that is taken from A to C along the blue path is +800 J. The work done on the gas along the red path ABC is −500 J. (a) How much energy must be added to the system by heat as it goes from A through B to C? (b) If the pressure at point A is five times that of point C, what is the work done on the system in going from C to D?

Figure P19.22

Chapter 19, Problem 22P, In Figure P19.22, the change in internal energy of a gas that is taken from A to C along the blue

(c) What is the energy exchanged with the surroundings by heat as the gas goes from C to A along the green path? (d) If the change in internal energy in going from point D to point A is +500 J, how much energy must be added to the system by heat as it goes from point C to point D?

(a)

Expert Solution
Check Mark
To determine

The energy which must be added to the system by heat which goes from A through B to C.

Answer to Problem 22P

The energy which must be added to the system by heat which goes from A through B to C is 1300J .

Explanation of Solution

Given info: The change in internal energy of a gas which is taken along the blue path from A to C is +800J , the work done on the gas along the red path ABC is 500J .

Write the equation of first law of thermodynamics.

(ΔEint)ABC=QABC+WABCQABC=(ΔEint)ABCWABC (1)

Here,

(ΔEint)ABC is the change in internal energy along the red path ABC.

QABC is the energy which must be added to the system by heat along the red path ABC.

WABC is the work done on the gas along the red path ABC.

Write the equation of conservation of energy.

(ΔEint)ABC=(ΔEint)AC

Here,

(ΔEint)ABC is the change in internal energy along the red path ABC.

(ΔEint)AC is the change in internal energy along the blue path AC.

Substitute +800J for (ΔEint)AC in the above equation to get the change in internal energy along ABC.

(ΔEint)ABC=+800J

Substitute +800J for (ΔEint)ABC and 500J for WABC in the equation (1) to get the energy QABC which must be added to the system by heat along ABC.

QABC=(+800J)(500J)=800J+500J=1300J

Conclusion:

Therefore, the energy which must be added to the system by heat which goes from A through B to C is 1300J

(b)

Expert Solution
Check Mark
To determine

The work done on the system from C to D if pressure at A is five times that of point C.

Answer to Problem 22P

The work done on the system from C to D if pressure at A is five times that of point C is 100J .

Explanation of Solution

Given info: The change in internal energy of a gas which is taken along the blue path from A to C is +800J , the work done on the gas along the red path ABC is 500J , The pressure at point A is five times that of point C.

Write the equation to calculate the work done on the system from C to D.

WCD=PCΔVCD (2)

Here,

WCD is the work done on the system from C to D.

PC is the pressure at point C.

ΔVCD is the change in volume of the gas.

Here,

The pressure at point A is five times that of point C.

PA=5PCPC=PA5

And the volume is,

ΔVAB=ΔVCDΔVCD=ΔVAB

Substitute (PA5) for PC and ΔVAB for ΔVCD in equation (2) to find the work done on the system from C to D.

WCD=(PA5)(ΔVAB)=15(PAΔVAB)=15WAB

Substitute 500J for WAB in the above equation to find the work done on the system from C to D.

WCD=15(500J)=500J5=100J

Conclusion:

Therefore, the work done on the system from C to D if pressure at A is five times that of point C is 100J .

(c)

Expert Solution
Check Mark
To determine

The energy exchanged with the surroundings by heat along the green path from C to A.

Answer to Problem 22P

The energy exchanged with the surroundings by heat along the green path from C to A is 900J .

Explanation of Solution

Given info: The change in internal energy of a gas which is taken along the blue path from A to C is +800J , the work done on the gas along the red path ABC is 500J .

Write the equation of first law of thermodynamics.

(ΔEint)CA=QCA+WCDAQCA=(ΔEint)CAWCDA (3)

Here,

(ΔEint)CA is the change in the internal energy along the green path from C to A.

QCA is the energy exchanged with the surroundings by heat along the green path from C to A.

WCDA is the work done on the gas along the green path from C to A.

Write the equation to calculate work done along the green path from C to A.

WCDA=WCD+WDA=WCD+PΔVCDA=WCD+P(VDVA)

Here, the volume VD and VA are equal. Thus, the difference ΔVCDA is zero.

Substitute 0 for ΔVCDA in the above equation.

WCDA=WCD+P(0)WCDA=WCD

Substitute WCD for WCDA in equation (3).

QCA=(ΔEint)CAWCD

Substitute 800J for (ΔEint)CA and 100J for WCD in the above equation to find the energy exchanged with the surroundings by heat along the green path from C to A.

QCA=(800J)(100J)=900J

Conclusion:

Therefore, the energy exchanged with the surroundings by heat along the green path from C to A is 900J .

(d)

Expert Solution
Check Mark
To determine

The energy which must be added to the system by heat which goes from C to D.

Answer to Problem 22P

The energy which must be added to the system by heat which goes from C to D

is 1400J .

Explanation of Solution

Given info: The change in internal energy of a gas which is taken along the blue path from A to C is +800J , the work done on the gas along the red path ABC is 500J .

Write the equation of first law of thermodynamics.

(ΔEint)CD=QCD+WCDQCD=(ΔEint)CDWCDQCD=[(ΔEint)CDA(ΔEint)DA]WCD

Here,

(ΔEint)CD is the change in internal energy along CD.

QCD is the energy which must be added to the system by heat which goes from C to D.

WCD is the work done on the gas along CD.

(ΔEint)CDA is the change in internal energy along CDA.

(ΔEint)DA is the change in internal energy along DA.

Substitute 800J for (ΔEint)CDA , 500J for (ΔEint)DA and WABC and 100J for WCD in the above equation to get the energy which must be added to the system by heat which goes from C to D.

(ΔEint)CD=[(800J)(500J)](100)=1300J100J=1400J

Conclusion:

Therefore, the energy which must be added to the system by heat which goes from C to D is 1400J .

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Chapter 19 Solutions

Physics for Scientists and Engineers with Modern Physics

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