Physics
Physics
3rd Edition
ISBN: 9780073512150
Author: Alan Giambattista, Betty Richardson, Robert C. Richardson Dr.
Publisher: McGraw-Hill Education
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Chapter 15, Problem 86P

(a)

To determine

The work done by engine during the 4 steps and the net work done.

(a)

Expert Solution
Check Mark

Answer to Problem 86P

The work done by engine during the 4 steps are 692J,0J,506J ,0J and the net work done is 186J.

Explanation of Solution

Write the expression for work done during process A.

WA=nRTln(VfVi)

Here, T is the temperature, WA is the work done, Vi is the volume, Vf is the final volume and R is the gas constant.

B and D are isochoric processes. Therefore, work done is zero.

Write the expression for work done during process C.

WC=nRTln(VfVi)

Here, T is the temperature, WC is the work done, Vi is the volume, Vf is the final volume and R is the gas constant.

Conclusion:

For the process A,

Substitute 373 K for T,0.020m3 for Vi, 0.025m3 for Vf, 1.00 mol for n and 8.314Jmol1K1 for R to get WA.

WA=(1.00mol)(8.314Jmol1K1)(373K)ln(0.025m30.020m3)=692J

For the process C,

Substitute 273 K for T, 0.025m3 for Vi, 0.020m3 for Vf, 1.00 mol for n and 8.314Jmol1K1 for R to get WC.

WC=(1.00mol)(8.314Jmol1K1)(273K)ln(0.020m30.025m3)=506J

The net work done is,

Wnet=(692J)+(0J)+(506J)+(0J)=186J

Therefore, work done by engine during the 4 steps are 692J,0J,506J ,0J and the net work done is 186J.

(b)

To determine

The efficiency of the engine.

(b)

Expert Solution
Check Mark

Answer to Problem 86P

The efficiency of the engine is 0.0670.

Explanation of Solution

Write the expression for efficiency of the engine.

e=WnetQin

Here, e is the efficiency and Qin is input heat.

Write the expression for input heat.

Qin=52nRΔT+WA

Here, ΔU is the change in internal energy and ΔT is the change intemperature.

Therefore, the input heat is,

Qin=52(1.00mol)(8.314Jmol1K1)(373K273K)+(692J)=2792J

Conclusion:

Substitute 186J for Wnet and 2792J for Qin to get e.

e=186J2792J=0.0670

Therefore, efficiency of the engine is 0.0670.

(c)

To determine

The ratio of efficiency of reversible engine and the given engine.

(c)

Expert Solution
Check Mark

Answer to Problem 86P

The ratio of efficiency of reversible engine and the given engine is 4.00.

Explanation of Solution

Write the expression for the efficiency of reversible engine.

er=1TCTH

Here, TC is the temperature of cold reservoir and TH is the temperature of the hot reservoir.

Conclusion:

Substitute 273 K for TC and 373 K for TH to get er.

er=1273K373K=0.268

The ratio is given by,

ere=0.2680.0670=4.00

The ratio of efficiency of reversible engine and the given engine is 4.00.

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