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Physics / Sears and Zemansky's University Physics with Modern Physics 13 / Chapter 20 / Problem 41P

Sears and Zemansky's University Physics with Modern Physics | 13th Edition | ISBN: 9780321696861 | Authors: Hugh D. Young; Roger A. Freedman; A. Lewis Ford

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Answer: CALC You build a heat engine that takes 1.00 mol

Chapter 20, Problem 41P

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QUESTION:

Problem 41P

CALC You build a heat engine that takes 1.00 mol of an ideal diatomic gas through the cycle shown in Fig. P20.41. (a) Show that segment ab is an isothermal compression. (b) During which segment(s) of the cycle is heat absorbed by the gas? During which segment(s) is heat rejected? How do you know? (c) Calculate the temperature at points a, b, and c. (d) Calculate the net heat exchanged with the surroundings and the net work done by the engine in one cycle. (e) Calculate the thermal efficiency of the engine.

Equation Transcription:

$p(Pa)$

$V({m}^{3})$

Text Transcription:

p(Pa)

V(m^3)

Questions & Answers

QUESTION:

Problem 41P

CALC You build a heat engine that takes 1.00 mol of an ideal diatomic gas through the cycle shown in Fig. P20.41. (a) Show that segment ab is an isothermal compression. (b) During which segment(s) of the cycle is heat absorbed by the gas? During which segment(s) is heat rejected? How do you know? (c) Calculate the temperature at points a, b, and c. (d) Calculate the net heat exchanged with the surroundings and the net work done by the engine in one cycle. (e) Calculate the thermal efficiency of the engine.

Equation Transcription:

$p(Pa)$

$V({m}^{3})$

Text Transcription:

p(Pa)

V(m^3)

ANSWER:

Solution 41P

Step 1:

We need to show process a-b is isothermal compression

Data given

${P}_{a}=\ 2\times{}1{0}^{5}\ Pa$

${P}_{b}=\ 4\times{}1{0}^{5}\ Pa$

${V}_{a}=\ 0.010\ {m}^{3}$

${V}_{b}=\ 0.005\ {m}^{3}$

${P}_{c}=\ 0.010\ {m}^{3}$

${V}_{c}=\ \ 4\times{}1{0}^{5}\ Pa$

From ideal gas equation we have

$PV=\ nRT\$

$nR\ =\ constant\$

Consider

${P}_{a}{V}_{a}=\ {T}_{a}$

$2\times{}1{0}^{5}\ Pa\times{}0.010\ {m}^{3}=\ 2\times{}1{0}^{3}K\$

Similarly

${P}_{b}{V}_{b}=\ {T}_{b}$

$\ 4\times{}1{0}^{5}\ Pa\times{}\ 0.005\ {m}^{3}=\ 2\times{}1{0}^{3}K\$

From the above calculations we have

${T}_{a}=\ {T}_{b}$

This shows the reaction from a to b is isothermal compression

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