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Physics / Chapter 18: Temperature, Heat, and the First Law of Thermodynamics / Section 18-8

A Closer Look at Heat and Work — Question 7

Section 18-8 · Physics pages 2964–2965

Question 7 A triangular cycle exposes both sign and unit conversion A gas follows the quasistatic cycle A (2 L, 100 kPa) to B (6 L, 100 kPa) to C (6 L, 300 kPa) and straight back to A. Use work positive when done by the gas. V p A B C Schematic pressure–volume paths; work is the signed area under a process curve. (a) Calculate work on each leg and the net cycle work. (b) Calculate the cycle work when the direction is reversed. Confirm both results by the enclosed triangular area.
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Question 7 - Solution A triangular cycle exposes both sign and unit conversion V p A B C (a) Method and calculation WAB = 100(6 −2) = 400 J, WBC = 0. WCA = 300 + 100 2 (2 −6) = −800 J. Wcycle = −400 J. (b) Interpretation and further calculation Atriangle = 1 2(6 −2)(300 −100) = 400 kPa L = 400 J. Wreversed = +400 J. The stated counterclockwise traversal requires 400 J net work input; reversing the path changes the sign of every line integral. Physical check Pressure in kPa and volume in litres already produce joules, since 103 × 10−3 = 1.

Question 7

A triangular cycle exposes both sign and unit conversion

A gas follows the quasistatic cycle A (2L,100kPa)(2\,\mathrm L,100\,\mathrm{kPa}) to B (6L,100kPa)(6\,\mathrm L,100\,\mathrm{kPa}) to C (6L,300kPa)(6\,\mathrm L,300\,\mathrm{kPa}) and straight back to A. Use work positive when done by the gas.

Diagram shown in the original SVG worksheet.

  1. Calculate work on each leg and the net cycle work.

  2. Calculate the cycle work when the direction is reversed. Confirm both results by the enclosed triangular area.

Question 7 - Solution

A triangular cycle exposes both sign and unit conversion

Diagram shown in the original SVG worksheet.

(a) Method and calculation

WAB=100(6−2)=400J,WBC=0.W_{AB}=100(6-2)=400\,\mathrm J,\quad W_{BC}=0.WCA=300+1002(2−6)=−800J.W_{CA}=\frac{300+100}{2}(2-6)=-800\,\mathrm J.Wcycle=−400J.W_{\mathrm{cycle}}=-400\,\mathrm J.

(b) Interpretation and further calculation

Atriangle=12(6−2)(300−100)=400kPaL=400J.A_{\mathrm{triangle}}=\frac12(6-2)(300-100)=400\,\mathrm{kPa\,L}=400\,\mathrm J.Wreversed=+400J.W_{\mathrm{reversed}}=+400\,\mathrm J.The stated counterclockwise traversal requires 400 J net work input; reversing the path changes the sign of every line integral.

Physical check

Pressure in kPa and volume in litres already produce joules, since 103×10−3=110^3\times10^{-3}=1.

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