A Closer Look at Heat and Work — Question 9

Section 18-8 · Physics pages 2968–2969

Question 9 A measured work value identifies an unknown pressure peak A quasistatic expansion starts at (1.00 L, 100 kPa), follows a straight line to (3.00 L, pm), then another straight line to (5.00 L, 100 kPa). The measured work done by the gas is 800 J. V p pm Schematic pressure–volume paths; work is the signed area under a process curve. (a) Infer pm using signed areas. (b) Find the work if the gas returns directly at 100 kPa and hence the net work for the closed loop. What would the same measured 800 J mean for a constant-pressure expansion over 4 L?
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Question 9 - Solution A measured work value identifies an unknown pressure peak V p pm (a) Method and calculation W = 100 + pm 2 (3 −1) + pm + 100 2 (5 −3). 800 = 200 + 2pm ⇒pm = 300 kPa. (b) Interpretation and further calculation Wreturn = 100(1 −5) = −400 J, Wloop = 400 J. pconstant = 800/4 = 200 kPa. Work constrains an area; without a specified path shape it cannot uniquely determine a peak pressure. Physical check The extra work above the return pressure equals the area of the triangular pressure excess.

Question 9

A measured work value identifies an unknown pressure peak

A quasistatic expansion starts at (1.00L,100kPa)(1.00\,\mathrm L,100\,\mathrm{kPa}), follows a straight line to (3.00L,pm)(3.00\,\mathrm L,p_m), then another straight line to (5.00L,100kPa)(5.00\,\mathrm L,100\,\mathrm{kPa}). The measured work done by the gas is 800 J.

Diagram shown in the original SVG worksheet.

  1. Infer pmp_m using signed areas.

  2. Find the work if the gas returns directly at 100 kPa and hence the net work for the closed loop. What would the same measured 800 J mean for a constant-pressure expansion over 4 L?

Question 9 - Solution

A measured work value identifies an unknown pressure peak

Diagram shown in the original SVG worksheet.

(a) Method and calculation

W=100+pm2(3−1)+pm+1002(5−3).W=\frac{100+p_m}{2}(3-1)+\frac{p_m+100}{2}(5-3).800=200+2pm⇒pm=300kPa.800=200+2p_m\ \Rightarrow\ p_m=300\,\mathrm{kPa}.

(b) Interpretation and further calculation

Wreturn=100(1−5)=−400J,Wloop=400J.W_{\mathrm{return}}=100(1-5)=-400\,\mathrm J,\quad W_{\mathrm{loop}}=400\,\mathrm J.pconstant=800/4=200kPa.p_{\mathrm{constant}}=800/4=200\,\mathrm{kPa}.Work constrains an area; without a specified path shape it cannot uniquely determine a peak pressure.

Physical check

The extra work above the return pressure equals the area of the triangular pressure excess.