A Closer Look at Heat and Work — Question 6

Section 18-8 · Physics pages 2962–2963

Question 6 A spring makes piston pressure grow linearly with volume A frictionless piston with area .0200 m2 begins against baseline pressure p0 = 100 kPa. A spring is initially uncompressed and has stiffness k = 2000 N/m. Quasistatic outward motion compresses the spring by x. Let final x = .300 m. k gas Schematic; labels, not drawing lengths, specify the data. (a) Derive p(V ) and calculate total gas work. (b) Find the portion stored in the spring and the final pressure. Verify the work using average pressure.
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Question 6 - Solution A spring makes piston pressure grow linearly with volume k gas (a) Method and calculation V −Vi = Ax, p = p0 + kx A = p0 + k A2 (V −Vi). W = ∫.300 0 (p0A + kx)dx = p0A(.300) + 1 2k(.300)2 = 690 J. (b) Interpretation and further calculation Espring = 1 2kx2 = 90.0 J, pf = 100000 + 2000(.300)/.0200 = 130000 Pa. W = pi + pf 2 ∆V = 115000(.00600) = 690 J. The pressure-average shortcut is exact because this pressure law is linear in volume. Physical check The spring contribution is one-half the final spring force times displacement, not the full final force times displacement.

Question 6

A spring makes piston pressure grow linearly with volume

A frictionless piston with area .0200m2.0200\,\mathrm{m^2} begins against baseline pressure p0=100kPap_0=100\,\mathrm{kPa}. A spring is initially uncompressed and has stiffness k=2000N/mk=2000\,\mathrm{N/m}. Quasistatic outward motion compresses the spring by xx. Let final x=.300mx=.300\,\mathrm m.

Diagram shown in the original SVG worksheet.

  1. Derive p(V)p(V) and calculate total gas work.

  2. Find the portion stored in the spring and the final pressure. Verify the work using average pressure.

Question 6 - Solution

A spring makes piston pressure grow linearly with volume

Diagram shown in the original SVG worksheet.

(a) Method and calculation

V−Vi=Ax,p=p0+kxA=p0+kA2(V−Vi).V-V_i=Ax,\quad p=p_0+\frac{kx}{A}=p_0+\frac{k}{A^2}(V-V_i).W=∫0.300(p0A+kx)dx=p0A(.300)+12k(.300)2=690J.W=\int_0^{.300}(p_0A+kx)dx=p_0A(.300)+\frac12k(.300)^2=690\,\mathrm J.

(b) Interpretation and further calculation

Espring=12kx2=90.0J,pf=100000+2000(.300)/.0200=130000Pa.E_{\mathrm{spring}}=\tfrac12kx^2=90.0\,\mathrm J,\quad p_f=100000+2000(.300)/.0200=130000\,\mathrm{Pa}.W=pi+pf2ΔV=115000(.00600)=690J.W=\frac{p_i+p_f}{2}\Delta V=115000(.00600)=690\,\mathrm J.The pressure-average shortcut is exact because this pressure law is linear in volume.

Physical check

The spring contribution is one-half the final spring force times displacement, not the full final force times displacement.