Ideal Gases — Question 5

Section 19-2 · Physics pages 3060–3061

Question 5 A spring-loaded piston requires force balance and the equation of state Treat the gas as ideal, pV = nRT, and use R = 8.314 J/(mol K). A vertical frictionless piston of area .0100 m2 holds .100 mol gas. At displacement x = 0, volume is .00200 m3 and the total baseline resisting pressure is 100 kPa. A spring adds force kx, with k = 1000 N/m. At final temperature 300 K, the piston is in mechanical equilibrium. spring gas Schematic; labels, not drawing lengths, specify the data. (a) Set up the equation for x and find the physically allowed nonnegative root. (b) Find final pressure and volume, and check both force balance and pV = nRT.
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Question 5 - Solution A spring-loaded piston requires force balance and the equation of state spring gas (a) Method and calculation p = 100000 + 1000x .0100 = 100000 + 100000x, V = .00200 + .0100x. (100000 + 100000x)(.00200 + .0100x) = 249.42. 1000x2 + 1200x −49.42 = 0 ⇒x ≈.0398594 m. (b) Interpretation and further calculation p ≈103985.94 Pa, V ≈.002398594 m3. pV = 249.42 J = nRT, (p −100000)A = kx. The other root gives a negative volume and must be discarded. Physical check Neither pressure nor volume is independently fixed: the mechanical constraint couples them.

Question 5

A spring-loaded piston requires force balance and the equation of state

Treat the gas as ideal, pV=nRTpV=nRT, and use R=8.314J/(molK)R=8.314\,\mathrm{J/(mol\,K)}. A vertical frictionless piston of area .0100 m2^2 holds .100 mol gas. At displacement x=0x=0, volume is .00200 m3^3 and the total baseline resisting pressure is 100 kPa. A spring adds force kxkx, with k=1000k=1000 N/m. At final temperature 300 K, the piston is in mechanical equilibrium.

Diagram shown in the original SVG worksheet.

  1. Set up the equation for xx and find the physically allowed nonnegative root.

  2. Find final pressure and volume, and check both force balance and pV=nRTpV=nRT.

Question 5 - Solution

A spring-loaded piston requires force balance and the equation of state

Diagram shown in the original SVG worksheet.

(a) Method and calculation

p=100000+1000x.0100=100000+100000x,V=.00200+.0100x.p=100000+\frac{1000x}{.0100}=100000+100000x,\quad V=.00200+.0100x.(100000+100000x)(.00200+.0100x)=249.42.(100000+100000x)(.00200+.0100x)=249.42.1000x2+1200x−49.42=0⇒x≈.0398594m.1000x^2+1200x-49.42=0\ \Rightarrow\ x\approx.0398594\,\mathrm m.

(b) Interpretation and further calculation

p≈103985.94Pa,V≈.002398594m3.p\approx103985.94\,\mathrm{Pa},\quad V\approx.002398594\,\mathrm{m^3}.pV=249.42J=nRT,(p−100000)A=kx.pV=249.42\,\mathrm J=nRT,\quad(p-100000)A=kx.The other root gives a negative volume and must be discarded.

Physical check

Neither pressure nor volume is independently fixed: the mechanical constraint couples them.