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Published byRoman Wadlow Modified over 3 years ago

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**W = PV Figuring out work: s The piston moves out a distance s: derive**

W = work done by gas P = pressure of gas V = change in volume of gas s

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**W = PV Figuring out work: s W = work done by gas P = pressure of gas**

V = change in volume of gas Example: What work done by an isobaric compression at 500 Pa from .85 m3 to .52 m3? s

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V P 500 Pa .5 m3 P-V diagrams:

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**How much work done by process AB?**

V P 500 Pa .5 m3 How much work done by process AB? A B 150 J

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Hugo First has a gas in a cylinder that expands from 200. liters to 500. liters at a pressure of 1200 Pa. What work did the gas do? (1000 liters = 1 m3) W = PV P = 1200 Pa, V = = .300 W = 360 J 360 J

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**Mr. Fyde compresses a cylinder from. 0350 m3 to**

Mr. Fyde compresses a cylinder from m3 to m3, and does 875 J of work. What was the average pressure? W = PV W = -875, V = = m3 P = Pa 62.5 kPa

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**How much work done by process EF?**

V P 500 Pa .5 m3 How much work done by process EF? E F W = PV, p= 200 Pa, V = = .6 m3 W = 120 J (work done by the gas) 120 J

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**How much work done by process BA?**

V P 500 Pa .5 m3 How much work done by process BA? B A W = PV, P = 300 Pa, V = = -.3 m3 W = -90 J (work done on the gas) -90. J

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**How much work done by process CD?**

V P 500 Pa .5 m3 D C W = PV, Pavg = 350 Pa, V = = .6 m3 W = 210 J (work done by the gas) 210 J

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**How much work done by process CD?**

V P 500 Pa .5 m3 How much work done by process CD? D C W = PV, p = 350 Pa, V =0 m3 W = 0 J (work done by the gas) waddayathink?

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B C V P 500 Pa .5 m3 Example: Total work done by process ABCDA? (2) D A 150 J

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Heat engines Turn heat to work

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Energy Flow c All heat engines waste a bit of heat

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**How much net work done by this cycle?**

V P 500 Pa .5 m3 How much net work done by this cycle? W = Area = LxW = (.3 m3)(100 Pa) = +30 J (CW) +30 J

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**How much net work done by this cycle?**

V P 500 Pa .5 m3 W = Area = LxW = (.3 m3)(400 Pa) = -120 J (ACW) -120 J

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**How much net work done by THIS cycle?**

V P 500 Pa .5 m3 W = Area = 1/2bh = 1/2 (.3 m3)(300 Pa) = +45 J (CW) +45 J

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**How much net work done by this cycle?**

V P 500 Pa .5 m3 W = Area = Lxh = (.6 m3)(300 Pa) = -180 J (ACW) -180 J

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Page 1 – Ideal Gas Law TOC PV = nRT P = pressure in Pa (Absolute, not gauge) V = volume in m 3 (Demo) n = moles of gas molecules R = 8.31 JK -1 (for.

Page 1 – Ideal Gas Law TOC PV = nRT P = pressure in Pa (Absolute, not gauge) V = volume in m 3 (Demo) n = moles of gas molecules R = 8.31 JK -1 (for.

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