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Figuring out work: The piston moves out a distance s: derive W = P  V W = work done by gas P = pressure of gas  V = change in volume of gas s.

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Presentation on theme: "Figuring out work: The piston moves out a distance s: derive W = P  V W = work done by gas P = pressure of gas  V = change in volume of gas s."— Presentation transcript:

1 Figuring out work: The piston moves out a distance s: derive W = P  V W = work done by gas P = pressure of gas  V = change in volume of gas s

2 Figuring out work: W = P  V 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 m 3 to.52 m 3 ? s

3 V P 500 Pa.5 m 3 P-V diagrams:

4 V P 500 Pa.5 m 3 How much work done by process AB? 150 J A B

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6 W = P  V P = 1200 Pa,  V = =.300 W = 360 J 360 J 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 m 3 )

7 W = P  V W = -875,  V = = m 3 P = Pa 62.5 kPa Mr. Fyde compresses a cylinder from.0350 m 3 to.0210 m 3, and does 875 J of work. What was the average pressure?

8 W = P  V, p= 200 Pa,  V = =.6 m 3 W = 120 J (work done by the gas) V P 500 Pa.5 m 3 How much work done by process EF? 120 J E F

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

10 W = P  V, P avg = 350 Pa,  V = =.6 m 3 W = 210 J (work done by the gas) V P 500 Pa.5 m 3 How much work done by process CD? 210 J C D

11 W = P  V, p = 350 Pa,  V =0 m 3 W = 0 J (work done by the gas) V P 500 Pa.5 m 3 How much work done by process CD? waddayathink? C D

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13 V P 500 Pa.5 m 3 Example: Total work done by process ABCDA? (2) A B C D 150 J

14 Heat engines Turn heat to work

15 Energy Flow c c All heat engines waste a bit of heat

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

18 W = Area = LxW = (.3 m 3 )(400 Pa) = -120 J (ACW) V P 500 Pa.5 m 3 How much net work done by this cycle? -120 J

19 W = Area = 1 / 2 bh = 1 / 2 (.3 m 3 )(300 Pa) = +45 J (CW) V P 500 Pa.5 m 3 How much net work done by THIS cycle? +45 J

20 W = Area = Lxh = (.6 m 3 )(300 Pa) = -180 J (ACW) V P 500 Pa.5 m 3 How much net work done by this cycle? -180 J


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