Elementary Principles 250

# Elementary Principles 250 - 7.11 1 m2 = 2.83 10 -3 m 2 10 4...

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7- 5 7.11 A == × 2 π 3 c m m cm m 22 2 2 bg 1 10 283 10 4 3 . (a) Downward force on piston: FP A m g d =+ = ×× + = atm piston+weight 5 2 2 2 2 1 atm 1.01325 10 N / m m atm 24.50 kg 9.81 m 1 N s k g m / s N 1 527 3 . Upward force on piston: FA P P ug × gas mN m 3 . di d i Equilibrium condition: FF P P ud =⇒ × = ⇒= × = × 2 83 10 527 186 10 186 10 3 00 55 .. . m N m P a 2 2 V nRT P 0 0 1 0677 ×⋅ = 1.40 g N mol N 303 K 1.01325 10 Pa 0.08206 L atm 28.02 g 1.86 10 Pa 1 atm mol K L 5 5 . (b) For any step, ΔΔ Δ Δ Δ Δ UE EQ W U Q W kp E E k p ++ = = = = 0 0 Step 1: QU W ≈⇒ = 0 Δ Step 2: Δ UQW =− As the gas temperature changes, the pressure remains constant, so that Vn R TP g = must vary. This implies that the piston moves, so that W is not zero. Overall: TT U Q W initial final = = Δ In step 1, the gas expands ⇒>⇒ < WU T Δ decreases (c) Downward force F d × += 100 101325 10 4 50 9 81 1 331 53 . d i b g N (units as in Part (a)) Final gas pressure P F A f × 331 10 116 10 3 5 N 2.83 m Nm 2 2 Since
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