A cylinder with a frictionless piston of diameter 10cm is filled with 0.15 m3 of air at 550K.ideal gas question help?
Heat is slowly added until its final temperature raises to 770K.Assuming that g = 9.81 m/s2, the frictionless piston carries a load of 79kg and the atmospheric pressure is 1.0132 x 105 Pa, find the work done (in kJ) by the air?
Topic:ideal gas please help! :(
sorry i meant 1.0132 x 10^5 pa
- NCSLv 72 months agoFavorite Answer
pV = nRT
so when p, n and R constant, V/T is constant.
so 0.15m³ / 550K = V / 770K → new V = 0.21 m³
pressure due to piston
p₀ = mg / A = 4mg / πd²
p₀ = 4 * 79kg * 9.81m/s² / π(0.10m)² = 9.8674e4 Pa
The pressure is constant, so the work done by the gas is
(1.0132e5 + 9.8674e4)Pa * (0.21 - 0.15)m³ = 1.20e4 J
or 12.0 kJ.
The work done by the air is the negative of this -- -12.0 J.
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- billrussell42Lv 72 months ago
1.0132 x 105 Pa ?? what does this mean ?