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Suppose a monatomic ideal gas is contained within a
verticalcylinder that is fitted with a movable piston. The piston
isfrictionless and has a negligible mass. The area of thepiston is
3.14 X 10^ -2 m^2, and the pressure outside the cylinderis 1.01 X
10^5 Pa. Heat (2093 J) is removed from thegas. Through what
distance does the piston drop?

Respuesta :

matthewadeyinka10

Answer:

From the conservation of energy law at steady state conditions given,

Heat removed from the gas Q, = Work done by the piston, W = PdV = PAdL

where

P = Pressure difference =  1.01 X 10^5 Pa

A=Piston area =3.14 X 10^ -2 m^2

dL = distance moved (dropped) by the piston ?

Q=2093J

2093 = 101000 x 0.0314 x dL

dL = 0.659====0.66m

Therefore, the distance dropped by the piston = 0.66m

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