Physics

Suppose a monatomic ideal gas is contained within a vertical cylinder that is fitted with a movable piston. The piston is frictionless and has a negligible mass. The area of the piston is 3.06 10-2 m2, and the pressure outside the cylinder is 1.02 105 Pa. Heat (2109 J) is removed from the gas. Through what distance does the piston drop?

Answers

The distance the piston drops can be calculated using the Ideal Gas Law. The Ideal Gas Law states that the pressure times the volume of a gas is equal to the number of moles times the temperature times the gas constant. In this case, the pressure times the volume of the gas is equal to the heat added or removed. Since the volume is the area of the piston multiplied by the distance the piston drops, the equation becomes: P1V1 = P2V2 1.02 x 105 x 3.06 x 10-2 x h = 2109 h = 2109 / (1.02 x 105 x 3.06 x 10-2) h = 0.691 m Therefore, the piston drops 0.691 m when 2109 J of heat is removed from the gas.

Answered by danielanderson

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