Posted by **Mary** on Wednesday, April 18, 2007 at 10:30pm.

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?

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Physics HELP!!!!!!!! - bobpursley, Tuesday, April 17, 2007 at 9:29am

PV=work

V= area *distance

You know P, area, and work. Solve for distance.

According to the formula provided

PV=work

P(area x distance)= work

solve for distance

distance = work - P/ area

this is where it get confusing for me

P= 1.02 x 10^5 Pa

A= 3.06 x 10^-2 m^2

W= ? I am unsure how to get he value of work

Work is heat, and heat is work. It was given.

Please check my working out. The system is saying it is wrong.

PV = W

PV = 2109J

V = 2109J/ 1.02 X 10^5Pa

V = 0.020676471m^3

Area x distance = 0.020676471m^3

distance = 0.020676471/ 3.06 x 10^-2m^2

distance = 0.6757 m

I dont see anything wrong.

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