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October 30, 2014

October 30, 2014

Posted by **Mary** on Thursday, April 19, 2007 at 9:06am.

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?

For Further Reading

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

For Further Reading

Physics, still don't get it! - bobpursley, Wednesday, April 18, 2007 at 10:38pm

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

Physics, still don't get it! - Mary, Wednesday, April 18, 2007 at 11:10pm

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

Physics, still don't get it! - bobpursley, Wednesday, April 18, 2007 at 11:16pm

I dont see anything wrong.

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