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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