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 don't see anything wrong.
Your calculations seem correct, and you have correctly applied the formulas to find the distance the piston drops. The equation PV = W is used to relate the pressure, volume, and work done on the gas. In this case, you are given the pressure, and the work done on the gas is equal to the heat removed, which is given as 2109 J.
To find the volume, you divide the work by the pressure: V = W / P = 2109 J / (1.02 x 10^5 Pa) = 0.020676471 m^3.
Then, you use the formula V = area x distance to find the distance the piston drops. Rearranging the formula, distance = V / area = 0.020676471 m^3 / (3.06 x 10^-2 m^2) = 0.6757 m.
Your calculations and reasoning are correct, so it is possible that there was a mistake or typo in the system when it indicated that your answer was wrong.