Posted by Mary on Saturday, April 21, 2007 at 1:23am.
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 3.06 102 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?
Let L be the original length of the cylinder of air. (You don't need to know it; it will cancel out). For a monatomic gas, Cp = (5/2)R, where R is the molar gas constant and Cp is the specific heat at constant pressure. This is a constant pressure heat transfer process.
The temperature change is
delta T = Q/(n Cp), and n = PV/RT
n is the number of moles. This leads to
delta T = (2/5) T/(PAL)
For the movement delta X of the piston, use the energy equation
Work = P A *delta X = Q  delta U
where delta U = Cv * delta T
and Cv = (3/2)RT (for a monatomic gas).
I get
delta X = (2/5) Q/(PA)
In your case, delta Q is 2109 J
Check my logic

Physics HELP!!!!!!!!  asdf, Sunday, December 11, 2016 at 9:04pm
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