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 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?

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
asd

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