 #1
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 3
Homework Statement
Homework Equations
F= ma
P= F/A
The Attempt at a Solution
a). Assuming the piston is in equilibrium, I'm applying newton's second law
F=ma=0
Equals zero because it is not moving
Note:
P=F/A which I rearrange to solve for the force for F= PA. This represents the pressure within (pushing upwards).
F=mg which represents the weight of the piston
P=F/A where F=P(atm)A for the outside pressure at room temperature
A=crosssection area of the piston
Now adding up all of the forces
+PAmgP(atm)A=0
Now solving for pressure within
P= (mg+P(atm)A)/A = mg/A +P(atm)
Where A is the area of the cylinder so,
/P= (mg)/(pi r^2)
b.) (del)U= (3/2)nR (del)T
There is no change in temperate so,
internal energy =0
therefore
(del)U=0=Q+W
c.)
Isothermal process, the gas expands and pressure falls while maintaining a constant temperature
final temperature =initial temperature
final internal energy 0 because there was no change in temperature
change in entropy
(del)S= Q/T
or
(del)S =Q=W
where W= P(del)V
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