- #1

Ashley1nOnly

- 132

- 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=cross-section area of the piston

Now adding up all of the forces

+PA-mg-P(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[/B]

#### Attachments

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