# Thermodynamics- Piston problem, pressure, internal energy

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]

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Chestermiller
Mentor
I think they expect you to assume that the pressure outside the cylinder is zero. Under these circumstances, your answer to part (a) would reduce to
$$p_0=\frac{Mg}{A}$$
In part (b) they are asking for the absolute value of the initial internal energy which, according to the standard equation for a monoatomic gas would be:
$$U_0=\frac{3}{2}RT_0$$ So, you were kind of right about this.

Regarding part (c): The weight of the piston is suddenly reduced from Mg to Mg/2. Do you think that the expansion that occurs as a result of this is spontaneous or reversible? Your determination that the final temperature and the internal energy do not change is correct. Your equation for determining the work is OK if P=Mg/(2A). But there is another way of determining the work that the gas does on the piston. From the ideal gas law, In terms of $V_0$, what is the final volume of the gas? What is the change in volume? What is the displacement of the piston? What is the change in potential energy of the piston if its weight is Mg/2? (This is also equal to the work)

In terms of $p_0$, $V_0$, and $T_0$, what is the number of moles of gas in the cylinder?

Your equation for the determination of the entropy change is incorrect. This equation only applies to a reversible path between the initial and final states of the system. But, the path for this process is not reversible. Can you think of a reversible path between the initial and final states that can be used to determine the entropy change?

Or

Del(S)=Rln(V(final)/V(initial))
This would work

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Chestermiller
Mentor
View attachment 216874

Or

Del(S)=Rln(V(final)/V(initial))
This would work
Excellent!!!! Don't forget to multiply by the number of moles.

Excellent!!!! Don't forget to multiply by the number of moles.

Del(S)=nRln(V(final)/V(initial))

Where
n =moles
V = volume
R= Gas constant

Chestermiller
Mentor
Del(S)=nRln(V(final)/V(initial))

Where
n =moles
V = volume
R= Gas constant
$$n=\frac{p_0V_0}{RT_0}$$

$$\frac{V_f}{V_i}=2$$