Need help understanding sceanario only, is it a closed reversible system?

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Homework Statement

@= alpha= thermal expansion coefficent= 5.1x10^-5
Kt= compression= 3.2x10^-5
B=beta= -(dE/dV)t = 0.5
Po=1 atm
To=293K
Vo=1L
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Solid held in a piston and cylinder at equilibrium is in a constant temperature bath at 293k. Two irreversible steps are done at constant temp: I) pressure is suddenly increased to 100 atm and this pressure is kept constant while solid is irreversibly compressed. II)The pressure is suddenly dropped to zero and the solid expands till V=Vo=1L at which point the pressure is quickly restored to 1 atm.

Homework Equations

V=Vo(1-Kt(P-Po)+@(T-To) solid equation of state

The Attempt at a Solution

I attempted to draw the graph came up w/ 5 irreversible steps. increasing pressure to 100 atm then going at constant pressure to a decreased volume then dropping the tempreture to zero then going to the initial volume then finally raising the pressure back to the initial pressure.

By doing all of this the system appears to be closed and reversible correct? So the delta S universe should equal zero but it isn't...I believe that is it reversible because of the fact that equilibrium is achieved by going up and down the isotherm but specifically in the question it has been done irreversibly. But according to the notes I've taken in class the delta S of the universe is not equal to zero and even though there is a tempreture bath, there is heat loss that escapes the bath...why is that so? The tempreture stays constant through out the whole process...nothing in this question makes sense T_T I'd really appreciate any kind of help or ideas. Thank you.

 

[nate808]

Thank you for your post. It seems like you have a good understanding of the problem and have attempted to solve it using the relevant equations. However, there are a few points that need to be clarified.

Firstly, in your attempt at a solution, you mention that the system appears to be closed and reversible. While it is true that the system is closed, it is not reversible. In fact, the problem explicitly states that the two steps are irreversible. This means that some energy is lost as heat during the process, which is why the entropy of the universe is not zero.

Secondly, you mention that the temperature stays constant throughout the process. This is not entirely accurate. While the temperature of the bath remains constant at 293K, the temperature of the system does change during the irreversible compression and expansion steps. This is because work is done on the system during compression, which increases its internal energy and therefore its temperature. Similarly, work is done by the system during expansion, which decreases its internal energy and temperature. This change in temperature also contributes to the increase in entropy of the universe.

Finally, it is important to note that the equation you have used, V=Vo(1-Kt(P-Po)+@(T-To), is only valid for small changes in volume and pressure. In this problem, the changes in pressure are quite large (from 1 atm to 100 atm) and the changes in volume are also significant. Therefore, the equation may not accurately describe the behavior of the system and a more complex equation of state may need to be used.

I hope this helps clarify some of the confusion and helps you solve the problem. If you have any further questions, please don't hesitate to ask. Good luck with your studies!

 

[Blueshift5]

Based on the given information, it appears that the system described is a closed, reversible system. This means that there is no exchange of matter or energy with the surroundings and that the system can return to its initial state after undergoing a reversible process.

In this scenario, the solid is held in a piston and cylinder at equilibrium and is in contact with a constant temperature bath. This means that the temperature of the system remains constant throughout the process. The two irreversible steps described involve changing the pressure and volume of the system, but the temperature remains constant.

Since the temperature is constant, the change in entropy of the system should be equal to zero. However, the change in entropy of the universe may not be equal to zero because there may be heat loss or gain from the surroundings during the irreversible processes. This can result in a non-zero change in entropy for the universe.

Overall, the system itself is closed and reversible, but the entire universe may not be in a state of thermodynamic equilibrium due to the irreversible processes that occur. It is important to consider the entire universe when analyzing thermodynamic systems and processes.