Thermodynamics: Reversible Compression of Solid Volume V

AI Thread Summary
In thermodynamics, a block of volume V is reversibly compressed isothermally from pressure P1 to P2 at temperature T. The internal energy of a solid can change during this process, even though it is not an ideal gas. The concept of reversibility implies that the entropy of the process does not increase, indicating that the process can be reversed without any net change in the system and surroundings. This characteristic allows for the deduction of specific thermodynamic properties related to efficiency and energy transfer. Understanding these principles is crucial for analyzing thermodynamic systems effectively.
mooneyes
Messages
10
Reaction score
0
Hi, this concerns thermodymanics.

A block of volume V is reversibly compressed from pressure P1 to pressure P2 isothermally at temperature T.

It goes on to ask about the heat expelled, but that's not my question.

It is a solid, obviously not an ideal gas, so I'm sure the internal energy can change, yes?
Also, what information can we deduce from the fact that it's a reversible process, if any?

Thanks.
 
Physics news on Phys.org
When a process is said to be 'reversible' (thermodynamically), then you're saying something very specific about the entropy of the process.
 
Ah ha! The entropy doesn't increase, I see. Thanks!
 
Thread 'Gauss' law seems to imply instantaneous electric field'
Imagine a charged sphere at the origin connected through an open switch to a vertical grounded wire. We wish to find an expression for the horizontal component of the electric field at a distance ##\mathbf{r}## from the sphere as it discharges. By using the Lorenz gauge condition: $$\nabla \cdot \mathbf{A} + \frac{1}{c^2}\frac{\partial \phi}{\partial t}=0\tag{1}$$ we find the following retarded solutions to the Maxwell equations If we assume that...
Maxwell’s equations imply the following wave equation for the electric field $$\nabla^2\mathbf{E}-\frac{1}{c^2}\frac{\partial^2\mathbf{E}}{\partial t^2} = \frac{1}{\varepsilon_0}\nabla\rho+\mu_0\frac{\partial\mathbf J}{\partial t}.\tag{1}$$ I wonder if eqn.##(1)## can be split into the following transverse part $$\nabla^2\mathbf{E}_T-\frac{1}{c^2}\frac{\partial^2\mathbf{E}_T}{\partial t^2} = \mu_0\frac{\partial\mathbf{J}_T}{\partial t}\tag{2}$$ and longitudinal part...
Back
Top