New answers tagged

0

If you can define a perfectly closed system which eventually arrives back at its starting state, that system is reversible. A reversible system is not irreversible, and can never "become" irreversable without changing the system. However, in the real world, there are limits which come into play. The first is that the systems are never actually isolated ...


1

Every deterministic system with chaotic trajectories is losing information on account of infinitely small deviations leading to completely different results. In your case, if the ball eventually settles into a stable orbit after some time, no matter where it started, then computing backwards would amplify any uncertainty so that after a few cycles, you know ...


3

Let's take the 1D equivalent of your problem for simplicity: a particle bouncing back and forth along a segment, reversing its velocity every time it hits the boundaries of the segment. If we know perfectly the initial state of the particle, i.e. its position and velocity at time $0$, $(x(0),v(0))$, we will know exactly what the motion of the particle will ...


0

If you assume that the ball is in the state you describe, and will continue to run through the same path over and over again, then you assume, that your system is deterministic, that means, from one position and momentum of the ball, you will be able to calculate its path for all the times that will come after that. In your special case: If the time goes ...


1

You need to consider the surroundings as well. If you go from state A to state B via a reversible process, the change in system's entropy exactly cancels out the opposite change in entropy for the surroundings; so overall there is no change in entropy. On the other hand, if it were an irreversible process, entropy change of the system (though same as the ...


1

There is something they forgot to mention in your notes (either from ignorance, or out of omission). The temperature within the system is spatially non-uniform during an irreversible process. So what value of the temperature are you supposed to use in the integral of dq/T? The Clausius inequality calls for the use of the temperature at the boundary ...


0

I think you have a typo in your question. A reversible process will have a smaller entropy change than an irreversible process. Your interpretation that the equality refers to a reversible process, while the inequality refers to in irreversible process is correct. Looking at the specific equations in that notes document, the integral in 8.31 applies to an ...


0

This is Clausius inequality. The term ds>dQ/T. Holds an impossible reaction which does not obey Clausius statement in 2nd law of thermodynamics. The things in your class thought is about this ds<0. Where you can add c that is entropy generation term. Hope it will help you


1

I would like to add to what Procyon said in his(her) answer, which is right on target. The first equation in the OP should be an equality, not an inequality. For an irreversible process, the temperature of the system is typically non-uniform, and when we write $\int{\frac{dQ}{T}}$ for the Clausius inequality, what we really mean is ...


1

Entropy is a property of the system. The change in entropy of a system as it traverses from an initial state(1) to a final state(2) is independent of the path by which the system is taken from state 1 to state 2. The path can be a reversible one, or even irreversible, the change in entropy is always the same as long as the initial and final states are the ...



Top 50 recent answers are included