Change in entropy in irreversible and reversible process

[Kenny Lee]

It is stated that the change in entropy of an isolated system during an irreversible process is > 0; while for a reversible process, it is = 0.
An isochoric process has a change in entropy given by : nCv(Pf/Pi). Does this mean that it is an irreversible process?
ARen't isochoric processes reversible? (I don't know if there are 'other' types of isochoric processes... I've only studied one, which is the one studied in high school).
Is it because the expression represents the change in entropy of a system and its surroundings; and hence, the initial statement still applies i.e. that the entropy of the isolated system is = 0 for a reversible process.
Or is it something else entirely?
I just need someone to clarify for me please. Won't take too long
Thanks.

 

[Chi Meson]

Isochoric means constant volume. Think of a container that is sealed shut and does not expand or contract. When you heat up (or cool down) the container, the process is isochoric. There is no work done in this process, so the energy that exchanges is entirely due to a spontaneous flow of heat. Whenever heat flows spontaneously, total entropy increases.

If you're talking a single isochoric "event," no it is not reversible. (If you put a cold can of helium in a hot basin of water, the heat that flows into the can will never flow sponateously back into the warmer water).

You can however include an isochoric event in a theoretical cycle of processes that involve a frictionless machine that moves it between hot and cold resevoirs such that total entropy does not change. This requires a "closed cycle" on a PV diagram and the assuption that no work is lost to friciton and no heat escapes during the transport of the gas between resevoirs. Not a realistic assumtion, but that's what "reversible" means.

 

[Kenny Lee]

How about an isothermal process? Taken from my textbook, an isothermal process is approximately reversible if you drop grains of sand, one by one, onto a piston above some volume of gas in a cylinder. The small change in pressure allows the system to find equilibrium quick enough.
There is still an entropy change in an isothermal process anyway, I mean according to the equation.
And what the heck, I just realized that my textbook's derivation for change in entropy (lemme try to write that down)
nCv ln (Tf/Ti) + nR ln (Vf/Vi)
is given for a "quasi-static, REVERSIBLE process"... and in the next section of the textbook, it says "in a reversible process, the total entropy of the system remains constant".
Help...

 

[Chi Meson]

In the famous "quasi-static" isothermal process, total entropy of the system (the gas) is unchanged when the system returns to the original state. In a single isothermal event (one compression or one expansion) heat must flow since work is done and internal energy remains constant. If heat flows, then entropy changes ($$\Delta S = Q/T$$).
The "grains of sand" isothermal process is reversible, meaning you can add one grain at a time, the gas will compress, work is done one the gas,heat will flow out, entropy of system and universe will both change; then you can reverse the process: remove one grain at a time, work is done by the gas, heat flows into the gas, entropy of system and universe return to its previous state.

So one "event" is not in it self a reversible process, but rather some processes can be reversed.

Again, not in the real world: the grains of sand would have to be stored at the height at which they are put on/taken off, otherwise outside work would be necessary; furthermore, perfect frictionless machines would be required to move the grains from their storage potisions to the piston. Nothing is truly reversible: another restatement of the 2nd Law.

 

[Kenny Lee]

Oh so entropy changes for a reversible process, and only if the process is reversed, then the entropy returns back to its initial value, and the total entropy is 'constant'. That makes sense yea?
Thanks 4 help btw

 

[GCT]

It is stated that the change in entropy of an isolated system during an irreversible process is > 0; while for a reversible process, it is = 0.
An isochoric process has a change in entropy given by : nCv(Pf/Pi). Does this mean that it is an irreversible process?
ARen't isochoric processes reversible? (I don't know if there are 'other' types of isochoric processes... I've only studied one, which is the one studied in high school).
Is it because the expression represents the change in entropy of a system and its surroundings; and hence, the initial statement still applies i.e. that the entropy of the isolated system is = 0 for a reversible process.
Or is it something else entirely?
I just need someone to clarify for me please. Won't take too long
Thanks.

where did you hear that the entropy change of the system for a reversible process is zero? The zero refers to the total entropy change, thus the entropy change of the system is the negative of the entropy change of the surroundings. The equation you mentioned for the isochoric process is actually derived from a reversible case, and it refers to the system.

 

[Kenny Lee]

Ah... okay. That makes sense.
It says so in my textbook that "an irreversible process of an isolated system is irreversible". I guess only if the system is isolated... so if we considered the universe as an 'isolated system', then the 'system's' entropy is constant if the process is reversible. Okay, I think I get it.

 

[GCT]

ahh, isolated...missed that, not absolutely sure about the difference at this point (currently taking p. chem).

 

[nema.prafull]

why entropy of a irreversible process is greater than zero,explain with example?

 

[nema.prafull]

what affect on entropy in case of reversible and irreversible process