- #1
zenterix
- 671
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- Homework Statement
- There is a basic derivation about entropy that is shown below that I still have doubts about even though I am able to do various calculations involving entropy.
- Relevant Equations
- Consider an isolated system and an irreversible process A from state 1 to state 2.
Then
$$q_{irrev}=0\tag{1}$$
Take the system from state 2 back to state 1 using a reversible process B.
My first question is: why can the system not be isolated for this reversible process to be possible?
Assume we have a non-isolated system in process B.
Process A and process B together form an irreversible cycle.
$$\oint dS<0\tag{2}$$
$$\oint dS=\int_1^2 dS_{irrev}+\int_2^1 dS_{rev}\tag{3}$$
$$=\int_2^1 dS_{rev}\tag{4}$$
$$=S_1-S_2<0\tag{5}$$
$$\implies S_2-S_1>0\tag{6}$$
From (3) to (4) we used the fact that ##dS_{irrev}=\frac{\delta q_{irrev}}{T}=0##.
(6) tells us that the entropy always increases for an irreversible process in an isolated system.
This is the basis for saying that entropy never decreases in the universe.
The universe is an isolated system so any irreversible process in it generates entropy increase.
Any reversible process in the universe is done such that
$$\delta q_{sys,rev}=-\delta q_{surr,rev}$$
$$d S_{sys,rev}=\frac{\delta q_{sys,rev}}{T}$$
$$d S_{surr,rev}=\frac{\delta q_{surr,rev}}{T}=-\frac{\delta q_{sys,rev}}{T}$$
$$\implies dS_{univ}=dS_{sys,rev}+dS_{surr,rev}=0$$
Thus,
$$dS_{univ}\geq 0$$
$$q_{irrev}=0\tag{1}$$
Take the system from state 2 back to state 1 using a reversible process B.
My first question is: why can the system not be isolated for this reversible process to be possible?
Assume we have a non-isolated system in process B.
Process A and process B together form an irreversible cycle.
$$\oint dS<0\tag{2}$$
$$\oint dS=\int_1^2 dS_{irrev}+\int_2^1 dS_{rev}\tag{3}$$
$$=\int_2^1 dS_{rev}\tag{4}$$
$$=S_1-S_2<0\tag{5}$$
$$\implies S_2-S_1>0\tag{6}$$
From (3) to (4) we used the fact that ##dS_{irrev}=\frac{\delta q_{irrev}}{T}=0##.
(6) tells us that the entropy always increases for an irreversible process in an isolated system.
This is the basis for saying that entropy never decreases in the universe.
The universe is an isolated system so any irreversible process in it generates entropy increase.
Any reversible process in the universe is done such that
$$\delta q_{sys,rev}=-\delta q_{surr,rev}$$
$$d S_{sys,rev}=\frac{\delta q_{sys,rev}}{T}$$
$$d S_{surr,rev}=\frac{\delta q_{surr,rev}}{T}=-\frac{\delta q_{sys,rev}}{T}$$
$$\implies dS_{univ}=dS_{sys,rev}+dS_{surr,rev}=0$$
Thus,
$$dS_{univ}\geq 0$$