In the textbook of thermodynamics by Zemansky, I came to free expansion. If $dQ=0$ (because there is no heat exchange between system and surroundings), entropy should be $0$ as $dS=dQ/T.$ Now, the book used the first law of thermodynamics, where it writes an expression for $dQ(=PdV),$ and then shows that $\Delta S= nR\ln(V_f/V_i).$ And obviously, this should be the case as the latter state is more chaotic. But, why does the formula $dS=dQ/T$ give $0$ entropy change? And, why do we write $dQ=PdV,$ which is true, but $dQ$ should be $0.$ I am not getting it.
$\begingroup$
$\endgroup$
2
-
$\begingroup$ physics.stackexchange.com/questions/527438/…. you can check out this stack question $\endgroup$– Naveen VCommented Feb 27, 2023 at 7:01
-
$\begingroup$ entropy for irreversible changes are >0 $\endgroup$– Naveen VCommented Feb 27, 2023 at 7:01
Add a comment
|
3 Answers
$\begingroup$
$\endgroup$
The formula dS = dQ/T is only valid if dQ is due to a reversible process. Since we are talking about a free expansion, it is certainly not reversible.
In general, you should start with the thermodynamic identity (dE = TdS - PdV) and solve for dS to see how it changes. In the case of a reversible process, dS = dQ/T can be used directly.
Look up the Clausius Inequality and reversible vs irreversible processes for more information.
$\begingroup$
$\endgroup$
Since entropy is a state function, meaning a change in entropy of a system is independent of the process between two equilibrium states, you can assume any convenient reversible path connecting two states, that satisfies the first law, and apply the entropy definition to calculate the entropy change. It doesn’t matter if the actual process did not involve heat. That’s what Zemansky did.
Hope this helps.
$\begingroup$
$\endgroup$
For a closed, rigid, insulated container, the heat exchanged between the container contents and its surroundings Q and the work done by the container contents on its surroundings W are both zero. So, from the 1st law of thermodynamics, the change in internal energy of the container contents (the ideal gas) is equal to zero. This means that its temperature change is also zero.
The equation dS=dQ/T applies only to a reversible process. So, to determine the entropy change of the gas for this irreversible free expansion, you must devise an alternate reversible path for the gas between the same initial and final thermodynamic equilibrium states and calculate the integral of dQ/T for that path. This will give the entropy change calculated in your book.
answered Feb 27, 2023 at 13:01