The definition of entropy contains the term $Q_\text{rev}$ which means the heat supplied or taken out reversibly. I thought yes it can be after all only the initial & final states are important as entropy is a state function irrespective of the process heat is transferred. However I was baffled when I first read Clausius' theorem where it is written that $dS \geq \dfrac{Q}{T}$. If $Q$ is transferred irreversibly, then $dS$ is greater than $\dfrac{Q_\text{irrev}}{T}$; if the heat transfer is reversible, then only $dS$ equals $\dfrac{Q_\text{rev}}{T}$. So, does that mean entropy depends on the process heat energy is transferred?? Then, how can it be a state function? Where am I mistaking? Please explain.
[Physics] Why must heat supplied in the definition of entropy be reversible? Can’t it be irreversible after all it is a state function
entropythermodynamics
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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 same. However, in order to calculate the change in entropy $S_{2}-S_{1}$, one has to connect a reversible path between the two states because $\displaystyle S_{2}-S_{1}=\int_{1}^{2}\frac{dQ_{rev}}{T}$, where $dQ_{rev}$ is an infinitesimal amount of heat transferred to the system in a reversible manner at the system temperature T. NOTE: $\displaystyle\int_{1}^{2}\frac{dQ_{irrev}}{T}$ is not the correct formula for calculating the change in entropy. It is always $\displaystyle\int_{1}^{2}\frac{dQ_{rev}}{T}$ irrespective of whether the path is reversible or irreversible.
However, my issue is, how do we know it is a state function? We were only able to say it is a state function when we went through a derivation using the Carnot cycle (a reversible process). It doesn't seem justified to say that is fact holds for any arbitrary process.
Entropy is usually thought of mathematically as a function of thermodynamic state variables.
"Entropy is state function" means that for given thermodynamic system (say gas in a cylinder with movable piston), all it suffices to determine value of entropy are values of all its state variables (say, internal energy and volume of the gas).
But this is indeed not clear from the original definition of entropy: entropy of state $X$ is often written or implied to be
$$ S(\mathbf X) = \int_{\mathbf X_0}^{\mathbf X} \frac{dQ}{T} $$ where $dQ$ is infinitesimal element of heat absorbed by the system and the integral is calculated along any continuous path connecting $X_0$ with $X$.
And you are right this does not make it clear why the expression does not depend on the kind of process that takes the state from $\mathbf X_0$ to $\mathbf X$.
The traditional way to show that the path does not matter and so the expression is a function of final $X$ only is based on the theoretical result that is true for the Carnot cycle with ideal gas. The result is that in such a cycle, sum of reduced heats is zero:
$$ \frac{\Delta Q_1}{T_1} + \frac{\Delta Q_2}{T_2} = 0. $$
This is then used to prove that for any reversible process that ends up at the same state it began, the integral along the closed path
$$ \oint \frac{dQ}{T} = 0. $$
This holds irrespective of the details of the path such as its shape or length. The usual way to do this proof is to replace the actual closed path (that is otherwise of arbitrary shape) by an alternating sequence of very tiny isotherms and adiabats that go very near the actual path. Those isotherms and adiabats can be thought of as part of fictive Carnot cycles and it can be shown that sum of reduced heats of those Carnot cycles approximates the above integral. But the sum of the reduced heats is 0 for those Carnot cycles. The smaller the Carnot cycles, the better the approximation to the original path, in the limit of zero size the isotherms and adiabats on the outer edge become the same as the original path.
After that is done, a mathematical theorem from multivariable calculus is used: if the loop integral is zero for any loop, the expression
$$ \int_{X_0}^{X} \frac{dQ}{T} $$
is a function of the points $X_0,X$ only, the actual path along which the integral is evaluated does not matter. One point ($X_0$) is usually set to be fixed and then the function is function of the final point ($X$) only.
Best Answer
Suppose you start with a system in some state $P_1, V_1, T_1$ and you add some quantity of heat $\Delta Q$ to it so the system changes to a different state $P_2, V_2, T_2$. The final state will depend on how you added the heat $\Delta Q$. Adding the heat $\Delta Q$ in a reversible process will result in different values for $P_2, V_2, T_2$ compared with adding the same amount of heat $\Delta Q$ in an irreversible process.
Entropy is indeed a state function, so if you know $P_2, V_2, T_2$ you can calculate the entropy change. Since reversible and irreversible processes will result in different values for $P_2, V_2, T_2$ they will also result in different values for the entropy change.