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Is entropy a state function11/15/2023 He gets right to the root cause/source of entropy. The second law of thermodynamics is best expressed in terms of a change in the thermodynamic variable known as entropy, which is represented by the symbol S.Entropy, like internal energy, is a state function. You should check out on YouTube, Professor Susskind's (of Stanford) lectures on Statistical Mechanics. In other words, if a sample is in the liquid phase at a particular tempertare, with entropy S(liquid-300K), it does not matter that it got to that state via melting or condensing first, and then heating or cooling. Heat is not a state function because it is only to transfer energy in or out of a system it depends on pathways. Enthalpy is the amount of heat released or absorbed at a constant pressure. The pathway or pathways that caused the particles to be in that particular state/condition is irrelevant and has no effect on their current condition. A state function is independent of pathways taken to get to a specific value, such as energy, temperature, enthalpy, and entropy. "W" is way greater for a gas sample of a substance compared to its liquid phase because there are many more ways (fewer restrictions on position and modes of motion) for the particles to assemble in the gas phase. The equation is S = k ln(W), where "W" is the number of different ways to arrange/have the particles in their current particular state, e.g. As with any other state function, the change in entropy is defined as the difference between the entropies of the final and initial states: S S f S i. The students are guided to use the idea that entropy is a state function to determine. This subtlety is discussed in a bit more detail in a footnote in Kerson Huang's book in page 16.Entropy is a state function because it is based purely on the probability (likelihood) of the particles of a system existing in their current "condition" (energy, arrangement, etc.). The initial and final states of the gas are identical for both processes. During isothermal and reversible expansion of the gas from volume v(1) to v(2), let there be absorption of heat q at temperature T. In effect 1/ T is an integrating factor which, when it multiplies the inexact differential Q, results in the exact differential Q / T dS. Consider a cylinder fitted with a frictionless and weightless piston, which contains a gas and is in contact with a large heat reservoir. Where $\delta Q_$ by a reversible process. This means that the net change in entropy during a complete cycle is zero, so that entropy is a function of state. We can take any path we want to calculate the entropy of a thermodynamic system. In fact, the Clausius statement of the 2nd law says this: Entropy, \(S\), is a state function, so it does not depend on the thermodynamic path. We now want to see what the machine-based statement of the second law enables us to deduce about the properties of S. S is, of course, the entropy function described in our entropy-based statement of the second law. We know that the entropy transferred to a system undergoing a reversible cycle is zero, and if you have intuition behind this, then you have intuition behind why entropy is a state function. For an ideal gas traversing a Carnot cycle, we have shown that. Now, since $S=k_Bln\Omega$, entropy is also a state function.ĮDIT: I've given some intuition behind the statistical mechanical definition of entropy being a state function, but what about the thermodynamic definition? Therefore $\Omega=\Omega(N,P,T.)$ is a state function. However, if you do not make the substitution you do not get zero so the two. Intuitively, you can think of the number of accessible energy states by particles in a system being influenced by the state of the system (e.g. The proof requires a substitution of p T n R V because when it is then differentiated with respect to T it equates to zero and so does C V T when it is differentiated with respect to V which shows it is an exact differential. The number of available energy levels is determined by quantities like temperature, which is also influenced by pressure and other thermodynamic parameters. The number of microstates is determined by the number of particles (of course) and the number of energy levels available to these particles. So can we say in other words that: 'state variable' is something that we take as independent variable, while 'state function' is something that depends on previously selected 'state variables' where this dependence is given in the equation of state for the particular thermodynamic system. Entropy is proportional to the number of microstates available to a system, $S=k_Bln\Omega$.
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