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12

First, you have system with some energy, named $U$ by physicists. You think you have all the information you need to characterize the system but then some guy comes near and says: "Whoa, that's bad, the volume of your system can change." You say: "No problem, we just add here $pV$. Our new energy is $H=U+pV$." "But hey," they say, "your temperature can ...


9

Short answer: Gibbs free energy $G = U + PV - TS$ combines internal energy $U$, pressure $P$, volume $V$, temperature $T$, and entropy $S$ into a single quantity that measures spontaneity. With that, I mean that processes that lower the Gibbs free energy of your system will spontaneously occur, and equilibrium is reached when the Gibbs free energy reaches ...


5

Sanaris's answer is a great, succinct list of what each term in the free energy expression stands for: I'm going to concentrate on the $T\,S$ term (which you likely find the most mysterious) and hopefully give a little more physical intuition. Let's also think of a chemical or other reaction, so that we can concretely talk about a system changing and thus ...


2

In the simplest sense of it, the Free Energy is the heat of the system minus the compulsory heat loss due to entropy. So, in short, it is the amount of "energy" left over in the system, after we consider losses due to entropy. So basically some amount of heat is wasted, and the remaining amount is useful. And this remaining amount is the Gibbs Free Energy. ...


2

There are two forces inluencing the spontaneity of a reaction: (1)The tendency of a system to attain a state of minimum energy and maximum orderedness, or stability. (2)The tendency of a system to attain a state of maximum energy and minimum orderedness , or entropy. If a system attains maximum stability, it attains mininum entropy; and if it attains ...


2

Consider the following proof by contradiction: Suppose you find a state with $E < -V_0$, then its kinetic energy becomes negative at every point $x$ (classically the velocity is imaginary at every point), which means the whole wavefunction (at all $x$) is an evanescent (exponentially decaying) wave, but such a solution is not a physically stable solution ...


1

Your boundaries are at $r=a$ and $r=b$. Notice that the potentials at these two surfaces are independent of $\theta$ (they are spherically symmetric). Look at a list of the first few Legendre Polynomials $P_{l}(\cos{\theta})$. For what value of $l$ does $P_{l}(\cos{\theta})$ not depend on $\theta$? Further, notice that $V(a) = V(r=a,\theta) = V_{0}$, and ...


1

is a thermodynamic potential that measures the "usefulness" or process-initiating work obtainable from a thermodynamic system at a constant temperature and pressure (isothermal, isobaric). Just as in mechanics, where potential energy is defined as capacity to do work, similarly different potentials have different meanings. The Gibbs free energy is the ...


1

This seems like a reasonable homework-like question, so I'll provide a hint. Realize that, for $x>0$, both the simple and semi oscillators have the same potential $V(x)$. For the simple oscillator, draw out the first few of the energy eigenfunctions $\psi_n(x)$. Now for the semi-harmonic oscillator, think about what the boundary condition on any ...


1

Thanks laying out your work so neatly in the question. I think the solution is the following $$\Delta KE= \int_{r_a}^{r_b}{ KQq \over r^2} dr$$ where $r_a$ is the initial position and $r_b$ is the final position (and I have added $q$ as the charge of the point charge). so, for example, if the point charge goes from $r$ to $2r$ we have two positive ...



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