New answers tagged entropy
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The formula is valid for units of energy
The multiplicity $Ω$ for q units of energy among N equally probable states is given by the expression
This is sometimes called the number of microstates for the system.
Organic life exists because it exchanges energy and diminishes entropy by using the environment it finds itself in. It is only in closed ...
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If we go with the definition of entropy as the number of microstates and a continuously expanding universe, the phase space is expanding leading to a growing number of microstates.
$S=k_bln(Ω)$
Energy densities do not come into the count. Take a totally expanded universe having become photons , even with very low energydensity , as the phase space expands ...
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There seem to be various misconceptions here:
Thermodynamic entropy is a microscopic quantity. Macroscopic complexity does not relate to entropy in any meaningful way. The difference in entropy between a human and a lump of bacteria with the same mass and temperature is negligible (and I have no idea who would “win”). In fact, if I had to maximise the ...
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The entropy is going to increase as time increases. The entropy was at zero or at least it is meaningful to put it at zero. As time increases the disorder increases(this happens initially due to the expansion of the universe) as this results in the dispersion of energy. Now the entropy is going to be high but it still not high enough. Thermal equilibrium is ...
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It's not really about complication, but about metabolism--entropy, after all, is just energy being converted to heat. A Greenland shark is a very complicated animal, but it has a slow metabolism because of the cold environment it inhabits. Conversely, E. coli is about as simple as an organism gets without being a virus, but it has a very fast metabolism--...
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Short answer:
The first curious thing is that the big bang is a very 'ordered', low entropy event. It only went downhill from there. Or rather up as entropy and chaos are concerned.
The second curios thing is that while overall chaos and entropy are constantly rising, it is still possible to create order in some places.
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The convention usually followed is to define $0 \ln 0 = 0$.
That is not so bad as it looks, you may calculate $\lim\limits_{x \rightarrow 0} x\ln x$.
Using L'Hopital's rules,
\begin{array}{lll}\lim\limits_{x \rightarrow 0} x \ln(x) & = & \lim\limits_{x\rightarrow 0} \frac{\ln(x)}{x^{-1}} \\ { } & = & \lim\limits_{x \rightarrow 0} \frac{x^{-...
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The usual explanation is that a physical system is expected to have a symmetric transition probability, i.e., "detailed balance", as noted by Yvan Velenik and jacob1729. Fundamentally this comes from PT invariance (I suggest this is the "subtle physical reason" you seek), which holds in most practical systems. If a system is microscopically irreversible as ...
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The basic desideratum when you want to determine a suitable probability measure to describe your isolated system at equilibrium is that it must describe a system at equilibrium!
Let's go back to the standard case of a classical Hamiltonian system. The reason the microcanonical measure is a plausible candidate to describe the equilibrium state of the system ...
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I think this is a good question and it gets at what statistical mechanics is actually doing when it assigns probabilities to states. In particular, it is not giving you the correct probability to be in any given state. This is even more clear in the fully deterministic case, where the pdf to be in state $x$ at time $t$ is given by $p(x,t) = \delta(x-r(t))$ ...
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I think that this is not a statistical system, in the sense that we study in statistical mechanics. Or, alternatively, this is not a system in equilibrium.
One of the basic ideas when we describe a system in statistical mechanics is that the system is macroscopic, with quantities that we can define as extensive and intensive, and these quantities are ...
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The amount of information carried by a system is given by entropy function $ H(X) $ which is defined as:$$ H(X) =-\sum p(x) \log_{2}(p(x)) $$ where $ p(x)$ is the probability of random variable $X$, where the sum is over support of $p(x) $.
For example: Now if you have biased coin with $p({\rm heads}) = 0.8$ and $p({\rm tails}) = 0.2$. You can calculate ...
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Quantum Gravity is the description of gravity according to the laws of quantum mechanics. It uses fundamental principles of quantum physics (like quantization for example) and applies it to gravity. Quantum gravity is needed in situations when you cannot just ignore quantum effects (even though you are talking about macroscopic objects). An example would be ...
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About your answer, have you found it to be true?
I would say that there is a problem with your reasoning because the hint is that you can increase the entropy.
My answer by following the same thought process as the book:
Since your system has minimum energy and the entropy is not a maximum then you can increase it's entropy by exchanging heat dQ from the ...
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Starting from the chemical potential denfinition:
\begin{equation}
\frac{\partial S(E,V,N)}{\partial N} = - \frac{\mu(E,V,N)}{T(E,V,N)},
\end{equation}
For two systems in contact $\mu_1 = \mu_2$ should be satisfied and for same temperatures of the system, the change in entropy until equilibrium will be (certainly positive, since 2nd therm. law):
\begin{...
2
It is true, for all processes whether reversible or not, that:
$$dU=TdS-PdV=dq+dw$$
as long as two of the four variables ($T, S, V, P$) can be defined for the system.
However, it is not true that $dq = T dS$ always; that equality only holds for reversible processes. Likewise, $dw = -PdV$ is only true for reversible processes. If the processes is ...
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No, dw = -pdV is ONLY for a reversible process and hence you can compare your two equations only if both are written for reversible processes for which TdS = dq. What you are doing wrong is comparing an equation which holds for all processes with an equation that is true only for reversible processes. That is why you are getting a contradiction
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The second law is actually an inequality,
$$ \mathrm{d}S \geq \frac{\delta Q}{T}.$$
The strict equality holds only for reversible processes.
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you have to prove e < 1 for all processes including irreversible processes to prove second law. your line of reasoning has nothing to say about irreversible changes. this is my first reaction. i too have been thinking about this for a long time. for another perspective, see my book 'The principles of thermodynamics' N.D. Hari Dass.
there is a flaw in my ...
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The value of $\Delta S_{\text{sys}}$
You are confusing the entropy change of the system with the total entropy change. Now since the process is cyclic, the total entropy change for the system will be zero ($\Delta S_{\text{sys}}=0$). It does not matter whether the process is reversible or irreversible because entropy is a state function and depends only on ...
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I would recommend Statistical Mechanics: Theory and Molecular Simulation from Tuckerman.
It is a good introductory book and it is easy to read. It complements the theory parts with chapters about the numerical algorithms that are used in statistical mechanics.
You will realize soon that only a handful of problems in statistical mechanics can be solved ...
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A good and readable book :
S. Lokanathan & R.S. Gambhir, Statistical and Thermal Physics an Introduction, Prentice-Hall India, 1991.
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There is a little known fact that is relevant here. The energies of microstates are only specified up to an arbitrary additive constant (as with all energies, which are always referenced to a zero of the energy). For example, it is common in statistical mechanics to reference all microstate energies relative to the energy of the lowest energy microstate. ...
1
This is a consequence of the Clausius theorem. The background to it is explained in this Wikipedia article.
We know that in a heat engine that goes through a cycle and returns to its original state, we have
$\oint \frac 1 T dQ \le 0$
with equality only if the cycle is reversible. This tells us that if we move from state $A$ to state $B$ in a reversible ...
1
Usually because that is what the second law of thermodynamics states. Entropy must always increase. Since things that create low entropy violate this law this means that it took a lot of energy to make these low entropy objects. Of course entropy can drop as long as the entropy of the universe keeps on increasing. Well I am not too sure but minimal entropy ...
1
Maybe the three forces(Electromagnetism, Strong and Weak nuclear forces) are fighting and helping entropy. For example the weak force works for entropy by bringing down object masses. For example the weak force causes Beta decay and this increases entropy
$$n \rightarrow p^+ + e^- + \bar \nu_e$$
However the weak force can fight entropy for example it can ...
1
This is a really good question, and, I might add, you are working with a really good book.
To understand what is happening here, you have to understand that the interface between the two reservoirs cannot be at both temperatures at the same time. There is only one temperature at the boundary of each reservoir, and that must be the temperature of the ...
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If the only thing that is happening is heat exchange (i.e., no work), then $dU = dq$ by the First Law. In this case, heat behaves like a state function (it is in fact equivalent to a state function). Which means that it does not matter whether the actual heat exchange is reversible or not -- we can proceed as if the process were reversible (because $dU = ...
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Those "related events" take place within the "machine" which causes the entropy decrease.
That "machine" is the human body.
The body is constantly consuming nutrition, fats, protein etc. - breaking down systematic low-entropy entities and released bound energy, causing an increase in entropy - in order to perform entropy-decreasing actions such as sorting ...
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ΔS=Sf-Si=mcln(Tf/Ti),where m: mass & c: specific heat of the material.Setting initial condition Ti=1 then ΔS=S=mclnT in general.Supposed a mass transfer Δm= mB-mA we count ΔS=(mBcB-mAcA)lnT for constant temperature and ΔS=(mBcB-mAcA)ln(T2/T1) for different temperatures.
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The system is an open system. You are inputting energy. It increases the overall entropy or "disorder" of the universe because now the energy that was used to shuffle the cards was converted into heat and light energy which is a disordered form of energy.
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Answers:
The relationship comes from the definition of a differential change in entropy
$$ds=\frac{\delta q_{rev}}{T}$$
Where $\delta q$ is a reversible transfer of heat at temperature $T$. So $q$ in your equation only applies to reversible heat transfer.
Entropy can either be expressed per unit mass, which is called its specific entropy $s$ or as total ...
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This question is extremely open ended, so I will just try to provide some intuition about a small part.
A central problem in your question seems to concern the idea of "information" and quantifying it in the first place. The question seems to be, if a wavefunction with some uncertainty (for example a particle in a big box) seems to have more information ...
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The formula given makes no assumptions about the sign of $\Delta H$. For a fixed amount of heat, the change in entropy is smaller at larger temperatures, be it an increase or a decrease. Comparing entropy to disorder is just an analogue and should not be taken too literally
1
It does not take the red curve, nor the blue. It does not take any curve
on that graph.
The P-V graph you show is a phase diagram. Each point represents a
equilibrium state of the system. In reversible expansion, the system is at each time at equilibrium. Then, you can draw its trajectory on the graph. However, in irreversible expansion the system goes out ...
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Here's my show at it. Vertical cylinder of gas with a piston of mass m siting on top.
Reversible process: Cylinder initially at temperature T1. Heat cylinder by putting it in contact with a continuous sequence of ideal reservoirs of gradually increasing temperatures, running from T1 to T2.
Irreversible process: Cylinder initially at temperature T1. ...
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In the derivation of equation (5.29) the steps go,
\begin{equation}
dS_{total}=dS+dS_{R}=dS+\frac{dU_{R}}{T}
\end{equation}
The next step replaces $dU_{R}$ by $-dU$. In other words, the derivation leading to equation (5.29) assumes $dU+dU_{R}=0$. However, suppose that a mechanical system does some non $P-V$ work $dW$ on the system plus reservoir,
\begin{...
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Suppose we measure the entropy change of a spontaneous process that takes place within an isolated system. We find that the entropy increases.
Spontaneous processes are regarded as irreversible in classical physics. This means that once a spontaneous process has come to its end, it never goes back to the initial state. If it were to go back then changes in ...
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