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How to correctly think about number of microstates of a system?

For a quantized system, i.e. one with discrete configurations, the quantity $W$ is simply a count of the number of microscopic states compatible with a particular macroscopic state. For a system with ...
ummg's user avatar
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2 votes

Why does the temperature of the working fluid have to be close to the temperature of the cold reservoir during a Carnot cycle?

The work $W_0$ produced in the Carnot cycle is the result of transporting a certain amount of entropy, say, $S_0$ from a higher temperature $T_h$ to a lower temperature $T_{\ell}$. Here $W_0=S_0(T_h-...
hyportnex's user avatar
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2 votes

Truesdell's formulation of second law for homogeneous systems in Rational Thermodynamics

The important idea here is not that there is some upper bound but that there is a constitutive upper bound. This idea is similar to Clausius's inequality but in a different guise. You surely learned ...
hyportnex's user avatar
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2 votes

'It is not possible to have a process in which the entropy of an isolated system is decreased'. can you please explain this statement?

Entropy of an isolated system can only increase (unlike pressure, temperature, volume, etc.) Thus, if we increase the volume of the gas, two things may happen: Reversible process - the entropy doesn'...
Roger V.'s user avatar
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2 votes
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Clausius inequality and negative value of entropy

I was wondering...if entropy is a "state function" then why does going through an irreversible cycle give a negative value for entropy (Clausius inequality), The Clausius inequality is not ...
Bob D's user avatar
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2 votes
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Find the ideal gas law from the internal energy

I am not totally sure if this is the most straightforward way, but you will receive the ideal gas law by varying $U$, that is $$\operatorname{d}U = \dfrac{\partial U}{\partial V} \operatorname{d}V + \...
Extraherby's user avatar
2 votes

Work obtained during isentropic expansion in the Carnot cycle versus the work done to gradually lower the outside pressure

What "forces" the gas to to some more work during the isentropic expansion? Nothing "forces" the gas to do work. During the isentropic expansion the external pressure is reduced &...
Bob D's user avatar
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1 vote

Second Law of Thermodynamics and heat flow

The cold water "gain" heat and the air in the dining room "loose" heat. But, why the temperature in the dining room remain the same? Should it be colder? It is cooler. But the ...
Bob D's user avatar
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1 vote

What's the significance of a quasi-static process?

Certainly, here is the text rewritten in Physics Stack Exchange markdown with math enclosed in $ $: The concept of a quasi-static process in thermodynamics is indeed central to understanding and ...
MrDBrane's user avatar
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1 vote

Does second law of thermodynamics imply the big bang?

Not really Although it's reasonable, there are many ways to avoid such a conclusion, e.g.: There was a major statistical fluctuation that caused a decrease in entropy in the past (unusually among ...
Allure's user avatar
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1 vote

Mixing entropy for ideal gases is decreasing

You are not comparing the same things. The entropy of the Sackur-Tetrode equation is: $$ S = \ln\frac{\Gamma}{N!} $$ while your subadditivity argument uses: $$ S = \ln\Gamma $$ Physically, your final ...
LPZ's user avatar
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1 vote
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'It is not possible to have a process in which the entropy of an isolated system is decreased'. can you please explain this statement?

Does that mean, if we increase the volume of a gas, entropy of gas increases, now when we decrease the volume of gas to its original volume, the entropy of gas comes back to its original state, but ...
march's user avatar
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1 vote

Does Einstein's Relativity contradict the Arrow of Time?

The passing from non-relativistic physics to Special Relativity does not change any of this. It's not relative in Relativity, there's an absolute relation there, too. It's just different. Rather: it ...
NinjaDarth's user avatar
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1 vote

Does Einstein's Relativity contradict the Arrow of Time?

Causality is conserved in special relativity. If $|s_{12}|^2= c^2 ((t_1-t_2)^2-(\tilde{r_1}-\tilde{r_2})^2 )$, then if $ |s_{12}|^2>0 $ the events are causally connected as $v<c$. Through a ...
Questioningmind's user avatar
1 vote
Accepted

Does Einstein's Relativity contradict the Arrow of Time?

No, Einstein's Theory of Relativity does not contradict the notion of "Arrow of Time". The concept of "Arrow of Time" means that time, contrary to the other physical dimensions, is ...
Mauro Giliberti's user avatar
1 vote

What is the number of quantum states compatible with isolated ideal gas macrostate $N,V,U$ and molecular mass $m$?

Preliminaries: Your question touches the connection of statistical mechanics with combinatorics and number theory, see e.g. Ref. 1. I think it is not possible to give a definite answer to your ...
Tobias Fünke's user avatar
1 vote

What is the number of quantum states compatible with isolated ideal gas macrostate $N,V,U$ and molecular mass $m$?

As a prolific answerer of your standing, the stuff I am about to suggest, is likely to already be things that you have already seen before. However, it is still worth pointing out explicitly. I do not ...
naturallyInconsistent's user avatar
1 vote

PROOF of Only Hot to Cold Heat Transfer

The laws of thermodynamics possess a probabilistic nature: they are expected to hold on average, but there is nothing to preclude their temporary violation when we go beyond the thermodynamic limit. ...
Alex's user avatar
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1 vote

Enthalpy of a Van der Waals gas continuation

If you know the initial and final states of the gas, it is easier to get the enthalpy change for a VDW gas by working with the internal energy change $\Delta U$ than $\Delta H$. This is because the ...
Chet Miller's user avatar
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1 vote

Enthalpy of a Van der Waals gas continuation

Your equation is wrong. It should read $$dH=\left(V-T\left(\frac{\partial V}{\partial T}\right)_P\right)dP$$ You may have to integrate this numerically because of the non-linearity with of the VDW ...
Chet Miller's user avatar
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1 vote

Why does the temperature of the working fluid have to be close to the temperature of the cold reservoir during a Carnot cycle?

Energy transfer down a gradient (e.g., in temperature) generates entropy. The Carnot cycle represents the (unachievable) ideal of zero entropy generation. Therefore, the Carnot cycle can’t incorporate ...
Chemomechanics's user avatar
1 vote

Enthalpy of a Van der Waals gas

Expand the enthalpy in $T$ and $V$: $$\begin{align}\require{cancel}dH&=\left(\frac{\partial H}{\partial T}\right)_V\cancelto{0}{dT}+\left(\frac{\partial H}{\partial V}\right)_TdV\\&=\left[T\...
Chemomechanics's user avatar
1 vote
Accepted

Enthalpy of a Van der Waals gas

Your equation for dS is for an ideal gas. For a real gas, it reads: $$dS=\frac{C_P}{T}dT-\left(\frac{\partial V}{\partial T}\right)_PdP$$
Chet Miller's user avatar
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1 vote

Definition of entropy as unavailable work

Entropy of a system is a measure of a part the internal energy that is not available for isothermal work. More specifically, the internal energy consists of several parts, thermal $TS$, volumetric $-...
hyportnex's user avatar
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