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45

The next approximation beyond the ideal gas is given by the Van der Waals fluid equation. It is a phenomenological law which takes into account the finite size of the molecules and their interactions with themselves. When you plot several Van der Vaals isotherms for a given substance, you observe that some of them show a phase transition from gas to liquid ...


38

You feel cold when heat is flowing from you to the surroundings, your body tries to burn more energy to keep up your temperature, so you shiver. Water conducts heat much more effectively than air (more than 100x as well) so even with water at the same temperature as air you will lose a lot more heat and feel cold. When your body is too hot it losses energy ...


21

Getting from gas to liquid is a matter of interparticle interaction winning over thermal agitation. There are several reasons why interparticle interactions are very weak in the case of helium atoms. On one hand, it is a noble gas and thus cannot form covalent bonds. On the other hand, it is very light hence highly non-polarizable: its Van der Waals ...


10

You seem to make the implicit assumption that your vessel is placed in an environment that does not emit any thermal radiation, i.e. is already at 0 K temperature. The temperature of your container will asymptotically decrease to 0 K but will never actually reach it. Assuming black-body radiation, fixed heat capacity $c$, and sufficient thermal ...


6

I will try to answer as many questions as I can. I won't presume to give you complete exhaustive answers, but maybe they will be nonetheless useful to you. What variables determine the range of temperatures over which matter is liquid? My understanding of thermodynamics is that matter changes from a solid to a gas when some thermal vibrations create an ...


5

You've discovered a famous problem in thermodynamics. In our case the piston will not move. The mechanical argument is right, while the maximum entropy argument is inconclusive. To see that $P_1=P_2$ is an equilibrium position you can also apply conservation of energy. Since there is no heat exchange, $$dU_{1,2} = -P_{1,2} dV_{1,2}$$ We require that ...


4

There are some issues with the experimental setup you proposed (apart from the fact that when its temperature is lowered the gas would become a liquid and then a solid - if it's not $^4$He: in that case it will stay a liquid). Let's see why. In the picture above, I've sketched your experimental setup. The black box must be impermeable to matter in order ...


4

It won't work because your perfect vacuum is permeated by the cosmic background radiation, which itself is only asymptotically reducing to zero with the expansion of the universe. Trying to exclude the cosmic background radiation backs you into the infinite steps that forms the basis of the third law again. Also, using a container results in quantum ...


4

Looking around, the root mean square speed of air at $20$ C is about $500 m/s$, and given that you have $\langle v^2 \rangle \propto \, T$ so that $v_{rms}(T) = \sqrt{\langle v^2\rangle}$ varies with $\sqrt{T}$ then have $$v_{rms}(15) = v_{rms}(20)\times \frac{\sqrt{15+273}}{\sqrt{20+273}} \approx 496 m/s$$ and $$v_{rms}(25) = v_{rms}(20)\times ...


4

As stated in the comment by Peter Diehr, the question is in principle no different whether you ask it for electromagnetic, gravitational or any other kind of wave. The wave's entropy is simply the conditional Shannon entropy of the specification needed to define the wave's full state given knowledge of its macroscopically measured variables. A theoretical ...


4

When you feel hot, you perspire so as to benefit by evaporative cooling. As the relative humidity gets closer to 100%, the sweat cannot evaporate and evaporative cooling becomes less effective. Liquid water is a much better conductor of heat than air (even humid air) is, so if the water is even a few degrees cooler than your body, you feel cold because the ...


4

Throttling the gas (Joule-Kelvin expansion) only lowers the temperature of the gas when the Joule–Thomson coefficient is positive. For Helium, that point (the "J–T inversion temperature") is reached at 43°K (source: Cryogenic Society of America; the wikipedia article gives an incorrect value of 51°K). Above that temperature, Joule-Kelvin expansion will ...


3

For the first question, it is the low boiling temperature, 4.21K for Helium-4 and 3.19K for Helium-3, that makes helium difficult to be liquefied. Hydrogen's boiling temperature at 1 atm is 20.27K, or about 4-5 times higher. For pre-cooling, one can take a look of entropy $$\delta S = \frac {dQ}{T}$$ We can see that, because T is very small, a slight ...


3

Simply Said: It doesn't attract itself enough to be solid. So much resistance to attraction that it is very hard to liquefy. The electrons prefer to repel because the electrons repel like the protons.


3

If the phase change occurs at the temperature of interest, then the system can give off a lot of heat without cooling down very much. Thus, melting ice is a great way to maintain something at a temperature around 0°C, and melting paraffin-18-Carbons is good if you are trying to maintain temperature around 20 °C. With a melting point of 28°C, the material ...


2

Liquid: molecules form bonds with neighboring molecules for most of the time, but there are enough energy for the bonds to break momentarily and be formed again with another molecule "These explanations seem hand-wavy to me." What is the level of your knowledge of physics? Are you aware of the quantum mechanical nature of atoms and molecules, at a ...


2

The answer is a definite yes and no. Gravitational waves have entropy in that we can think of them travelling from their source to our detector as a channel that sends units of information in the sense of the Shannon formula. The ringing of our detector is then the reception of that information. The Shannon formula $S=-k\sum_n p_n log(p_n)$ would give a ...


2

For the hypothetical case of a thermally perfectly insulated system, I'm sure you can work out yourself from the specific heat and the enthalpy of fusion for water. Given that the enthalpy of fusion (330 kJ/kg) and the specific heat of ice (2 kJ/kg-K) have a ratio of 165 K, and you need the entire ice bucket to stay below the melting point, and your 20:1 ...


2

It would be more scientific to have compared a covered with an identical non-covered bowl, with temperature measurements over an identical period of cooling time. But covering a bowl of hot soup with a plate does indeed reduce heat losses and slows down cooling of the soup, because: Radiative losses are usually small at these temperatures but the plate ...


2

' if we employ any other process that is not adiabatic, say an isochoric process, to take the system from one temperature to another, we shall need a series of reservoirs in the temperature range $T_1$ to $T_2$ to ensure that at each stage the process is quasi static ' This means if we want to have a reversible process between two temperatures, then we ...


2

Your scenario is actually a classic transient conduction problem tackled in undergraduate engineering heat transfer, so we can handle this scenario easily. I took the figure below and adapted a derivation from a popular heat transfer textbook by Incropera and DeWitt: In this figure a warm object is placed in a tank filled with a known liquid (the ...


1

You are describing two different mechanisms of cooling the human body: 1) When we sweat our body produces fluids that tranfer heat from us through their evaporation (fluids gets the heat from our body and evaporate). One can easily understand that the more the surrounding enviroment has a highly humidity value , the more this heat- transfer gets ...


1

There's an important difference between liquid water and water vapor. The difference is 539 calories per gram to be exact. That's why being in humid air makes you hotter - the water vapor has all this latent heat it can release upon you by condensing. Equivalently, it prevents your sweat from evaporating and absorbing this energy. Liquid water doesn't have ...


1

Your question seems to be about heat transfer through convection. The formula that describes this phenomenon is: $dQ/dt=h*A*(To-Tenv)$ where $dQ/dt$ is the heat transferred per unit time, $A$ is the area of the object, $h$ is the heat transfer coefficient, $Ta$ is the object's surface temperature and $Tenv$ is the fluid temperature (temperature of air ...


1

A thermodynamic process is called reversible if an infinitesimal change of the external conditions is capable of reversing the process. A thermodynamic process is called quasi-static if it is a dense succession of equilibrium states. Roughly speaking, a quasi-static process is one that can be represented by a continuous curve in the thermodynamic ...


1

A significant factor in this is that helium is hard to produce. There is very little helium in the atmosphere because it is so light that individual molecules have a good chance of achieving escape velocity at atmospheric temperatures so it disappears into space fairly quickly. Also because it is very uncreative it doesn't get bound as chemical compounds, ...



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