4

The properties and behaviour of a gas (almost) never depend on its nuclear energy, so we (almost) never consider it. For example if I am trying to design a steam engine then the kinetic energy of the gas molecules is vitally important because it determines the gas temperature and hence how the gas is going to behave in my steam engine. By contract the ...


4

Yes we are assuming this. The gas laws deal with gases in equilibrium, and with macroscopic samples of gas (say of linear dimensions thousands of times greater than the mean separation of the molecules). On this scale any pressure differences will cause bulk movements in the gas so that the pressures quickly equalise.


3

Here is an answer to (1) only, with the additional assumption that you pick the half of the box in advance. The half you pick can be any shape you want: you just need to select half the volume of the box. The regioun you select does not even have to be connected. An ideal gas consists of noninteracting particles which just whizz about. The position of ...


3

We are not interested in the force just while the molecule is in contact with the wall, but with the mean force all the time that the molecule is banging backward and forwards. Therefore we divide the molecule's change in momentum on hitting a wall by the time between one mid-collision with the wall and the next mid-collision. You might care to think of a ...


3

The ideal gas law takes no account of weather patterns over land and water; it takes no account of atmospheric circulation. So the connection you attempt to draw between the gas law and weather reporting is invalid. Furthermore, the gas law's relationship between temperature and pressure you cite requires that the volume be held constant. No such rule ...


3

A real gas will condense if you get close enough to $T=0$, no matter how low the pressure, so in practice there won't be a conflict. But this is an unsatisfying answer. The basis of the third law is that a system at absolute zero has zero entropy, because there is only one possible lowest-energy state. In the case of an ideal gas at absolute zero, all its ...


2

Part 1 The ideal gas law and first law are independent laws. However, the combination of the two laws shows that the change in internal energy of an ideal gas depends only on charge in temperature, as shown in part 2 below. Part 2 For the version of the first law that you are using when the gas does work (expands) work is negative (reduces internal energy). ...


2

Short answer Insanely small. Also insanely small. Longer answer. To get some definite numbers here, I'll assume that the temperature is 25°C, and the pressure in the cube is 1 standard atmosphere. I'll also assume we're dividing the cube into left & right halves, and we want to know the odds of all the gas molecules being in the left side. Using ...


1

A good answer has been posted, and accepted. Yet, I have a couple of things to say about the subject matter that might shed additional light on this, as well as other "science". What we teach in science classrooms is quite analytical and mathematical, and it is mostly quite appropriate and useful. However, what is mostly glossed over is the fact ...


1

Because for a monoatomic gas at pressures and temperatures such that the ideal gas law is a good approximation of its behavior, nuclear reactions do not take place and hence nuclear effects do not enter into the equation of state. Furthermore, at those pressures and temperatures, the motions of the gas atoms are so slow relative to the speed of light that ...


1

You can determine such properties by combining the ideal gas law with the first and/or second laws. For example, by combining the ideal gas law and the first law of thermodynamics, you can show that, for any process, the change in specific internal energy is a function of temperature only according to $$\Delta u=c_{V}\Delta T$$ For the derivation see here: $\...


1

I'll give this a go! My answer is that, no, the temperature of the gas will not increase. Qualitatively, as you accelerate the chamber (let us say from right to left), the right wall of the chamber that is normal to the acceleration vector, will strike numerous gas particles. Gas pressure can be related to the change in momentum of a gas particle striking ...


1

As pointed out by @Chet Miller, during the acceleration phase part of the gas gets compressed, but at the same time part of the gas gets expanded, so that the net effect upon reaching constant velocity should be no change in temperature. The following explains why the motion of the gas at constant velocity should not cause a change in temperature of the gas. ...


1

Suppose you have a container with volume $V$. If the particles were point-like, then each one's center can be anywhere within a volume $V$. On the other hand, if the particles have radius $r$, and one particle has its center at the origin, then a second particle can't have its center within $2r$ of the origin. If it were any closer, it would overlap with the ...


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