# Tag Info

21

If gas A and gas B are of different density, then the situation sketched is not in equilibrium: the water level on the side of the light gas will be higher. There, the containers are moving down, and you have to push your containers through this net difference in level. You do need to put in energy here, which is probably the piece that you are trying to ...

10

The notation df denotes differential of function f. The differential df is a map $$df:\mathbb{R}\rightarrow \Omega^1(\mathbb{R})$$ where Ω1(ℝ) is the set of linear maps from ℝ to ℝ. The linear map corresponding to point p∈ℝ is often written as dfp $$df(p)=df_p$$ Note that in less formal settings p ...

10

If your an engineer and your looking for a material to spin your turbine, your goal should be to maximize pressure and minimize temperature. Well no. Temperature is generally a constraint of your heat source, and your goal is to maximize profit. That goal does, however, map to physical properties in logical ways, but it's much more complicated that ...

7

The ideal gas law is derived from a model (the ideal gas), and like every other model it applies where it's underling assumptions are good approximations to reality. So, important assumptions for the idea gas law: Point particles In the ideal gas, the particles occupy no volume. A real gas in which the atoms of molecules occupy a vanishing fraction of the ...

7

It's a steady state. If there were a pressure gradient, there would be net force on the gas (ignoring gravity). There's no net force here because the air isn't accelerating. Thus the pressure is constant. The number density varies across the box inversely to the temperature so the ideal gas law holds.

6

dmckee gives some good qualitative considerations, but we can also develop rules for when the ideal gas law is and isn't appropriate. To start: The law applies perfectly in the case of a gas when $P\rightarrow 0$. The law does not apply to liquids. Between these two states is a gray area. In that case you should look at the compressibility factor, ...

6

The above equation solves for the average kinetic energy of a gaseous particle at a given temperature. k is known as Boltzman's constant, $k_B = 1.3806503 × 10^{-23}\frac{m^2kg}{s^2K}$ and is equal to the ideal gas constant divided by Avagadro's number, $\frac{R}{N_A}$. So where does the equation come from? The short answer: The equation above is ...

5

Photons are radiation so their equation of state is $$p = \frac{\rho}{3}$$ where $\rho$ is the energy density. So we have $$p = \frac{mc^2}{3V} = \frac{1\times 9\times 10^{16}\,\,{\rm J}}{0.003\,{\rm m}^3} = 3 \times 10^{19}\,\,{\rm Pa}$$ It's a huge pressure. Not a surprising fact because the actual mass of photons we can produce is negligible. One ...

5

According to the second law, thermal equilibrium between two systems means that they both have the same temperature $T$. The fact 2 that $PV$ coincide whenever two gases are at thermal equilibrium (and, I assume, for the same $n$) means that $PV$ is only a function of $T$. In othor words, there is a function $g()$ such that \begin{align}PV&=g(T)& ... 5 The Equipartition theorem states that each degree of freedom has an average energy of 1/2KT. This is valid at large enough temperatures where quantum mechanics does not play a role. A = 3/2 KT(3 degrees of freesom) B = 3/2 KT C = 1/2 KT (1 vibrational degree of freedom in a di-atomic molecule) D = KT (2 axis of rotation, the third has very low Moment of ... 5 This follows from the equipartition theorem. The equipartition theorem states that in thermal equilibrium, the average energy of each degree of freedom (each independent way the system can move) is k_B T/2, where T is the temperature and k_B (or just k) is called the Boltzmann constant. There are three independent directions in which a gas particle ... 5 In addition to Bernhard's answer, just because three gases (Gas A,B and air - which is itself a mixture of nitrogen, oxygen, and other gases) have different densities, it does not mean they will remain seperated when in a container. In fact, as entropy of the system increases over time, Gas A, B and air will make an even (if heterogeneous) mixture. 4 Do exactly what the passage says to do. Consider it in the wall's frame of reference. Say that in the box frame, the wall is moving right at 10 m/s and a particle approaches it at 100 m/s. In the wall frame, the wall is stationary and a particle approaches it at 90 m/s. The particle bounces and is now going 90 m/s to the left. Transforming back to the ... 4 Pressure and volume have an inverse relationship when n and T are constant. How do you imagine the pressure in the balloon is increased? Either n or T must increase, or V must decrease. Additionally, balloons are roughly constant-pressure systems. The rubber membrane is a very weak elastic, so the internal pressure of the balloon is at almost ... 4 I am a student so please point out in gory detail anything I did wrong. For a process to be quasistatic, the time scales of evolving the system should be larger than the relaxation time. Relaxation time is the time needed for the system to return to equilibrium. We have an adiabatic process, so equilibrium must be preserved at each point, that is to say ... 3 You need an equation for the density of the gas as a function of temperature and pressure. Assuming the tyre is full of air, this is reasonably close to an ideal gas so the molar volume is given by:V_m = \frac{RT}{P} where R is the ideal gas constant and and the average molecule weight of air (20% oxygen, 80% nitrogen) is about 14.4. From this you can ...

3

If one defines an ideal gas as a gas made up of non-interacting entities, then the answer is yes, the gas may indeed have internal structure. The equation of state for such a gas will still be $pV = nRT$, but its energy will not be $E = (3/2) RT$ per mole. The exact form of the energy equation will depend on the internal energies of the molecules. The ...

3

Think about this: why is the pressure increasing? If it's because you're blowing air into the balloon (which is the usual way to increase the air pressure), then what you're actually doing is raising $n$. And it makes sense that an increase in $n$ should be correlated with an increase in $p$ (or $V$, or both). Boyle's law doesn't apply in this case because ...

3

If we assume that the gas in the basic container is at equilibrium, then it has a temperature that we can take to be $T_1$. If the gas being added is also at equilibrium it too will have a temperature that we can call $T_2$. If the two temperatures are the same, then the final temperature will be the same as the initial one. On the other hand, if the two ...

3

You were onto it in the comments, so I might be late to offer anything new here. The pressure is irrelevant in this problem; it's a trick, I guess. A reversible adiabatic process is one in which there's no heat flow in or out of the gas, so all of the work done in the expansion/compression goes into the temperature change. Just calculate the change in energy ...

3

I'm guessing you mean for a fixed n moles of ideal gas floating freely in the atmosphere. I also assume a short term heat injection (not a source that holds a particular temperature). Temperature is directly related to kinetic energy. So the first response of the free floating ideal gas will be to become more excited. This will create a slow pressure ...

3

Bravo on some wonderfully clear and careful thoughts on your problem! I think this paper will help you: E. T. Jaynes, "Gibbs vs Boltzmann Entropies", Am. J. Phys. 33, number 5, pp 391-398, 1965 as well as many other of his works in this field And a summary of my answer below is: Statistical correlation between a system constituents' states and the use of ...

3

You're trying to ask a question about real life (would the tank melt) with a model that approximates away the very thing that you need (adiabatic). With the assumptions you have made, you are in fact using the correct equations. The adiabatic assumption is only valid if it is thermally insulated which is not the case in real life. There will be losses of ...

3

It does not work for the following reason. Let's look at the right side, where the containers float to the top. When a new container enters at the bottom (from water to gas B), it pushes some gas B away, so it can occupy the space. The pushed away gas B has nowhere to go, but up. Gaining height means potential energy. That is energy used, to make the ...

3

As it turns out, you can use it for the vast majority of the time. The ideal gas law is considered "ideal" because it assumes interactions between the gas molecules only occur due to collisions and no long-range forces are present. If the fluid you are modelling is non-polar and doesn't have things like van der Waals forces between molecules, then the ideal ...

3

In the limit of very high temperature all gases become ideal (assuming they don't ionise, dissociate, etc) but this regime is far above the Boyle temperature. Around the Boyle temperature the long range attractive forces are still significant and cause non-ideal behaviour. It's just that there is a sweet spot where the attractive forces are balanced by the ...

3

There are many problems : $1.$ As pointed out by Olin, gas cannot exist as a gas at $0 K$. $2.$ In ideal gases, interaction between molecules are absent. Hence, there is no potential energy. Remember that Potential energy always has an additive arbitrary constant. $3.$ As pointed by Wojciech, you would need (to take}energy to cool that ...

3

There is a huge difference with the usual Joules experiment. Here the system is not isolated during the gas expansion into the evacuated chamber. Two things enter in competition in principle when a setup as the one considered in the article you mention is considered: Whether the gas has to work to expand itself in the larger chamber Is there anything from ...

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