# Tag Info

## Hot answers tagged ideal-gas

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

Because you are doing work to compress the gas, and the energy has to go somewhere. The molecules speed up because they collide with the wall moving forward--- if you move a wall forward, a ball which bounces off the wall reflects going faster by twice the speed of the wall, because if you move along with the wall, it reflects at the same speed. Answers 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

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.

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 ...

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 ...

6

There's actually not one simple answer to your question, which is why you are a bit confused. To specify your problem fully, you must specify exactly how and whether the gas swaps heat with its surroundings and how or even whether it is compressed. You should always refer to the full gas law $P\,V=n\,R\,T$ when reasoning. Common situations that are ...

5

As the comments to the question have stated, in real gasses ( contrasted to ideal gasses which just bounce around elastically) there exist both elastic and inelastic scatterings controlled by quantum mechanical interactions. Photons are generated leading to what we call Black Body radiation and an isolated gas volume will lose energy according to the ...

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

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

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 ...

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 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 The next line right beneath the equation you copied says: ...where \delta=\frac{V-V_0}{V_0} is the condensation (e.g. relative volume variation). Is it too difficult to read one more sentence? 4 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 ... 4 You have to realize first that Charles' law is the change in volume with respect to temperature for constant pressure while Boyle's law is the change in volume with respect to pressure for constant temperature. So when you combine them, you need to account for these If I take a gas of volume V_1, pressure P_1 and temperature T_1 and let it change have ... 4\rho=\frac{n*M}{V}$$With n the number of moles of the gas, M the molar mass of the gas, and V the volume of the gas. n and M are constant, so:$$\rho_1=\rho_0*\frac{V_0}{V_1}$$For V we have:$$V=\frac{n r T}{p}$$n and R are constant, so:$$V_1=V_0*\frac{T_1}{T_0}*\frac{p_0}{p_1}$$So:$$\rho_1=\rho_0*\frac{T_0}{T_1}*\frac{p_1}{p_0}$$... 4 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 ... 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 ... 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 In hydrodynamics, conservation means that what flows into the control volume is equivalent to the flow out of the control volume. With respect to momentum, we mean precisely that any change in momentum of the fluid within a control volume is due to the net flow of fluid into the volume and the action of external forces on the fluid within the volume ... 3 Ahh, I spent quite some time reading this problem, the problem with applying Dalton's Law of Partial Pressures is that we shouldn't be multiplying moles of CO2 with the total Pressure, rather we should multiply the mole fraction of CO2 with the total Pressure, in this case however, since the initial quantity/moles of oxygen is not known, it is not ... 3 Yes you are right. The 10kg piston acts as a force over the area of the piston, increasing the pressure and decreasing the volume of the gas inside. When the set up is tilted, the force no longer acts on the gas, but sideways, so the pressure equalizes. 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 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 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 ...

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