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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 ...

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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 ...

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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 ...

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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 ...

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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 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, ... 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 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 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. 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 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 ... 3 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 ... 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 An ideal gas can have internal degrees of freedom, and they contribute to the specific heat, but the ideal gas law is still obeyed so long as the following conditions are satisfied: The deBroglie wavelength of the particle at the typical thermal energy kT is significantly smaller than the interparticle separation (so that the particle phase space may be ... 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 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 ... 2 First of all -- we are talking about elastic collisions. I strongly recommend you to read the Wikipedia page. Let me denote: v -- speed of the ball before collision m -- mass of the ball V -- speed of the wall M -- mass of the wall Substituting this into solution, presented on the Wikipedia page (which is, let me stress, derived from the ... 2 Yes, the CF is a way of comparing RGs and IGs kept at same temperature and pressure. It is a comparison of volumes, as stated in the question. Hence the name "compressibility factor". The CF equation can be better stated as: P_{obs}V_{m,obs}=ZRT_{obs}, where the m stands for "molar", and "obs" means "observed". This is in contrast with ... 2 The easiest way would be to use statistical physics. Gases are effectively described with a one-particle distribution function:$$f \sim \exp \left[-\frac{E}{kT} \right]$$which leads to a spatial part of f (concentration) in a uniform gravity field to be$$n_i \sim \exp \left[-\frac{m_i g z}{kT} \right]$$That's the easiest way to show that "the ratio ... 2 Assuming you're using degrees Fahrenheit 160F is only (about) 70C so you won't be boiling the water. The increase in pressure will come partly from heating the air in the bottles and partly from the increased vapour pressure of the water. The pressure of an ideal gas (air is near ideal in this temperature and pressure range) is:$$ P = \frac{nRT}{V} $$... 2 An aspect of this that isn't covered by the other answers is the following: you say "we are assuming that an externally applied force is pulling the wall out. It's not the gas that is pushing the wall out." But imagine for a moment that the wall is not moving. The gas is pushing on the wall, so in order for it to remain stationary there must be another ... 2 Have a look at the Wikipedia article on adiabatic processes. The key equation you need is:$$ PV^\gamma = K $$where K is a constant and the value of \gamma depends on what type of gas you have: for air \gamma = 7/5. As with any expansion the work done is the integral of PdV, so:$$ W = \int_{V_1}^{V_2} PdV = K \int_{V_1}^{V_2} V^{-\gamma}dV $$So ... 2 The answer by Claudius is correct with one assumption: That is that the process is reversible. In other words the process must take place so slowly that the system is always in equilibrium. Why is this important? Because the definition of work is really$$W = -\int_{V_1}^{V_2} p_{ext} dV$$Note the p_{ext}. That's the external pressure because doing ... 2 Electrically charged particles interact via their fields and so there is, in general, wide range interaction throughout the gas. The electromagnetic interactions between particles of the gas/plasma can give raise to effects which are significantly different from neutral gas, such as e.g waves. So to what extend the ideas gas law can be considered to "hold" ... 2 E(V,T) is an unspecified function. You don't need to know the form to get the derivation correct. The equation:$$dE=\left(\frac{\delta E}{\delta V}\right)_T dV+\left(\frac{\delta E}{\delta T}\right)_V dT$$is called the differential form of the total derivative and can be written for any function regardless of the form; it's simply a definition. As you ... 2 I think the last line does not follow from the previous steps. It is used to show how \gamma comes in place, so I extrapolated a bit and show the next few steps: Since$$ \frac{C_V}{Nk_B} = \frac{C_V}{C_p-C_V} = \frac{\frac{C_V}{C_V}}{\frac{C_p}{C_V}-\frac{C_V}{C_V}}=\frac{1}{\gamma-1} $$Therefore,$$ \frac{C_V}{Nk_B} (pdV+Vdp)= \frac{1}{\gamma-1} ...

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To define the instantaneous state of the gas (assuming its homogeneous and at equilibrium at any instant) you need two thermodynamic state variables besides the number of moles (or composition in general). Since both T and P are variable in your case, you need another variable besides V. Lets say you know the total energy you put in U. Then you can use a ...

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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 ...

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