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The practical thing that you are missing here is that the air compressor is cooled constantly so that it doesn't melt during operation (can't be adiabatic). So the upper limit on the temperature is driven by the operating temperature of the compressor. Then the other posters comments about heat transfer out of the tank apply.

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Start with the ideal gas law: $$PV=nRT$$ Let's compare the volume of air in the tyre with a similar volume outside the tyre under normal atm conditions. We know that $V$, $R$ and $T$ are constant so: $$\frac{P_{atm}}{n_{atm}}=\frac{P_{tyre}}{n_{tyre}}$$ We know that $$n \propto m \propto \rho$$ So we can say: $$\frac{P_{atm}}{\rho_{atm}}=\frac{P_{tyre}}{... 0 Foundations Given a state property P, the definition of its partial molar value in a mixture at a given temperature T and total pressure p is below.$$ \check{P}_i \equiv \left( \frac{\partial P}{\partial n_j} \right)_{T, p, n_j \neq n_i} $$The definition measures the change in the total intensive property of the system when the number of moles of ... 0 Would you feel better if they said that it applies to all extensive properties with units involving energy, or, more precisely, with its most fundamental units involving Newtons? So Joules is Newton-meters. But, in the case of volume, if you want to say that it is energy (Joules=N-m) divided by pressure (N/m-s^2), the N's cancel out. Proof regarding ... 1$$V\bigg(\frac {PV}{Nk} \bigg)^{f/2}=\frac{P^{f/2} V^{1+f/2}}{constant}$$2 You did not put P^{f/2} after the first step. From there you will get the right answer (then you will take both sides of the equation to the 2/f power). 1 Review the equipartition theorem. It tells us that each degree of freedom quadratic in coordinate or momentum contributes \frac{1}{2} kT per element. Ideal gas particles have no internal structure, so there is no contribution from internal mechanisms. The energy of the particle does not depend on location, so there is no dependence on coordinate, and ... 0 Use an inverted measuring cylinder instead of a graduated beaker: it's more precise: Once the cylinder is near full, adjust its position so that the liquid meniscus inside and outside of the cylinder coincide exactly. That means pressure inside the cylinder is equal to atmospheric pressure. The ratio of volume collected and time elapsed is the volumetric ... 1 Interesting question! Like you say, temperature is related to molecular kinetic energy as$$e_t = \frac{3}{2} \frac{k T}{m} ,$$where m is the molecular mass, k is Boltzmann's constant, and e_t the specific translational internal energy (in J/kg). The latter comes from the kinetic energy from random thermal translational motion of the gas molecules. ... 2 Another way of phrasing your question might be how to determine how much air flows through the engine rather than around it. Super Sonic Started Under normal ramjet operation this is actually fairly easy to calculate. We can multiply the volumetric flow rate by the density:$$\dot m = \rho_1 V_1 A_1 Where $\dot m$ is the mass flow rate, $\rho$ is ...

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The incoming air slows down and compresses due to its speed relative to the engine and the shape of the inlet. In order to slow it down, a force must be applied to the air. This force comes in part from the inlet hardware, and in part from the high pressure air in the combustion chamber. It is this incoming air that creates the pressure in the combustion ...

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