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Whoever taught you that that equation is valid only for an isothermal atmosphere was just wrong. That equation is for a fluid in hydrostatic equilibrium. Our atmosphere is close to being in hydrostatic equilibrium, as are the oceans, the interior of the Earth, and the interior of our Sun. That vertical motions obviously do exist in the atmosphere, the ...

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Assuming we can treat the air in the room as an ideal gas, it will obey the ideal gas equation of state: $$PV = nRT \tag{1}$$ where $n$ is the number of moles of the gas. The question tells us that the pressure is constant, and obviously the volume of the room is constant, so the only things that can vary are $T$ and $n$. The question tells us that $T$ ...

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The growth of internal energy or change in internal energy can be determined by the following equations $\Delta U = W + \Delta Q$ or $U = nc\Delta T$, with c being specific heat at constant volume Now, W is zero since the volume hasn't changed ($W = \int P dV$), so all that is needed is to determine change in heat energy. Heat energy added is equal to ...

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The Helium atoms are so small they diffuse rapidly through the rubber walls of the balloon, and as soon as the volume has decreased to the point were the bouyant force is less than the weight of the balloon the balloon sinks to the floor. If you're interested, some Googling found this school project investigating the helium diffusion through balloons. They ...

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If the temperature of the gas is kept constant during the compression then the bulk modulus of an ideal gas is just equal to the pressure. The definition of the bulk modulus is: $$K = -V\frac{dP}{dV} \tag{1}$$ For an ideal gas $PV = RT$, so $P = RT/V$. If the temperature is constant this gives: $$\frac{dP}{dV} = -\frac{RT}{V^2} \tag{2}$$ and ...

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Static pressure in a compressible flow depends on the density but not the speed (not directly). Speed and geometry may affect the density. For isentropic flow (neglecting gravitational potential): $${p \over \rho^\gamma} = constant, \gamma = {c_p \over c_v}$$ which could be turned into this: $${p \over p_0} = ({1 \over 1+{(\gamma-1) \over ... 1 The density is directly given by \begin{equation*} \begin{split} \rho(z)&=\frac{\int dpAe^{-\beta(p^2/2m+mgz)}}{\int dp\int_0^hdzAe^{-\beta(p^2/2m+mgz)}}\\ &=\frac{e^{-\beta mgz}}{\int_0^hdze^{-\beta mgz}}=\frac{\beta mg}{1-e^{-\beta mgh}}e^{-\beta mgz} \end{split} \end{equation*} since the momentum is homogeneous throughout the system. ... 1 The force would be proportional to the area, so the force on the hole would be 10^(-5) N 2 The Van der Waals equation of state contains two constants a and b, and for an ideal gas these are zero. So the question you need to answer is whether your experimental data shows these constants to be significantly different from zero. So you need to perform some form of curve fitting to calculate a and b, then determine the statistical error in ... 0 The basic idea you're reaching for here is called hydrostatic equilibrium. That is, gravitational forces are balanced by pressure gradients. In a context like geophysical fluid dynamics, you'll encounter hydrostatic equilibrium written as$${\partial P\over\partial z} = -\rho g, $$but as you point out when you bring up Bernoulli's equation, that won't work ... 2 Your overall method for determining the equilibrium for a star is correct. We use a combination of an assumed 'polytropic relation':$$P = K \rho^{1+\frac{1}{n}}$$and the requirement that this balances the gravitational pressure at a radius r (for a spherical shell of thickness dr):$$4\pi r^2\frac{dP(r)}{dr} = -\frac{GM(r)}{r^2}4\pi r^2\rho(r) where ...

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