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I propose the following order-of-magnitude, very rough line of reasoning. The rate of momentum escaping from the aperture is $$\frac{dp}{dt}=\rho\pi(D_2/2)^2 v^2,$$ all in the horizontal direction. I assume that a fraction $(1-\cos\theta)$ of the horizontal momentum will be lost by the fluid since the change of direction of its motion. This fraction of ...

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For solids, as well as for liquids and gases, you can define a specific heat as $$c_V \equiv \left(\frac{\partial U}{\partial T}\right)_V$$ and $$c_P \equiv \left(\frac{\partial U}{\partial T}\right)_P$$ For solids, however, $V$ is basically a constant, so that with excellent approximation $$c_P \approx c_V \equiv c$$ and we can now define an ...

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The equation for enthalpy is valid for 'all' gases under normal conditions. It is because all the equations of thermodynamics (except ones which have ideal gas laws substituted in them) were experimentally confirmed under normal conditions. That is the reason why even enthalpy is an extremely useful tool in calculating the thermodynamic properties in any ...

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The ideal gas equation you try to use for understanding does not provide a full description. To completely describe a thermodynamical system, you need the relevant thermodynamic potential, which here is the internal energy if I understood well your formulation (is a bit vague). Processes like the isobaric expansion occuring here, cannot be explained through ...

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You are correct. In a real process, one would modify this formula to include the so-called polytropic exponent $n$ such that $PV^n=const$. This reflect that the process is not perfectly isentropic. For a fixed final volume this means the final temperature and pressure will be higher than in the ideal case, and more work need to be expended.

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The idea of the question is to find the temperature at which the average interparticle spacing is equal to the average de Broglie wavelength. Both of these are averages because the atoms of the ideal gas are not evenly spaced and the velocity (and therefore de Broglie wavelength) of the ideal gas atoms follows the Maxwell-Boltzmann distribution. So this is ...

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"The size of a single gas particle" is a bad term for what you calculate. A better term would be: "The volume one of the gas particles would occupy, if the total volume were distributed equally among all gas particles" And this translates loosely to "The volume, in which you find one gas particle on average" If you then imagine every particle sitting at ...

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Why would the temperature of the gas in the balloon be fixed? Note that it will depend, among other things, on the thermal conductivity of the rubber of the balloon: if the thermal conductivity were zero, the gas in the balloon would expand isentropically, rather than isothermally.

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Mass conservation assuming stationary conditions: $$\rho(x,t)v(x,t)=\rho(0,t)v(0,t):=\rho_0(t)v_0(t),$$ or equivalently $\rho(x,t)q(x,t)=\rho_0(t)q_0(t)$ since $q=vA$, with $A$ a constant cross section. Therefore $$q(x,t)=\frac{\rho_0(t)q_0(t)}{\rho(x,t)}.$$ Then, we use the initial conditions $$q_0(t)=-\frac{V}{\rho_0(t)}\frac{d \rho_0(t)}{dt},$$ and you ...

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Effectively ideal gases are pretty easy to come by. Air at SATP would be a good example. To be a little more precise, what you're looking for is that the mean free path $\lambda \gg \sigma$, the diameter of the molecules. Basically this just means that molecules spend most of their time far away from each other. You can find $\lambda$ by: ...

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