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Is you are adding gas at constant volume. Then , $$W = P\Delta V + V\Delta P.$$ Since $\Delta V = 0\;_,$ $$W = V\Delta P$$ Then extra gas you put in container increase pressure, increasing Work and hence increasing internal Energy. $$\Delta U = Q + W = Q + V\Delta P$$ Now, $Q$ here is heat inside system which is a measure of Temperature and not ...

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Expansion A gas is a collection of atoms or molecules that are constantly moving and colliding with each other. Credit for this great visualization goes to Greg L at the English language Wikipedia. If the bounding box were to disappear, the particles would escape in the direction they were previously travelling - which for lots and lots of particles ...

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The state of a substance depends very much on its temperature. There can be a change of state due to expansion that is how gas liquefiers work. A gas is compressed and then is allowed to expands rapidly. In expanding the gas does work against the surroundings and in separating the gas molecules so the average kinetic energy of its molecules drops ie it ...

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For both hydrogen and chlorine E = 3/2 kT is only true at very low temperatures since they are diatomic gases. In general you get a contribution of 1/2 kT to the energy for every quadratic degree of freedom. For a monatomic gas that is 3 translational degrees of freedom, hence 3*1/2 kT. For a diatomic gas there are in addition 2 rotational, 1 vibrational ...

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The answer you are looking for is that $$\rho \propto 1/V$$ and $$m/\rho = 1/V$$ with this knowledge you should be able to see how your final equation relates to $$pV = nRT$$

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Any gas behaves as an ideal gas under high temperature an low pressure. Atmospheric oxygen is at 25deg celcius which is greater than it's critical temperature which is -150 deg celcius. And the pressure is somewhere around 159 mm of Hg. Hence atmospheric gases is said to behave like that of an ideal gas.

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The state equation says: P V = R T For an isothermal process, T is constant by definition R is a physical constant For an expansion process V is increasing So, P must necessarily be decreasing for the gas within the system Assuming a constant piston area of A, the force on the piston is: F (on piston from internal gas) = P (internal gas) A ...

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The pressure of the environment is not held constant. The gas is allowed to expand very gradually by, say, removing tiny weights from the top of the piston. Along the entire path of the expansion process, the pressure of the surroundings is held slightly below the pressure of the gas so that the gas is always only slightly removed from being at ...

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The 'balloon in a jar' demonstration just illustrates the principle of lung ventilation, but really shouldn't accepted too closely in modeling the dynamics of breathing. In a real lung the only gas-filled space is the space that communicates with the upper ways, and terminates at the other extreme with sac-like structures called alveoli. The space outside ...

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I think the referenced experiment is interesting but not an exact model of the lungs. But in the "close enough" realm, what's happening is that the movement of the diaphragm (and the ribcage) happens as various muscles act against the external air pressure. Once there's additional volume inside (and assuming no collapsed lung, the interior of your lungs, ...

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The number of moles in the tank at any time is $$n=\frac{pV}{RT}$$ The rate of change of the number of moles in the tank is: $$\frac{dn}{dt}=-\frac{p}{RT}Q$$where p/RT is the molar density at time t. If you combine these two equations, you will get an ODE for dp/dt which you can solve for p as a function of time. The tank will stop leaking when $p = p_0$.

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