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The answer lies in something called the virial theorem. You are correct, a cloud that is in equilibrium will have a relationship between the temperature and pressure in its interior and the gravitational "weight" pressing inwards. This relationship is encapsulated in the virial theorem, which says (ignoring complications like rotation and magnetic fields) ...

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As gas clouds collapse, they increase in internal energy (measured by temperature). This is part of what causes their pressure to increase. As they increase in temperature, though, they also increase the amount of radiation they emit. As they emit radiation, their internal energy decreases and thus their pressure also decreases, allowing for further ...

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Its because if we divide the container in two halves then the volume of the gas will also get half. But the pressure applied on the walls of both the containers will be same.

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When a fluid is squeezed, as in a cylinder by a piston, work is done on the fluid. This work 1) elevates the pressure (pressure energy), and 2) the temperature (heat energy). (If the cylinder is insulated, this is called "adiabatic".) In an ideal gas, these are all related by the ideal gas law, which says roughly that volume times pressure equals heat.

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The technical answer First I need to explain entropy. Suppose you have any system which you can only see at a certain granularity: we say that you see its "macrostate" but this could be any set of "microstates" which all "look alike". Since particle-interactions tend to multiply and distribute our uncertainty about a system, we could imagine under certain ...

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The pressure energy is the energy in/of a fluid due to the applied pressure (force per area). So if you have a static fluid in an enclosed container, the energy of the system is only due to the pressure; if the fluid is moving along a flow, then the energy of the system is the kinetic energy as well as the pressure. Because of the unit breakdown you have ...

3

For a uniform, spherical distribution of mass (cloud of gas and dust) of radius $R$ and mass $M$ in absence of magnetic, radiation fields etc, we have $dm = 4\pi \rho r^2 dr$ and the potential energy of a spherical shell of inner radius $r$ and outer $r + d r$ is $dU = -G\frac{m(r)dm}{r}$, $m(r) = \frac{4}{3}\rho r^3$, and a simple integration yields, ...

3

From what I've gathered, I think my initial guess is correct. Air tries to maintain a constant pressure. According to the ideal gas law, there are two ways to maintain the same amount of pressure with an increasing volume: 1) increase the amount of gas, and 2) increase the temperature of the gas.

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The pressure is caused by the weight of the water above. The compressibility (or lack thereof) of water is irrelevant to the pressure. Try this experiment. Put your hand on the table. Now put a brick on your hand and feel the pressure. Then add a second, third, etc brick. You will feel the pressure increase, but you will not see the bricks being ...

3

From the ideal gas equation, $$P=\frac{nRT}{V}$$ Now assuming the gas is uniformly distributed over space (has constant density for a given temperature), halving the number of moles will divide the volume by the same amount. Essentially, if we divide the number of moles by any number, we will end up dividing the volume by the same number to maintain ...

3

To derive the Bernoulli equation for inviscid fluids, the plan is to rewrite the Euler equation in such a way that we have gradients. I'll write the Euler equation with gravity here $$\frac{\partial \vec{u}}{\partial t} + \vec{u} \cdot \vec{\nabla} \vec{u} = -\frac{1}{\rho} \vec{\nabla} p + \vec{g}.$$ Recall $g = - \vec{\nabla} \Psi$, and $\vec{u} \cdot ... 3$P_1$and$P_2$are defined in terms of a reference pressure and a contribution due to the weight of the liquid column: $$P_1= P_0 + \rho g h_1 \quad P_2=P_0+\rho g h_2$$ Note that the$\rho g h = \rho g V/A = F/A$is the pressure exerted by the weight of the liquid column. Taking the difference$P_1-P_2=\rho g (h_1-h_2)$cancels the reference pressure ... 2 This is how brake boosters on cars work. The check valve will obviously leak slowly over time, but it can last several minutes, perhaps hours, and a better-designed system could likely last longer depending on your application. There's no such thing as "vacuum" in some tangible sense. Vacuum is just a relatively lower pressure, and pressure tends to go from ... 2 As Mikael Kuisma remarked, gas particle do accumulate at lower altitude. Consider two volumes$V_u$and$V_l$that are vertically thin as compared to their horizontal extend, which are separated by a distance$H$. A tube of negligible volume connects$V_l$to$V_u$. A gas particle of mass$m$in volume$V_u$has a potential energy of$mgH$as compared to a ... 2 When the cup is turned upside down, the water wants to fall out. The air-filled cavity is therefore stretched a bit as the gravity pulls down the water. The air expands a bit. This reduces the air pressure inside the cup, since increasing volume reduces pressure. This is hinted in the ideal gas equation: $$pV=nRT$$ Soon this lower pressure pulls upwards ... 2 1) The relation$\frac{dF}{dS}=d\left(\frac{F}{S}\right)$is certainly incorrect as @Floris has mentioned in the comment. As the simplest counter-example, consider a linear function,$F(S) = \alpha \, S$, with$\alpha \neq 0$as a proportionality constant. Then, one could easily see that$$\frac{dF}{dS}= \alpha \neq 0 = d\left(\frac{F}{S}\right) = ... 2 You are correct that your chest muscles are in fact pulling the lungs "open," which creates a pressure differential and draws air into the lungs. When the muscles relax, the chest cavity collapses to its original state, expelling the air (not 100% of it!). You may have heard of a "collapsed lung" injury. What happens there is that the lung is ripped loose ... 2 Be very careful. What you're doing is very dangerous. Large pressures and forces are present here and you're lucky that the ordinary (1/8" thick?) glass panes didn't shatter and throw glass shards at you and your family. Look at what you've made. All that water trapped between the glass panes has no strength so it just wants to go down in the direction of ... 2 Background There is a good reference1 on the physics of sound/shock waves in solids (look at Chapter XI). I found the following (on page 688) very interesting and relevant to your question: In a solid or liquid, a shock wave with a strength of even a hundred thousand atmospheres is regarded as weak. Such a wave differs little from an acoustic wave: it ... 1 Knowing a specific density is rather handy, because it allows us to relate the volume and the mass of a substance. But to calculate the pressure, the density isn't necessary. What is needed is the mass. If the fluids mix, the volume (and therefore the density) may change, but the mass does not. The total mass remains constant. If prior to mixing, the ... 1 You seem to understand the$\rho g h$term, so I will explain the pressure energy in terms of that. Really the$P$term and the$\rho g h$term are very similar. For now lets just ignore the$P$term and focus on the$\rho gh + \frac{1}{2} \rho v^2 = \textrm{constant}$. This is basically just conservation of energy. It says that if you throw an object up ... 1 Pressure "energy" P is simply the measured pressure anywhere in the system. For example, if we have a venturi whose entry and exit cross section are 1m^2 and velocity 1m/s, but it has a throat of cross section 0.1m/s, the gas will speed up through the throat, reaching a velocity of 10m/s. The measured pressure on a pressure gauge with an inlet ... 1 If the valve is totally effective, then yes, the pressure will remain low. The pressure of a gas is a measure of how much force its particles exert on the container it's in, over a given area. The pump removes many of the particles, so the force they collectively exert goes down; it stays down as long as no new particles enter. 1 There is not just one particle in the the box or container. There will be many particles rebounding every time on the wall under consideration. That's why we used$F_{avg}$and not just$F$. It's because$F$means the force applied by just one single particle and$F_{avg}$means the total force applied by all the particles over that period of time$ \Delta ...

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The ideal gas law is the time averaged steady state of this system. Consider it on a very long timescale: if you wait several round trips, what is the average impulse imparted in $\Delta t$? This of course includes the interior transit time. Another way of seeing this is that at any instant, most of the gas particles are not pushing on the balloon surface ...

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What is the relation between pressure and velocity of a fluid in a closed pipe flow? Bernoulli's equation: By continuity equation velocity at all points is the same. Then shouldn't the pressure be same at both points? No the pressure won't be the same at all points in the pipe. Considering Pascal's Law, a change in pressure at any point in an ...

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Answering isn't really possible since we don't know the exact conditions involved. A couple notes, though: First, the bag is sealed at the factory, not where you bought it, so it's the pressure difference between your new location and the factory that matters. Second, it doesn't take a lot of pressure difference to make a flimsy chip bag expand. You can ...

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By the Ideal gas law, $PV=nRT$, or "pressure times volume equals the number of molecules times a constant times temperature". So, all else being the same, as the temperature goes up, the pressure goes up in an exact ratio. However, all else does not have to be the same. So, for instance, if you reduce the number of molecules in a container ($n$), the ...

1

Of course, they are relate to each other but that doesn't mean they are the same things. Temperature is the average kinetic energy of the molecules while pressure is the force they exert perpendicularly on any surface. Of course, more the temperature, more would be the pressure. While the former is related to the energy, the later is related to the ...

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