Tag Info

0

The significance is that it is the work done by the fluid. Moreover, $$W=\int_{V_i}^{V_f} P\,dV$$ where $W$ denotes the work done by the system during the whole of the reversible process. This relation appears regularly in thermodynamics, usually as the first law of thermodynamics $$dU=\delta Q+\delta W=\delta Q-pdV$$ (the minus sign represents ...

0

In the conical container, the downforce is certainly the same. It can be found via calculus, taking into account the following: Since the inside surface of the container is touching the fluid, the calculation is a surface integral. The force on a surface due to a pressure is exerted perpendicular to the surface. The downward component will have to ...

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 ...

0

Flow always flows from high total pressure to low total pressure. And if its gets obstruction at any point then static pressure at that point gets high

0

As OP wished, assuming just ideal gas, we have for N ideal gas particles $$H(\{r_i\},\{p_i\}) = \sum_i^N (mgz_i + \frac{p_i^2}{2m})$$ We get a Boltzmann factor $$P(\{r_i\},\{p_i\}) = \Pi_i^N {e}^{-mgz_i/k_BT - \frac{p_i^2}{2m}/k_BT}$$ We can calculate the density $$\rho(r) = \frac{< \sum^N_i \delta(r_i-r) >}{< 1 >}$$, where the notation ...

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 ...

0

As mentioned in one of the comments, the gas particles do accumulate at the bottom of the cylinder to some extent. This is essentially why the atmosphere becomes "thinner" as the altitude increases. The exact circumstances of how the gas would behave are also dependent on the particles' kinetic energy and the thermal environment. If the thermally conductive ...

1

$p=\frac{1}{3}\rho$ is the well-known equation of state of a photon gas. It may be derived by looking at the ultra-relativistic limit of the energy momentum tensor for a bunch of particles.$^1$ $p=-\rho$ follows from the fact that the energy momentum tensor of $\Lambda$-style dark energy is proportional to the metric. Thus, at a point and in the proper ...

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 ... 0 Air pressure is equal to the weight of the air column per unit area above you. However, air pressure is not like a stack of blocks above you. Like water, air is a fluid, and any local pressure differences will cause rapid movement of the air to equalize the pressure. That means that air pressure isn't directional; the air will be pushed into any cavities ... 0 In all the gas laws, it was assumed that the vector sum of momentum of all those molecules is zero.That is a pretty strong assumption in that case. But here obviously, its not zero. Due to this reason, the force on 'The walls of a container' per unit area, depends on the 'wall' you are choosing, which also depends on the alignment of initial velocity vector ... -1 Please see Ideal Gas Law You can use relation PV=Nk_{B}T where k_{B} is Boltzmann constant=1.38*10^-23 J/K For volume of 1m^3 and T=300K you will get P=k_{B}*300=4*10^-21 Pa. For comparison, ultra high vacuum is defined as pressure lower than 10^-7 Pa. 1 On the phase diagram of methane, you can see that at RT (20°C), methane can only be gas (or super-critical if pressure is enough). The pressure can be calculated with the ideal gas equation\frac{pV}{T}=nR$You need to calculate quantity of n (in moles, given the volume, density of liquid methane, and the weight of the molecule). R is a constant, V is ... 1 According to these lecture notes, the Coriolis parameter at mid-latitudes is on the order of$f_c = 1\times10^-4 \text{s}^{-1}$and this needs to be multiplied by a wind speed to get a force. This is the first important note -- Coriolis forces do not create wind/motion, they merely change the direction of it. For a pressure force, let's look at a ... 1 Background Let us assume we have a function,$f_{s}(\mathbf{x},\mathbf{v},t)$, which defines the number of particles of species$s$in the following way: $$dN = f_{s}\left( \mathbf{x}, \mathbf{v}, t \right) \ d^{3}x \ d^{3}v$$ which tells us that$f_{s}(\mathbf{x},\mathbf{v},t)$is the particle distribution function of species$s$that defines a ... 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 ... 4 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. 0 Credits given to all answers posted. They helped me figured this out. Thanks a lot. Temperature is heavily linked with Kinetic Energy. Pressure is heavily linked with number of Collisions per Time AND Kinetic Energy. Example: A gas is hot when the molecules posses high Kinetic Energy and collides with the measuring device with great force. A gas is ... 0 A gas is hot when the molecules collided with your measuring device. Not quite. Gas heats your measuring device when the collisions are mostly such that the colliding gas molecule has more kinetic energy than the colliding measuring device molecule. It's instructive to think colliding molecules as sumo wrestlers: The molecule which has more ... 1 When a liquid or solid evaporates, it turns into a gas. In a closed container, pressure builds as gas accumulates. There are two competing processes. In the solid or liquid, the higher energy atoms at the surface fly off. In the gas, the slower atoms stick to the surface and condense. The number of atoms available to condense is proportional to the gas ... 0 To measure somethings means to compare it with an etalon or a measurement instrument, made by the help of an etalon (or the combination of etalons). To measure the pressure of a gas inside a volume one take for example a barometer and measures the pressure difference to the outer room. The measured pressure inside the volume is the result of the hitting of ... 0 Pressure is a measure of force per unit area exerted on the 'measuring device', while the temperature is a measure of kinetic energy of the individual molecules of the gas. Thus, high pressure can arise when there are either many slow moving molecules with low kinetic energy colliding with the container, or a few fast moving molecules colliding with the ... 0 An example of a difference where the pressure of a reasonably dilute gas depends on something else other than the kinetic energy of the particles is actually just the air on Earth. A classic exercise in statistical mechanics is to consider an ideal gas subject to gravity and find how the pressure varies with altitude. Of course, in reality the temperature ... 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 ... 1 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 ... 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 $$\frac{dF}{dS}= \alpha \neq 0 = d\left(\frac{F}{S}\right) = ... 2 When the cup is turned upside down, the water wants to fall out. The air-filled cavity is therefore stretched a bit as the gravity pulls down the water. The air expands a bit. This reduces the air pressure inside the cup, since increasing volume reduces pressure. This is hinted in the ideal gas equation:$$pV=nRT$$Soon this lower pressure pulls upwards ... 1 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 ... 1 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 ... 0 Parameters The surface gravity of Mars is ~0.376 g, where g ~ 9.81$m/s^{2}$for Earth. The surface pressure of the atmosphere on Mars is ~0.636 kPa, which is roughly 0.63% of Earth's atmospheric pressure (i.e., ~101.325 kPa at sea level). The density of air at STP on Earth is ~1.2$kg/m^{3}$, compared to Mars at ~0.020$kg/m^{3}$. Background Typical ... 0 Air pressure is the force that the air exerts on the container per unit area. You have to think about how this force on the container is being exerted. This force is being exerted through the air molecules bouncing off of the container walls, both inside and out. The pressure on the walls of the container is then going to depend on the rate at which the ... 0 The trapped air is compressed - it was squashed in there by air pressure in the first place. When you closed the lid, you kept it in its compressed state. Think about those films you see of them squeezing people into the Tokyo Underground carriages. When the doors slide closed, the people are still squashed, aren't they? 1 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 ... 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 ...

3

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 ...

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 ...

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.

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 ...

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.

0

Just like when you "enlarge" an accordion the air rush inside. In both cases, muscles pull the air bag, in order to suck air.

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, ...

36

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) ...

1

The secret is to evacuate the heat, mainly by radiation. But for this you need dust or "metals", since H and He alone radiates very unefficiently. Paradoxically it is not so easy to collapse completely enough. ( BTW for dark mater there is no possible radiation to dissipate energy, which keeps it fuzzy and a lot less concentrated than ordinary mater.)

13

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 ...

0

First thing, the units of 'atm' - atmosphere is an absolute measure - not a relative or gauge pressure. So while you can have fractions of an atmosphere (a vacuum), the lowest pressure one can achieve is zero atmospheres - a 'hard' vacuum. Secondly when dealing with balloons, the pressure-size relation is nonlinear and elasticity actually decreases as they ...

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 ...

Top 50 recent answers are included