How will accelerating a container with an ideal gas in it affect the conditions? My initial thoughts are that the ideal gas will collect at the opposite end of acceleration, which means that the volume drops. According to the formula $PV = nRT$, either the pressure or the temperature would have to raise as well. Firstly, which quantity would rise if I just squeezed a gas in normal conditions? I don't see how I can differentiate between squeezing isothermally and squeezing isobarically, so would they both rise?  Upon further thought however, we are already experiencing 9.8 metres per second squared acceleration, and that has minimal effect at low heights. So does this mean at low accelerations the afore-mentioned effect can be ignored?
121866
I was thinking about a situation where some gas is enclosed inside a container and kept in a train at rest. The train accelerates, gains a maximum speed and then suddenly stops. Would the temperature of the gas change during this process?  I know that temperature is directly proportional to average kinetic energy of gas molecules but would the train during this process change the kinetic energy of the gas molecules inside the closed container?
21470
I am wondering, air makes sound propagate. So in a vacuum there is no sound, but what is the relation of pressure to sound volume? Is it linear?  If I have a source of sound and let's say I am in a room with an air pressure half of the pressure outside, would the sound be 0.5 lighter? Or does air pressure have no impact on the sound volume?  Is there a comprehensible formula for that?
48508
When you invert a water-bottle in a container, the water rises and then stops at a particular level --- as soon as it touches the hole of the inverted bottle. This will happen no matter how long your water-bottle is. I understand this happens, because once the water level touches the hole, air from outside cant go inside and therefore there is nothing to displace the water that falls out of the container.  Now according to the laws of pressure ---- the pressure at the water level must be same everywhere --- whether it's inside the water bottle or outside. And that must be equal to the atmospheric pressure. Therefore the pressure of the water column + air column inside the inverted bottle must be equal to the atmospheric pressure.  What I dont understand is, no matter how long a bottle you take, the water level will always stop at the hole. So that means that no matter how long a bottle you take, the pressure of the water column + air column inside the water bottle will be equal to the atmospheric pressure. How could this be possible?  Also I'd like to let you know that, if you pierce the upper part of the bottle with a small pin, then the water level rises and overflows out of the container. I'm assuming air from outside rushes in and pushes the water out.
115376
If we had an airtight container that was large enough, say extended out to the Karman line, would there still be a pressure gradient? If we assume that the pressure at the bottom was $1\ \mathrm{atm}$, would it have the same exact pressure gradient at 6 miles up on the outside? This is assuming that the gravity is still $9.8\ \mathrm{m/s^2}$.
73561
On John Baez's website, , he discusses the problem of how entropy increases when a cloud of ideal gas collapses gravitationally (no black holes - keeping it simple).  But he doesn't give away the answer - instead he implies that, I presume, either gravitons or photons are released during collapse and these increase the overall system entropy.  Can someone please answer which it is, gravitons or photons?  If gravitons (or gravity waves), then does this mean that pre-20th century physicists, with no notion of gravity waves, would have concluded that gravitational attraction violates the 2nd law?  And if photons, then what about if the gas particles were totally neutral - wouldn't they not emit any EM radiation in that case?  Or am I wrong that gravitons or photons are the answer? Thanks!
63710
How is negative pressure created in a fluid system?Isn't it counter intuitive that we are reducing the pressure of a system containing no molecules(zero pressure) to a lower value?
72052
I understand that the average energy of each degree of freedom in a thermodynamic system is $\frac12kT$. And so, for an ideal monatomic gas, there are three degrees of freedom associated with the translational components of each atom, which gives $E$ = $\frac32kT$ for each atom in this system.  For an ideal diatomic gas, there are the three degrees of freedom from the translational components plus two more degrees of freedom associated with the rotational components of each molecule. So we end up with $E$ = $\frac52kT$ for each molecule.  My question here is why did we ignore the rotational degree(s) of freedom in case of a monatomic gas ? Like why isn't there another degree of freedom for the rotational component of a single atom ? Also, if that is neglected for some reason, then when does it become significant and can no longer be ignored ? For example, shouldn't it be included in calculations under extreme conditions like temperature of millions of degrees ?
50252
Consider an adiabatic system as follows. It consists of a gas in a container and a piston.  Initially, the system is at equilibrium and the gas inside it occupies a volume $V_i$ at a pressure $p_i$ which is equal to the outside pressure. Suddenly, the outside pressure changes and reduces to $p_{atm}$. The piston moves to equalize the pressure and the gas expands isothermally to obtain equilibrium. The gas now occupies a volume $V_f$ at a pressure of $p_f$ which is equal to $p_{atm}$  Now, two textbooks I have define the work done from two different viewpoints.  1: From the viewpoints of the surroundings:   The work done on the system by the surroundings equals $-p_{atm}\Delta V$. Since $p_{atm}$ is pretty much constant for the whole of the whole of the process, we can say that the work done equals:  $$W_1 = -p_{atm}(V_f - V_i)\tag1$$  2: From the viewpoint of the system:  We can write the internal pressure of the system as a function of its volume: $p_{in}(V)$. As during the expansion, the internal pressure changes, the work done by the system equals   $$W_2 = \int_{V_i}^{V_f}p_{in}(V)dV$$  Now, I don't know which definition to use. The work is done by the system (from definition 2) is done on the surroundings. But what about the negative work that the surroundings did on the system? Where did that energy go? Maybe, the two definitions express the same thing: the work done by the system. The negative sign in definition 1 signifies that the work is done by the system. But that would mean that   $$W_1 = W_2$$  We can simplify $W_2$ as follows. Clearly,   $$p_{in}(V) = \frac{p_iV_i}{V}$$  $$W_2 = \int_{V_i}^{V_f}\frac{p_iV_i}{V}dV = p_iV_i \ln{\frac{V_f}{V_i}} \tag2$$  The internal pressure when the volume is equal to $V_f$ is $p_{atm}$  $$\implies p_{atm} = p_{in}(V_f) = \frac{p_iV_i}{V_f}$$  $$\implies V_f = \frac{p_iV_i}{p_{atm}}$$  Putting this into  $(1), (2)$ gives us,   $$W_2 = p_iV_i\ln{\frac{p_{i}}{p_{atm}}}$$  $$W_1 = -p_{atm}\left(\frac{p_iV_i}{p_{atm}} - V_i\right) = -V_i(p_{atm} - p_i)$$  Consider some values. Let $p_i = 5 \ \rm{Pa}, V_i = 1 \ \rm{m^3}, p_{atm} = 1 \ \rm{Pa}$.  $$W_1 = -1\cdot(1-5) = 4$$  $$W_2 = 1\cdot1\cdot\ln{\frac{5}{1}} = 1.609$$  Where am I making a mistake?
62459
There is an experiment we learned from high school that demonstrated how atmosphere pressure worked.  Fill a cup of water and put a cardboard on top of it, then turn it upside-down, the water will not fall out. The explanation said this was because the atmosphere pressure was greater than the water pressure, which holds the water up.  I believed this explanation once, until I found some points that confused me:  1. Is the water pressure in the cup really smaller than the atmosphere pressure?  This is what we have been taught through our life. However, consider an object in the water under sea level, it experiences the pressure from water plus atmosphere pressure. So the water under sea level must be greater than the atmosphere pressure. Even it is contained in a cup, the pressure wouldn't change. Is this True?  I read . If the cup is fully filled, the compressibility of water is much greater than that of the air, also the surface tension of the water keeps the air out of the cup. So the water is held in the cup. This explains the problem. But I still want to ask if the water pressure smaller or greater than the atmosphere pressure in this situation.  2. When the cup is half filled with water, why it still holds?  I saw most of articles or opinions against this. They all agree that the water would not fall only if the cup is entirely filled. But I did the experiment myself, the water stayed still in the cup even it is not fully filled. Actually even with little amount of water, as long as it covers the open of the cup and the cardboard on it, the water stays in the cup.  Even the compressibility of water is much smaller, the air inside the cup provide enough compressibility, how come it still holds?
82508
At higher temperatures (for an ideal gas), the Maxwell-Boltzmann distribution is spread more widely and has a lower maximum. At lower temperatures, the spread is much more narrow and the peak is much higher. Why is this so? I was thinking that this is to maintain constant area under the curves, but the integration of the curve gives total energy—and total energy should not be constant since temperature is increasing.  So why does the distribution curve change thus? I’m looking for an explanation that’s intuitive or that involves math not surpassing first-year university level calculus. I would appreciate it if it could be explained without the Maxwell-Boltzmann distribution equation, since at this level I’m really treating this qualitatively.
92896
I was wondering about this:  If there is a pressurized container, like a tank of compressed air at some pressure that is greater than the ambient air pressure, and this tank of air has a hole in it, what is the velocity of the escaping air through the hole?  Is there a formula for this?
84991
A particle experiences an acceleration described by  $$ a=kx^{-2} $$ where x is the displacement from the origin and k is an arbitrary constant.  To what value does the velocity v of the particle converge to as x approaches infinity if the particle starts at some point x0?  If I approach this problem with energy, then $$ W = \int F \mathrm{d}x $$ $$  = \int_{x_0}^\infty mkx^{-2} \mathrm{d}x $$ $$  = mk(-\infty^{-1}+{x_0}^{-1}) $$ $$ W = K = mk{x_0}^{-1} $$ $$ \frac{1}{2}mv^2 = mk{x_0}^{-1} $$ $$ v = \sqrt{2k{x_0}^{-1}} $$  How would I solve this problem with pure kinematics? (there appears to be some sort of cyclical dependency where acceleration affects velocity, velocity affects displacement, and displacement affects acceleration)  Likewise, two particles experience accelerations described by $$ a_1=k_1x^{-2} $$ $$and $$ $$ a_2=-k_2x^{-2} $$ where x is the distance between the two particles  What two velocities do the particles reach as x approaches infinity if the two particles are initially separated by some x0?
93276
I was thinking that since atmospheric pressure is 760mmHg what would happen if I shrink a mercury barometer until it's shorter that 760mm in height. What would happen? Would the barometer retain it's ratio of mercury height? Will the mercury fully fill the barometer? Is 760mmHg a constant and not affected by the size of the barometer (or diameter)? Thanks in advance.
49968
For an ideal gas, why is the specific heat capacity at constant volume lower than the specific heat capacity at constant pressure?
35406
I understand that air friction cools off an object at low speeds. For example, if you blow on a spoon of hot soup, it cools off. Or if you swing a hot frying pan in the air, it cools off faster.  But at higher speeds, the situation changes in the opposite. For example, consider a meteor falling on Earth. It is so fast that it heats up to such a high temperature that it burns into ashes.  What is the critical speed in which an object starts to heat up?  For example, consider a spherical object with radius of 1 meter. Let its density be 1 g/cm3 if needed. And let the air pressure be 1 atm. Assume that the temperature of the object is 400 K and the temperature of the air is 300 K. Also assume that the specific heat constant of the material is 1 cal/(gr.K). What is the critical speed for this object after which its temperature starts to rise above 500 K?
1711
When I was in high school, my teacher did an experiment to show the power of atmospheric pressure.  Experiment: Prepare a glass bottle, fill with water, put a glass plate on the bottle, make sure there is no air in the bottle, just water. Hold the glass bottle and plate, and invert them. Slowly, release the hand which hold the plate. Because the atmospheric pressure, the glass plate will not fall down.    Question 1: As we know, if we didn't put water into the glass bottle, just air in the glass bottle the glass plate will fall down. So, If we use other liquid instead of water, this liquid has smaller density than water, like alcohol, the plate will not fall down. But, if we continue to chose the smaller density liquid until the liquid has density like air the plate will fall down. So, if there is a threshold density of liquid make the plate between falling  down and not falling down?  Question 2: If we put the bottle and plate into water slowly, when the plate and part of bottle are in the water but part of the bottle still above the water, will the plate fall down?  Question 3: Continuing question 2, when both bottle and plate are in the water I know the plate will fall down. So, how does the pressure of water and atmosphere change?  Is there a good description of this phenomena?  Why does the trick continue to work with lower density fluids and why does placing the plate in the water cause it to fall?
14344
We know, the rockets in space use Newton's 3rd law to increase their velocity and hence move. What I don't understand is how it is possible in space aka vacuum-state without air? From what I know, Joule's "Free Expansion of Gas" says that free-expansion compresses the gas and is therefore "affected" by vacuum so it can't make the rocket move as the gas will have zero press/force. Could someone please explain me how rockets do really work and the above-mentioned statement?   Actually, please have a look at this site:   Not: The site appears to include some conspiracy theory thingummies, but made me wonder anyway.