50

When you pour the hot water in, the air inside the thermos is still quite cold (ambient temperature, approx.) But then when you shake it up the cold air is heated by the hot liquid. Gases expand considerably when heated, approximately acc. the Ideal Gas Law: $$pV=nRT$$ This causes a modest (and harmless) pressure increase in the flask, which is what you ...


39

There is another effect here which is significant, as follows. Warm water wants to evaporate, but in a flask-shaped container, the evaporation can take place only at the free surface of the water in the flask. Furthermore, as soon as the boundary layer of air right next to the warm water becomes saturated with vapor, the diffusion of water vapor into the air ...


14

The analogy is good, but you appear to misunderstand it. Water pressure is not a function of the surface area of the piping, any more so than voltage is a function of the surface area of the wiring. The water pressure is a function of the strength of the pump which is charging the pipework (assuming the quantity of water available to charge the pipework in ...


7

@Steve's answer is really very very good, and I can't improve upon it. But I want to add my thoughts anyway. I find this analogy very useful, not just for helping understand why the currents in various legs of a circuit are what they are, but also for building intuition about what voltage actually is and where it comes from. Water under high pressure really ...


6

You can see it like this: if you consider the universe as a whole, "of course" each process is "spontaneous" in the sense that the total entropy of the universe entropy always increases. After all, if something happens somewhere in your universe, then it must be allowed to happen: otherwise, it just would not happen! However, what you ...


5

You have assumed that the fluid beneath the puck is exactly supporting the weight of the puck and the fluid above the puck, so of course you will find the puck to be supported. You have assumed an incompressible fluid, so in this case the puck will not move, because if it did the fluid would have to compress as it has nowhere to go.


4

Converting kilograms to pounds, even though useful in a grocery store, is not a valid operation from point of view of dimensions. One is mass, the other is force. Newtons to pounds will be ok and give you the right answer.


4

Unless you're pulling hard enough to cause cavitation, which I rather doubt, he's wrong. The primary reason for propulsion is good old action-reaction. You push water backwards; momentum conservation pushes your body forwards. To the extent that water "rushes in" to the displaced area in front of you, that has little to no effect. See the famous ...


4

The pressure reduces in going through a porous plug because of friction. So you need more pressure on the upstream part than on the downstream part. The work done $$ W=P_{\rm upstream} V_{\rm upstream} - P_{\rm downstream} V_{\rm downstream}$$ on forcing a gas through the plug goes into the internal energy, so, as no heat enters or leaves the system, the ...


4

Let's assume that the total amount of water is so large that submerging the sphere raises the overall water height by only a negligible amount. Completely submerging a weightless sphere of volume $V$ is then equivalent to elevating a mass of water $m=\rho V$, where $\rho$ is the water density, by a distance $r$, where $r$ is the sphere radius. The reason is ...


3

There may be constant area before and after the plug, but, after the plug, the velocity of the gas is higher than before the plug. So the mass flow rates before and after the plug are the same. What happens in the plug is that there is a pressure drop. This is caused by irreversible viscous shear between the plug and the flowing gas (just like the pressure ...


3

As I can make out from your questions you are assuming that water cannot pass from the sides of the puck. In reality there are 2 cases possible. If we assume that the gap between the puck and tube is water proof then your derivation is correct. All such pucks will remain in such position. To think intuitively: imagine you somehow lift the puck by a small ...


3

Dimensions of Pascal Pascal is the SI unit for pressure, which is force per unit area. In SI units, this corresponds to $N/m^2$. To get to SI base units, note that Newton's 2nd law can be used. Thus, the following are the base SI units for a Pascal: $Pa = \frac{N}{m^2} = \frac{ma}{L^2} = \frac{kg-m/s^2}{m^2} = \frac{kg}{m-s^2}$ Dimensions of psi psi is ...


3

Convert Pa in $\rm{kg\over {m\sec^2}}$ to $\rm{lbm \over {ft\sec^2}}$ then use $\rm{1\ lbf = 32.174\ {{lbm\ ft} \over sec^2}}$ to obtain pressure in $\rm{lbf \over ft^2}$, then convert to $\rm{lbf \over in^2}$. Note: in English units $\rm{1\ lbf = 32.174 \ {{lbm\ ft} \over sec^2}}$.


3

In an ideal case in which particles don't interact, or in the limit of dilute gas, yes. Because the bottle contains no $O_2$ then the gas is more likely to enter the bottle than to leave it (because... there is no $O_2$ inside) so you have a net diffusive flow inside the bottle. That is, until the bottle reaches a (partial) pressure so high that the outgoing ...


3

Imagine we take the submerged (or floating) object away and fill the volume $V$ that it displaced with liquid instead. If the buoyancy force were greater than the weight of the volume $V$ of liquid then that volume of liquid would rise. If the buoyancy force were less than the weight of the volume $V$ of liquid then that volume of liquid would sink. In order ...


3

The thermal coefficient of expansion of a substance is the proportional increase in volume for a $1$ Kelvin rise in temperature. For an ideal gas we know that $PV = nRT$ so if pressure $P$ is constant we can express volume $V$ as a function of temperature $T$: $\displaystyle V(T) = \frac {nR} P T$ So near $T=300$, for example, we have $\displaystyle V(300) = ...


3

The two definitions you cite do not mean that non-spontaneous processes decrease entropy, if that is what you are implying. In other words, the fact that all spontaneous (natural) processes increase entropy does not mean all non-spontaneous processes should decrease entropy. The refrigeration/heat pump cycle is a non-spontaneous transfer of heat from cold to ...


3

It depends what exactly you mean by 'space'. In our solar system (interplanetary medium) the particle density is about $10-100/cm^3$ so the atoms/molecules would be a few millimeters apart. In interstellar space it is similar but the density can be less or more depending on where you are (see https://en.wikipedia.org/wiki/Interstellar_medium#...


2

He cannot be a liquid at room temperature. You would have to have very cold liquid He to start with then let the temperature increase and evaporate the He in the container. I assume the container is not initially completely filled with He. (If completely filled, as the liquid He heats up it could rupture the container, since the density decreases with ...


2

The other answers are right--this is just to explicitly point out the reason that it feels worse when 30 people stand on your hand than when your body is subjected to an equivalent amount of hydrostatic pressure. The 30 people squash your hand like a ball of dough in a tortilla press, squeezing parts of it out the sides. Hydrostatic pressure, on the other ...


2

No. The atmosphere is thin enough that the variation of gravity from bottom to top is a small correction on top of larger sources of variation. Atmospheric pressure at a point can be regarded as due to the weight of the air stacked on top of that point, and though the air expands with increasing temperature, its weight doesn’t change. If temperature ...


2

It would be easier if you treat pascals as "Newtons per square meter". PSI is "Pounds per square inch", right? Well, both inches and meters are units of length, and both newtons and pounds are units of force. So, try to convert the newtons to pounds and meters to inches using a simple conversion factor. Dont lay out all the units kg m * s^...


2

Pressure has the ability to do work and change the energy of a system. From thermodynamics: \begin{equation}\tag{1} dE = T \, dS - p\, dV. \end{equation} You could also define an energy density associated to pressure (at constant entropy): \begin{equation}\tag{2} \frac{\partial E}{\partial V} = -\, p. \end{equation} Since from special relativity energy is ...


2

I appreciate that this is a little counterintuitive but the diagram doesn't lie. As you keep adding heat to the vessel (of constant volume), pressure goes up and you move through the vapour-liquid area, until get into the liquid only area. This has to be understood by means of the effect pressure has on so-called 'non-permanent' gases, which can be liquefied ...


2

My first question is whether there is any justification for this postulate or whether it is simply an empirical law? The justification is in the unpacking of what the writer means by "simple compressible". I could write the fundamental relation for a thermodynamic system as $$dU=T\,dS+\sum_i\mu_i N_i-P\,dV+\sigma\,dA+E\,dp\cdots,$$ where $U$ is ...


2

You are correct that the chemical potential $\mu=\left(\frac{\partial G}{\partial N}\right)_{T,P}$, being the conjugate thermodynamic variable to matter, is the true arbiter of whether matter will shift phases. If the chemical potential of the gas phase is lower than that of the condensed phase, for example, then evaporation is spontaneous. But "...


2

But the work done is the same As you realized, this is not true. Work done is a path dependent process. You are taking a different path and you expect the amount to work done to change. It is hard to say exactly what has happened with the amount of work done, but that is because you have not specified exactly how you made the process irreversible. For ...


2

This means that the heat transferred to get from state a to state b must be less in the irreversible case than in the reversible case. But the work done is the same. Though it depends on the details of the process, you are correct that less heat is transferred to the system in the case of the irreversible expansion process connecting the same two ...


2

Two types of processes: There are thermodynamics processes that change the state of some variables like temperature and entropy. In statistical mechanics there are microscopic processes which move particles between microstates, this change may move the system closer to the equilibrium macrostate or not. There is no such thing as a spontaneous ...


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