Volume VS Pressure

Adding 1 liter gives an absolute pressure of 2 bar (approx), so the differential (i.e. gauge) pressure is 1 bar.

Another liter would give 3 bar abs., 2 bar gauge - and so on.

so thinking say sea level "Air"

ambient rounded off 14 Psi

2 liters in the container = 14 psi differential

3 liters in the container = 28 psi

4 liters in the container = 42 psi

5 liters in = 56 psi

etc..

Correct?

Yes.

That's only true when the temperature stays constant.

So in reality its a bit more complicated.

Yes.

but close enough

Depends. Look up the Ideal Gas Law to understand more.

The formula is PV = nRT

Note that T (temperature) is directly related to pressure and volume (PV), so temperature is a real factor to consider.

Really?

So say 10 liters in the 1 liter container, roughly 130 Psi or so

Now if it is say 0 deg f vs 70 f when filled whats the pressure now?

Well done.

Thanks. I finally started getting my notices again, after somehow they went away two weeks ago.

A GA from me also.

I have no idea whether it is correct or not but it's worth a GA for effort alone.

I would almost vote it down because now James did the homework!

Not sure, but he did a very good job demonstrating how to use PV = nRT and what it means.

He sure did!

Oh yes - required course for EE's back in my days at PSU - Mech Eng 101 - thermodynamics and the gas laws. Looks familiar even after 46 years.

That seems a complicated way to calculate it, but more important, IMHO it's not the right answer! Assuming (as your post) that it behaves as an ideal gas, if pressure = P₁ at absolute temperature T₁, at temperature T₂, pressure P₂ = P₁*T₂/T₁. 70°F = 530°R, 0°F = 460°R so T₂/T₁ = 0.87.

Following the OP in not being too worried about great accuracy, if pressure = 130 psia at 70°F, at 0°F it is 130*0.87 = 113 psia = 98 psig.

If pressure = 130 psig at 70°F, at 0°F it is (130 + 14.7)*0.87 = 126 psia = 111 psig.

It took me a while to spot it, but where you went wrong is you used T₂ = 273.15K. But that's 0°C, it should 0°F, as you had it earlier.

I humbly stand corrected. I did use 273.15 as final temperature on the last calculation.

Can't you work it out from the formula in #7? (BTW T is absolute temperature)

What is changing in the thought experiment presented is not V, it is n. V is held constant while n is doubled/increased. It's that whole "standard cubic foot is a measure of mass/moles, not volume" business, but with liters instead of cubic feet.

A slightly different angle on this.

Pressure measured at the container surface is a result of molecules of gas colliding and bouncing off the internal wall. At a temperature above absolute zero gas molecules are moving around at considerable speed.

Although there are an astronomical number of molecules in a litre container of gas ( look up Avagadro's number ) most of the space is still empty for our purposes*

Doubling the number of molecules simply doubles the number of molecules hitting the inner container wall over any given time period so doubles the pressure

Although the "rules" here are statistical the huge number of molecules of gas means that predictions are sound.

There are simplifications here - for instance a gas will resist the introduction of more molecules into the container ( hence the need for a compressor or pump ) so we are ignoring the forces between molecules but for an ideal gas the picture gives a reasonably good insight of whats going on.

*No offtopic quantum counter debate please!

I like this explanation. I was told at school by my science teacher that pressure is due to the weight of the column of air above us. I argued that if a ( non compressible ) box full of air was taken up Everest it would still have 1Bar inside. Similarly if it was taken down a deep mine etc. However this explanation does account for pressure difference at altitude but i can't figure out how to do this with your explanation. Can you enlighten me further?

Jim

The problem is essentially equivalent to taking a volume of 2 liters (or 3 or 4 or ..N*V₁, where V₁ is the final volume of 1 liter and N is whatever multiple of that volume one wants to compare) and compressing it down to 1 liter while keeping the temperature constant (or waiting until the temperature reaches the initial temperature). The solution can easily be obtained by the use of Boyle's law.

According to Boyle's law PV = c where c is a constant. So for the problem at hand then

P₀V₀ = P₁V₁

Where P₀ = initial pressure, V₀ = 2 V₁ and V₁ = 1 liter

Solve for P₁:

P₁ =P₀V₀ / V₁ = P₀ (2*V₁ )/ V₁ = 2*P₀

or more generally to fit N liters of air into 1 liter:

P₁ =P₀V₀ / V₁ = P₀ (N*V₁ )/ V₁ = N*P₀

So the pressure required to hold N liters of ambient air in a 1 liter volume would be:

P₁ = N*P_ambient

very informative thx all.

Question. Is the use of the word ambient appropriate? It is the atomic weight of the injected air that is an essential part of the equation, it seems. In another words, in order to inject the second volume of air in the container, it will have to be pressurized in a compressor, and will by definition no longer be ambient. It will also be heated by that pressurization, so will need to be cooled before injection in order not to impact the temp constant in the container. So here is the question. Does the act of injecting a volume of 15 degree C 2bar air into a volume of 15 degree C 1 bar in the container raise the temperature? Does the equation account for that?

IN the isothermal limit it does not (obviously), however, in the adiabatic limit it surely does, since the air cannot emit heat out of an adiabatic bottle.

Consider a cylinder with piston of volume 3V, a valve, a bottle of volume V. The system is considered to be adiabatic, and consists of all the part. The initial state of the system: 2bar in cylinder (on the left, say), and 1 bar in the bottle (on the right). Total system at is at 15C. When valve is opened at time t=0, gas from cylinder enters the bottle, and as long as there is only gas of the same composition, we ignore back diffusion. We also have both volumes sufficiently large that each side contains one mole of gas. (22.4 L on the right), (11.2 L on the left). At the final state of such a system, there will be 33.6 L of gas, n= 2, P{2/1}, T>15C, but are we sure? Note that piston is not allowed to move at all after opening of valve. As the gas in the cylinder expands to occupy three times its original volume it has now apparently 2/3 atmospheres, and the gas that was in the bottle now has expanded from 1 atm to 1.5 times its original volume (as if the other container were at vacuum). The bottle would contribute 1/1.5 atm, so that the combination of the two gives 4/3 atm. Neither air mass is heated, since one absorbs compression work, and the other one releases, within the reversible limit.

I am so sleepy after typing this, I will have to take a walk.

James thanks have a nap.

Pressure and volume are inversely proportional to each other. This means that as the pressure decreases, the volume increases, and as the pressure increases, the volume decreases. One way to think of this is if you push on a gas by decreasing its volume, it pushes back by increasing its pressure.

Pressure and volume have an inverse relationship when n and T are constant. How we imagine the pressure in the balloon is increased? Either n or T must increase, or V must decrease.

Additionally, balloons are roughly constant-pressure systems. The rubber membrane is a very weak elastic, so the internal pressure of the balloon is at almost constant pressure, just above atmospheric. When you squeeze a balloon, you usually don't change the pressure or volume much, because the rubber just expand in an area where you aren't pressing.

If it is constant pressure, how come there is a super-sonic shock wave (pop) when you puncture the ballon that is inflated? Granted the pressure is not something like 10 bar, but it is more than 1 bar.

The maximum that can be exerted by my lungs at STP is about 0.25 Bar. Obviously this is enough to get that effect, imagine the noise from an exploding hot water bottle or pressure vessel.

Jim

From TORNADO on CR4

Thank you! Cheers!