lets say my tire is on the car. and i inflate the tire to exactally 40psi. now, when i take the tire off of the car, wouldent the PSI go down just a bit since it has no load?

and if so, does anyone know around about how much PSI fluctuates with/without load?

 

bump

 

from what i have experienced. i would change tires on the tire machine, put about 33-35 psi in them. put them on the car drive around for a while, and when i would check the pressure couple times a month. it would be 33-35 psi.

but not sure if that answers your Q.

 

If I understand your question, the PSI should remain the same within reason. The force produced by the car on the tire causes the PSI to rise in that area, and the increased pressure (instantly) deflects the tire elsewhere so the pressure you measure is relatively unchanged. If you then apply pressure to the top, the tire will again deform somewhere else causing equal pressure. That is to say, the volume of the tire does not change. The only way you would see a significant rise in PSI is if you applied force over the whole surface of the tire (like a tire under 100ft of water), added or expanded the gas inside, or changed the volume of the tire.

The practical application of this is pretty minimal, since the force on the tire is compounded by the motion of the car at speed, not just at rest and on the jack. Also, heat will do much more to your PSI than direct pressure since heat affects the whole system.

 

Why not just check the pressure with tires on the car. Then jack up the car and check the pressure again?

 

I've checked this when swapping tires. The pressure is the same with the weight of the car on it as it is without the weight of the car.

 

Air is considered an ideal gas for most appllcations (not combustion, high velocity, etc.).

PV = mRT

Set it up for your initial and final conditions. The force doesn't change R, T (static situation), or m (closed system). The V may change a minute amount due to the different strains the tire experiences (practially is doesn't). Therefore P is constant.

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by GSpeedR &raquo;</TD></TR><TR><TD CLASS="quote">PV = mRT</TD></TR></TABLE>

I believe the constant is usually designated as "n", not "m".

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by nsxtasy &raquo;</TD></TR><TR><TD CLASS="quote">

I believe the constant is usually designated as "n", not "m".
</TD></TR></TABLE>

It's both:

pv = RT

Where p is absolute pressure, v is specific volume, and T is absolute temperature. Using v = V/m where m is the mass of the gas, the equation becomes pV = mRT

In the Molar Form:

One mole of gas is a quantity of gas whose mass in grams is numerically equal to the molecular mass (M) of the gas.

n (kmol) = m (kg)/M(kg/kmol) = m/M (kmol)

n (lb/lbmol)=m (lb)/M(lb/lbmol)=m/M (lb/lbmol)

From pV = mRT, substituting for m gives

pV = nMRT

Since n, the number of moles, is just equal to the mass m divided by the molecular weight M, you can substitute in the m/M for n in the ideal gas law and solve for M.

 

Depends where you read it or who your professor is... some use Number of moles, or just Moles. same thing either way... also seen it represented as Z in my thermo course.

 

I usually use "Pv =RT" where v is the specific volume, but I figured the V/m would be more recognizable. I despise moles in general, so I try to avoid them.