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What Is The Weight Of Air In A Tyre?

predator
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predator Grand Master

predator

Posted
Posted

This is maybe an interesting one.. :sorcerer:

What is the formula for calculating the weight for the volume of air contained in say an 'average' 17"x 8" rim with say an 255/35R17 tyre compressed at 28psi?

Anybody?

ok, we got pV = nRT

where

p is the absolute pressure [Pa],

V is the volume [m3] of the vessel containing n\, moles of gas,

n is the amount of substance of gas [mol],

R is the gas constant [8.314 472 m3·Pa·K−1·mol−1],

T is the temperature in kelvin [K].

Now to work it out.. sau is my 'paper'

1psi = 6,894.76 pascals

p = 193,053.28 Pa @ 28psi

T = 298.15K @25degC

TBC

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Rekin Grand Master

Rekin

Posted
Posted

ah crud this brings painful memories of Foundation Chemistry in Uni >.<

If i studied more for that subject i could have helped you haha.

All i can tell you is;

Oxygen Standard atomic weight: 15.9994(3) g·mol−1 (grams/mol)

I think you have all the values now, just use teh formula?

adam-__- Grand Master

adam-__-

Posted
Posted (edited)

Hey, to use that formula, you need your pressure to be in kPa. Not just Pa. And V needs to be in litres.

Cheers,

Adam

Year 12 Chemistry exam in two weeks time :thumbsup:

Edited by adam-__-
predator Grand Master

predator

Posted
  • Author
Posted

I reckon you could be on the mark, I still haven't got around to working it out properly, but I'm figuring it's not too much..

as air in atmosphere = 14.5psi.. .air in tyre = 28psi = 1.93 ratio x some volume equations of air in a tyre and there you go.

Just curious as to what a rim with a tyre and air weighs, compared to a bare rim weight. Silly I know.

simpletool Mentor

simpletool

Posted
Posted

OK yeah, it's PV = nRT.

If you use:

P = Kpa

then V = L

if you use:

P = pa

then V = cubic metres (1000L)

anyway....now that we can all sleep.......

I got 164 grams for a 235/45/17.

Assuming that "air" is 70% nitrogen and 30% oxygen.

So 1 mole of air = 28 x 0.7 + 32 x 0.3 = 29.2 g/mol

And I got the volume as 46L, at 300 kelvin and 3 atmospheres. Or ~30psi (relative)

So that gave me number of moles, n = (304 Kpa x 46L)/(8.314 x 300) = 5.6 moles.

then 5.6 moles x 29.2 g/mole = 163.5 grams

daisu Mentor

daisu

Posted
Posted

I was going to have a go at you about 30 psi being closer to 2 atmospheres, but then i remembered that 30psi tyre pressure is compared to atmospheric pressure (already 14.7psi), so yeah approx 30psi(gauge) would be about 3 atmosphere.

What I'm curious about though, is how did you calculate the volume of air?

adam-__- Grand Master

adam-__-

Posted
Posted
I was going to have a go at you about 30 psi being closer to 2 atmospheres, but then i remembered that 30psi tyre pressure is compared to atmospheric pressure (already 14.7psi), so yeah approx 30psi(gauge) would be about 3 atmosphere.

What I'm curious about though, is how did you calculate the volume of air?

The way I did it was, I got the area of the entire front of tyre and wheel (side facing the road), subtracted area of rims, multiplied the area calculated by width of tyre and then convert cm^3 into litres.

Hope that made sense. :)

Cheers,

Adam

51jay Mentor

51jay

Posted
Posted
The way I did it was, I got the area of the entire front of tyre and wheel (side facing the road), subtracted area of rims, multiplied the area calculated by width of tyre and then convert cm^3 into litres.

Hope that made sense. :thumbsup:

Cheers,

Adam

Easy way...... pop wheel with deflated tyre on bathroom scales, note reading

inflate tyre to desired pressure, place on scales, note reading

Subtract ist reading from 2nd

I didn't do chemistry :rolleyes:

djr81 Grand Master

djr81

Posted
Posted

God I hate chemistry.

Anyway, density of air at, say atmospheric pressure & 50 degrees C is 1.09kg/m3.

So three bar will obvisouly be three times this.

Work out your tyre volume & you have an answer.

By the way the wheel and tyre will be on the heavier side of 20kg....

wonkynutz Experienced

wonkynutz

Posted
Posted
Easy way...... pop wheel with deflated tyre on bathroom scales, note reading

inflate tyre to desired pressure, place on scales, note reading

Subtract ist reading from 2nd

I didn't do chemistry :)

Close, but no cigar. The "deflated" tyre will still contain air at atmospheric pressure. The correct way of doing it is:

1/ Weigh tyre.

2/ Weigh wheel.

3/ Add 1 and 2 to get weight of tyre and wheel combo.

4/ Fit tyre to wheel. Inflate tyre to desired pressure. Weigh.

5/ Subtract 4 from 3 to get weight of air in tyre.

5/ Smoke cigar. Get lung cancer.

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