Tire pressure changing due to temperature

  • #1

Homework Statement


Before embarking on a car trip from Michigan to Florida to escape the winter cold, you inflate the tires of your car to a manufacturer-suggested pressure of 33.5 psi while the outside temperature is 26.1 °F and then make sure your valve caps are airtight. You arrive in Florida two days later, and the temperature outside is a pleasant 79.1 °F.

a) What is the new pressure in your tires, in SI units?
b) If you let air out of the tires to bring the pressure back to the recommended 33.5 psi, what percentage of the original mass of air in the tires will be released?


Homework Equations


Gay-Lussac's Law p1/T1=p2/T2

The Attempt at a Solution



Part a)
The rubber tire keeps the volume constant while varying the pressure within depending on the temperature. From this information, I deduced that I needed to use Gay-Lussac's Law to determine the new pressure.

First I converted everything to SI units.

T1 = 26.1 °F = 269.9 K
T2 = 79.1 °F = 299.31 K


p1=33.5 psi = 230974.3692 Pa ≈ 231 kPa

Solving Gay-Lussac's Law to solve for the final pressure...

p1T2/T1=p2

Finally

(230974.3692 Pa)(299.31K)/(269.9K)=p2=256142.7879Pa ≈ 256 kPa

Part b)
If we let out (256,142.7879Pa - 230,974.3692Pa =) 25,168.4187Pa to bring the pressure back to what we started, then releasing 25,168.4187 Pa of pressure decreases the tires mass by some percentage x.

Since Increasing the pressure to 256,142.7879 Pa doesn't change the mass of the air inside until we let some out, we can consider the percentage change in the pressure

[(25,168.4187 Pa)/(256,142.7879 Pa)]*100%=9.826%

9.826% of mass of air has been released.


My homework is an online assignment with 11 questions. I can submit each of my answers in two fields, one containing the numeric value, and the other contains the units(unless given to me). I can submit my answer and it will tell me if I am correct or incorrect.

I got part b correct but not part a. I tried expressing the pressure in Pascals, kiloPascals, and psi (just to check), and I tried using the exact value(calculator's limit), and 3 significant figures, I tried expressing in scientific notation, but it still says I'm incorrect. Is it not looking for the value that I found as the new pressure? I got part b correct so I would assume my answer to part a is correct also.

Any help(or insight on ideal gases in general) is appreciated.

Thanks
 

Answers and Replies

  • #2
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I think you did it right. They ask for the value in SI-units, try expressing Pascals in SI units. Not sure if it will work, but it might :-)
 
  • #3
BvU
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I agree p2 is 256 kPa. Perhaps they want to see it in Pa ?
 
  • #4
I had initially just did a google search to see what the SI unit for pressure was and I realized that the information I was reading said that Pascals was an SI derived unit. The SI unit for pressure is N/m2.

So I tried inputting the following as my answer...

256142.7883 N/m2

256142.7883 Pa

256143 N/m2

256143 Pa

256000 N/m2

256000 Pa

Still says they are all incorrect. :-/

Maybe this McGraw Hill textbook/online-coursework thing has mistakes in it? That would be the third one that they have and this is barely the second assignment. I don't want to be presumptuous but that would be very disappointing.
 
  • #5
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It's still not in SI units. What about the SI units for Newton? [kg*m/s^2]
 
  • #6
ah but of course.

N = kg*m/s2

Pa = (N/m2) = ((kg*m)/s2)/m2 = kg/m*s2

Right?

Ahh now I'm sure that's right and it says it is still wrong.

I tried

256000 kg/m*s2 I typed "kg/m*(s^2)" into the "units" field
256143 kg/m*s2
 
  • #7
also tried kg/m/s/s as the units but still nothing
 
  • #8
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Hmm, then it's probably their mistake, I can't see, what else the problem might be.

But "kg/m*(s^2)" might be misunderstood by a computer, you would put brackets like this:
kg/(m*s^2) or kg/(m*s*s)

The same goes for the last thing you tried, kg/m/s/s

Though I don't exactly remember how this works with McGraw.. But if kg/(m*s^2) doesn't work, perhaps they have special method for writing units? You might want to google that?

But spending too much time on that is probably bothersome and wasted, though I know the feeling of wanting to see "correct answer" :-)
 
  • #9
though I know the feeling of wanting to see "correct answer" :-)

Like you wouldn't believe :-)

Well the homework assignment has a link right next to the "unit" field which tells you the accepted format of the units(that's where I got the m/s/s from). I'll try the brackets but I'll email the proffessor to see if he knows of any corrections we should know about. I have until Sunday at 11:59 to complete these. I am now on problem 4 of 11 so I'll just finish the rest.

So expect more homework related questions from me this semester. Thanks, this is the 2nd maybe 3rd question you have helped me with hjelmgart
 
  • #10
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That 33.5 psi is gage pressure, not absolute pressure. You are supposed to use absolute pressure in the ideal gas law. You need to add 14.7 to the gage pressure to get absolute pressure. This is the cause of your difficulties.

Chet
 
  • #11
That 33.5 psi is gage pressure, not absolute pressure. You are supposed to use absolute pressure in the ideal gas law. You need to add 14.7 to the gage pressure to get absolute pressure. This is the cause of your difficulties.

Chet

Hmm...

I converted the 33.5 psi to 48.2 psi and solved for the final pressure by

Converting 48.2 psi to 332,327.302 Pa then...

p1T2/T1=p2

(332,327.3 Pa)(299.31 K)/(269.9 K)=p2=368,539.773 Pa

So I submitted these as my answer

368,540 kg/(m*s*s)

369,000 kg/(m*s*s)


No luck :-(
 
  • #12
127
17
Try using 14.5 psi instead as that corresponds to 1 bar, it's hard to know, which they have used to calculate it, but it's worth a shot.

If that also doesn't work, perhaps you need to calculate the atmospheric pressure depending on the temperature as well :-)
(that was a bad joke, I doubt you have to do that)

That 33.5 psi is gage pressure, not absolute pressure. You are supposed to use absolute pressure in the ideal gas law. You need to add 14.7 to the gage pressure to get absolute pressure. This is the cause of your difficulties.

Chet

Out of pure interest: how did you see/notice that? Experience with pumping tires, or just common sense, as they didn't write absolute pressure?
 
  • #13
I tried using 14.5 psi, p2 pressure came out to 367010 kg/(m*s*s)....

Nope

Tried with 3 sig figs...

No dice
 
  • #14
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Hmm...

I converted the 33.5 psi to 48.2 psi and solved for the final pressure by

Converting 48.2 psi to 332,327.302 Pa then...

p1T2/T1=p2

(332,327.3 Pa)(299.31 K)/(269.9 K)=p2=368,539.773 Pa

So I submitted these as my answer

368,540 kg/(m*s*s)

369,000 kg/(m*s*s)


No luck :-(
Did you convert that final pressure back to gage pressure? I get 267 kPa.
 
Last edited:
  • #15
Did you convert that final pressure back to gage pressure? I get 267 kPa.

So basically I have to take the gauge pressure, set everything to absolute temperature to do all my calculations, once complete, then since they ask for the pressure inside the tire, I change back to gauge pressure.

I ultimately subtract 14.5psi*(6.894 kPa/psi)=101.35 kPa from the p2 I calculated, 368.5 kPa.

368.5 kPa - 101.35 kPa = 267.15 kPa

THAT WORKED!

Definitely glad to know now that I fully understand how to do this calculation, it may be something that is going to be discussed today. Thanks for clearing that up for me.
 
  • #16
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I am impressed. Have also learned something!
Answer for b) being approved wrong-footed me into accepting the pressures.
Not right because b) answer only depends on T ratio, no matter what p value.
 
  • #17
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Some of us are easier to be fooled than others :P

I am still confused how to see, that it is not the absolute pressure, that was given in the information though.
 
  • #19
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Some of us are easier to be fooled than others :P

I am still confused how to see, that it is not the absolute pressure, that was given in the information though.
You just have to know that, by convention, when they give the tire pressure, it's the gage pressure. When you use a tire pressure gage to measure the pressure, by the way it is constructed, it measures the difference between the pressure inside the tire and the atmospheric pressure.
 
  • #20
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You just have to know that, by convention, when they give the tire pressure, it's the gage pressure. When you use a tire pressure gage to measure the pressure, by the way it is constructed, it measures the difference between the pressure inside the tire and the atmospheric pressure.

Okay I see, then I guess if I had ever used such instrument, I would have known it, for instance if I had car. :)
 
  • #21
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Okay I see, then I guess if I had ever used such instrument, I would have known it, for instance if I had car. :)
If you know someone who has a car, they usually have a tire pressure gage in the glove compartment. Otherwise, stores like KMart (and possibly even your local supermarket) have sections with car products that often carry tire pressure gages. That way, you don't even have to buy one.

Chet
 
  • #22
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I begin to love this thread... ;-)
 
  • #23
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If you know someone who has a car, they usually have a tire pressure gage in the glove compartment. Otherwise, stores like KMart (and possibly even your local supermarket) have sections with car products that often carry tire pressure gages. That way, you don't even have to buy one.

Chet

I see I have used several instruments in the lab to do it, but I'm fairly certain they measure absolute pressure. Or perhaps they don't but the computer simply convert it to absolute pressure, although here we typically want a lower pressure to stabilise a gas, plasma or whatever (at least in the things I've done).

I do wonder though, how can there be any advantage in measuring pressure relative to atmospheric pressure?
As I see it, there is some insecurity about doing so, unless you also measure the atmospheric pressure?
So why not just measure absolute pressure instead?


Now since we are talking pressure, another question popped to mind. A way of measuring pressure is by heating a wire inside a gas, and then measure the loss of energy to the gas, which naturally decreases as the pressure falls, as less gas is in contact with the wire.

This makes good sense, however, when reaching some quite low pressures, wouldn't you reach quite a low heat capacity as well?
Then contribution from electrons to the heat capacity would suddenly have an influence, and your measurements could become inaccurate at very low temperatures, unless you of course incorporate this in your calculation of pressure relative to the change in heat capacity.

So when there is some limit in which the measuring is accurate for instance to about 10 mTorr.
Would the reason then be, as I suggested, that for instance electron contributions to heat capacity makes the measurements inaccurate? Or what exactly could the origin for such a limit be?
 
Last edited:
  • #24
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1. Tire pressure in Barg (psig?) is more useful than absolute pressure. Has to do with the function of a tire: staying in shape. Pabs_inside-Pabs_outside is relevant! Cars stay on the ground. Planes have to worry a lot more. Accuracy requiremensts aren't that severe. Indeed, weather conditions influences are a few percent due to pressure and up to 10% due to temperature (as we've seen). Operating temperatures of tires are much higher than environment. You can try that out without having to buy a car.
Manufacturer recommended tire pressures are for 'cold' tires. They don't even specify a standard T (and almost don't have to specify a standard P_outside)

2. Vacuum pressure measurement is far removed from the current thread subject. Almost a science by itself.
 
  • #25
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1. Ah that made perfectly sense.

2. yeah I know that, but you know when a question surfaces, while writing another, you just write that down as well. May take it to another thread perhaps. Thanks anyway!
 

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