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Change In Tyre Air Pressure
hello...CAN SOMEONE PLEASE TELL ME A ANSWER TO A QUESTION I HAVE..HOW MUCH DOES TYRE AIR PRESSURE CHANGE WITH THE TEMPERATURE (A) APPROX 0.2BAR/15C (B) APPROX 0.1BAR/15C (C) APPROX 0.1BAR/10C (D) APPROX 0.2BAR/ 10C Advance thanks..
Answers
Best Answer
No best answer has yet been selected by slocjo. Once a best answer has been selected, it will be shown here.
For more on marking an answer as the "Best Answer", please visit our FAQ.Your question has a lack of data or parameters such as:
Temperature increase or decrease
The range of temperature you are talking about (from/to)
Size of tyres in question
Whether the vehicle is static or in motion (or a combination)
What the starting pressure is and whether it is standard for that tyre
Temperature increase or decrease
The range of temperature you are talking about (from/to)
Size of tyres in question
Whether the vehicle is static or in motion (or a combination)
What the starting pressure is and whether it is standard for that tyre
Zacs, I think you've misunderstood the question but there is still data missing.
I think the question is a multiple choice and you have to choose the correct answer from the 4 given.
Let's assume that the tyre is a normal car tyre, starting temp 20deg (=293 Kelvin) at a pressure of about 30psi = 2 bar and that the volume of the tyre doesn't change with increase in pressure.
We can now use Gay-Lussac's Law p1/T1 = p2/T2
p2 = (p1 x T2)/T1
for a 15 deg increase in temp we get p2 = (2 x 308)/293 = approx 2.1bar
the increase in pressure
is 2.1 - 2.0 = 0.1 bar for a 15 deg increase in temp.
Now see which of these options this agrees with and there's your answer.
I think the question is a multiple choice and you have to choose the correct answer from the 4 given.
Let's assume that the tyre is a normal car tyre, starting temp 20deg (=293 Kelvin) at a pressure of about 30psi = 2 bar and that the volume of the tyre doesn't change with increase in pressure.
We can now use Gay-Lussac's Law p1/T1 = p2/T2
p2 = (p1 x T2)/T1
for a 15 deg increase in temp we get p2 = (2 x 308)/293 = approx 2.1bar
the increase in pressure
is 2.1 - 2.0 = 0.1 bar for a 15 deg increase in temp.
Now see which of these options this agrees with and there's your answer.
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