What is the Tire Pressure After Temperature and Volume Changes?

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SUMMARY

The discussion focuses on calculating tire pressure changes due to temperature and volume variations. Initially, the tire pressure is determined using the ideal gas law, yielding a pressure of 400 kPa after compressing the air to 28% of its original volume at 40°C. Following a temperature increase to 85°C and a 2% increase in tire volume, the new absolute pressure is calculated to be approximately 449 kPa. The calculations utilize the relationships between pressure, volume, and temperature as outlined in the ideal gas law.

PREREQUISITES
  • Understanding of the ideal gas law (PV=nRT)
  • Knowledge of temperature conversion (Celsius to Kelvin)
  • Familiarity with pressure units (atmospheric pressure to pascals)
  • Basic algebra for solving equations
NEXT STEPS
  • Study the ideal gas law and its applications in real-world scenarios
  • Learn about the effects of temperature and volume changes on gas pressure
  • Explore advanced thermodynamics concepts related to gases
  • Investigate practical applications of tire pressure monitoring systems
USEFUL FOR

Automotive engineers, physics students, and anyone interested in understanding the principles of gas behavior in relation to temperature and pressure changes in tires.

atelaphobia
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here's another one that stumped me

An automobile tire is inflated with air originally at 10C and normal atmospheric pressure. During the process the air is compressed to 28% of its original volume and the temperature is increased to 40C.
a) what is the tire pressure?
b) after the car is driven at high speed the tire air temperature rises to 85C and the interior volume of the tire increases by 2%. what is the new tire pressure (absolute) in pascals?

the only reason why i don't know how to start this problem is because i don't know what is the initial pressure. all it says is "normal atmospheric pressure" :confused:
 
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ok i got the first part
a)

Pinitial/T initial=Pfinal/Tfinal

normal pressure= 1atm= 101.325kPa
convert temps into Kelvins
T initial= 283.15
Tfinal= 313.15

SO...

101.325/283.15=P(.28)/313.15

P= 400kPa

but now i don't know how to get the second one
I know the answer for b is 449 kPa
 
Last edited:
i think i solved the second part but tell me if this makes sense at all!

the temp rises to 85C (359.15 K) and tire increases volume by 2%

so using the numbers from part A with the same equation

400/313.15=P/358.15
solve for P=457.726 BUT it's saying the volume of the tire increased by 2%
so... 2% of 457.726 is 9.154

since we got to incorporate the 2% factor i subtracted 457.726-9.154
gave me 448.57 which is 449kPa

now did i do that right??
 

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