Solving Gas Law Question: Tire Pressure at 14°C & 2.2atm

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SUMMARY

The discussion revolves around calculating the temperature of air in a car tire after its pressure increases from 2.0 atm to 2.2 atm while the external temperature remains at 14°C. Participants reference Charles's Law, Boyle's Law, and the Combined Gas Law to approach the problem. It is clarified that while atm is a unit of pressure, the conversion to kPa is unnecessary for solving the problem since the factors will cancel out. The focus is on applying the gas laws correctly to find the unknown temperature in the tire.

PREREQUISITES
  • Understanding of Charles's Law (V1/T1 = V2/T2)
  • Familiarity with Boyle's Law (P1V1 = P2V2)
  • Knowledge of the Combined Gas Law (V1P1/T1 = V2P2/T2)
  • Basic concepts of pressure units, specifically atm and kPa
NEXT STEPS
  • Study the application of the Combined Gas Law in real-world scenarios
  • Learn how to convert between different pressure units, including atm and kPa
  • Explore the implications of temperature changes on gas pressure in closed systems
  • Review practical examples of gas law applications in automotive contexts
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Students studying physics or chemistry, automotive engineers, and anyone interested in understanding gas laws and their applications in real-life situations.

iluvu
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The question given were:

The tire pressure for a car that has not been driven is 2.0atm when sitting outside at 14 degrees celsius. After the car was on the highway the temperature of the tire increased even though the air temperature remained at 14 degrees celsius. The tire pressure increased to 2.2atm. Calculate the temperature of the air in the tire in degrees celsius.

The gas laws that the teacher ever told us is

charles law: V1/T1=V2/T2
Bogle's law is P1V1=P2/V2 and
Combine law, which is V1*P1/T1=V2*P2/T2

But in the question, there is atm...which stands for Atmosphere...I know that Atomosphere is a unit of pressure, but isn't the unit for pressure (kPa)?

I'm so stuck...=( someone please help me...
 
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1 atm is approximately 1.013 \times 10^5 Pa. But you do not need to perform this conversion to solve the problem because this factor will eventually cancel off.

What quantities remain the same before and after the car was driven?
 
Last edited:
1 kPa is 1000 Pa
 

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