Gauge Pressure of Tire After Driving

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SUMMARY

The gauge pressure of an automobile tire after driving can be calculated using the ideal gas law, specifically the equation p1V1/T1 = p2V2/T2. Initially, the tire has a gauge pressure of 1.65 atm at a temperature of 3.0°C (276 K) and a volume of 1.50x10^-2 m³. After driving, the temperature increases to 47.0°C (320 K) and the volume expands to 1.65x10^-2 m³. The correct gauge pressure, accounting for atmospheric pressure, is determined to be 1.74 atm.

PREREQUISITES
  • Understanding of the ideal gas law and its application
  • Knowledge of gauge pressure vs. absolute pressure
  • Basic thermodynamics concepts, particularly relating to temperature and volume changes
  • Familiarity with unit conversions, particularly between Celsius and Kelvin
NEXT STEPS
  • Study the ideal gas law in detail, focusing on its applications in real-world scenarios
  • Learn about the differences between gauge pressure and absolute pressure
  • Explore how temperature affects gas volume and pressure in closed systems
  • Investigate the effects of tire pressure on vehicle performance and safety
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Automotive engineers, physics students, and anyone interested in understanding tire pressure dynamics and thermodynamic principles in automotive applications.

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Homework Statement



An automobile tire has a volume of 1.50x10^-2 m^3 on a cold day when the temperature of the air in the tire is 3.0 C and atmospheric pressure is 1.02 atm. Under these conditions the gauge pressure is measured to be 1.65 atm. After the car is driven on the highway for 30 min, the temperature of the air in the tires has risen to 47.0 C and the volume has risen to 1.65x10^-2 m^3.

What then is the gauge pressure?

T1 = 276 K
V1 = 1.50 * 10^-2
p1 = 1.65

T2 = 320 K
V2 = 1.65 * 10^-2



Homework Equations



p1V1/T1 = p2V2/T2

The Attempt at a Solution



p2 = p1 * V1/V2 * T2/T1

p2 = 1.65 * (1.50 * 10^-2)/(1.65 * 10^-2) * 320/276

p2 = 1.74

The answer only needs to be to three significant digits. This answer is close to the correct answer, but it's still not correct. I know that somehow the atmospheric pressure of 1.02 atm is significant, but I can't figure out what to do with it.
 
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The gauge pressure is the net pressure acting on the tyre. I think the pressure that must be used in the equation is actually the gross pressure in the tyre, e.g., p_1+p_a.

Does this give a better answer?
 

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