Ideal Gas Law Application: Finding the Temperature of Air in a Tire

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Homework Help Overview

The problem involves applying the Ideal Gas Law to determine the temperature of air in a tire after a change in gauge pressure. The context includes initial and final pressures, as well as an initial temperature, with the assumption that air behaves as an ideal gas and that tire volume remains constant.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the application of the Ideal Gas Law and the significance of gauge pressure versus absolute pressure. There are attempts to clarify the correct formula to use for calculating the final temperature based on the pressures provided.

Discussion Status

Some participants have provided guidance on adjusting the pressures to account for atmospheric pressure. There is an ongoing exploration of the implications of using gauge pressure and the appropriate temperature scale for calculations. Multiple interpretations of the problem setup are being considered.

Contextual Notes

Participants note the importance of understanding gauge pressure and its relationship to absolute pressure. There is also a discussion about the limitations of using Celsius in the calculations due to the nature of temperature scales.

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[SOLVED] Ideal Gas Law Application

Homework Statement



One Sunday morning a family takes an automobile trip to Grandma's. At the start of the trip, the temperature is 288K (15C), and the gauge pressure in the tires is 32lb/in^2 (psi). (The gauge pressure is the excess over 14.5 psi, the exterior air pressure.) After an hour's ride over an interstate highway, the gauge pressure in the tires is 38 psi. What is the temperature of the air in the tires, assuming that air behaves as an ideal gas? Neglect any changes in volume of the tires.



Homework Equations


PV=nRT


The Attempt at a Solution


3. After working around with the universal gas law, the formula I used is: T2=(P2*T1)/P1

T2=(38/32)*288=342K

342-273=69C

The textbook gives the answer 52C. What am I doing wrong? Did I use the wrong units?
 
Last edited:
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They already told you. Gauge pressure is pressure above 14.5 psi. Try solving T2=(38+14.5)*T1/(32+14.5).
 
Dick said:
They already told you. Gauge pressure is pressure above 14.5 psi. Try solving T2=(38+14.5)*T1/(32+14.5).

Hmmm, I missed that part. What does it mean when the gauge pressure is above 14.5psi?

I forgot, but why can't you use the celsius scale to do the problem?
 
Last edited:
What they meant is that real pressure=14.5psi+gauge pressure. It's the pressure you would read with a tire gauge, for example. If it reads 0psi that just means that the pressure in the tire is the same as atmospheric pressure. You can't use degrees C because 0C=273K, 1C=274K, but 274/273 is NOT equal to 1/0. You tell me what went wrong.
 

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