How Does Temperature Affect Tire Pressure?

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Homework Help Overview

The discussion revolves around the relationship between temperature and tire pressure, specifically in the context of an automobile tire inflated at a certain temperature and pressure, which is then subjected to changes in volume and temperature. The problem involves applying the ideal gas law in a scenario where the volume is reduced and the temperature is increased.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of the ideal gas law, questioning the accuracy of their calculations and the assumptions regarding temperature and pressure conversions. There are attempts to estimate the expected pressure based on changes in volume and temperature.

Discussion Status

The discussion is ongoing, with participants providing insights and calculations. Some guidance has been offered regarding the need to convert temperatures to Kelvin, and there is an exploration of the expected pressure based on the changes in volume. However, there is no explicit consensus on the final answer yet.

Contextual Notes

Participants note the requirement to express the final answer in pascals, which may influence their calculations. There is also mention of significant figures in relation to the data provided.

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Homework Statement


An automobile tire is inflated with air originally at 10.0°C and normal atmospheric pressure. During the process, the air is compressed to 26.0% of its original volume and the temperature is increased to 32.0°C.
(a) What is the tire pressure?


Homework Equations



p1V1/t1=p2v2/t2

The Attempt at a Solution


i know this is simple but I am just not getting the right answer I keep using p1V1/t1=p2v2/t2 but wrong answers come out. i used 1 atm for normal atmospheric pressure
 
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Remember that T must be in Kelvin since you are dividing by it.
 
yes it is in kelvin...i got 419903.5 is that right?
 
First estimate the answer!
Temperature hasn't changed much in absolute terms
volume has gone down by a factor of 4 - so we want a pressure of around 4 atm.

V1 = 1
V2 = 0.26
P1 = 1atm
T1 = 283K
P2 = ?
T2 = 305K

1/283 = 0.26 P /305
So P = (305 /283) / 0.26 = 4.1 atm or around 420KPa

ps. That's an impressive number of significant figures given the data!
 
it asks for it in pa that's y
 

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