What is the tire pressure at -20°F compared to 90°F, assuming the starting pressure was 32 psi?
Ideal Gas Law notes:
PV=nRT
P: Pressure
V: Volume
n: Moles of Gas
T: Temperature
R: Gas Constant
PV/nT = nRT/nT
P1 V1/ n1 T1 = P2 V2/ n2 T2
1. First I used a conversion calculator at http://www.infoplease.com/pages/unitconversion.html?qty=-20&fromopts=tempF&toopts=tempK&unittype=temperature&grp=&convert.x=40&convert.y=8
ReplyDeleteBut I found the equations to solve for converting Fahrenheit, Celsius, and Kalvins.
Formulas:
Fahrenheit to Kalvins (F-32) * 5/9 + 273.15
Celsis to Fahrenheit (C*9/5) + 32
Kalvins to Fahrenheit (K-273.15) * 9/5 + 32
Kalvisn to Celsis K-273.15
I converted Fahrenheit to Kalvins for 90 degrees Fahrenheit = 305.2 Kalvins
Then I divided 32psi/ 305.2 = .105psi/kalvins
2. For the second question I first converted Fahrenheit to Kalvins -20 degrees Fahrenheit is equal to 244.3 Kalvins.
Next I put the Kalvins of 90 degrees and -20 degrees back into the Ideal Gas Law Equation.
32psi(244.26 K)/305.2 K = Xpsi(305.26K)/ 305.2 K
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X= 25.6 psi
25.6 is the pressure.