A bicycle tire is filled to a pressure of 110. psi at a temperature of 30.0°C. At what temperature will the air pressure in the tire decrease to 105 psi? Assume that the volume of the tire remains constant.

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The calculated temperature is 289.37 K

According to Gay-Law, Lussac's an ideal gas's pressure is inversely proportional to its absolute temperature when the gas' volume is constant. In other words, according to Gay-Law, Lussac's the pressure of a certain quantity of gas at a fixed volume is directly proportional to the gas's kelvin temperature.

For every 10 degrees that the temperature drops, the inflation pressure in tires typically decreases by 1 to 2 psi.

This is an illustration of Gay-law, Lussac's which states that at constant volume, pressure and Kelvin temperature are directly related. P1/T1=P2/T2

P1=7.48 atm

K is the equation to employ.

P2=7.14 atm

T2=?

T1=30.0C+273.15

   =303.15 

Solve T2 by rearranging the equation to isolate it.

P1/T1=P2/T2 

T2= T1T2/P1 

7.14 x 303.15/7.48

 = 289.37 K 

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