Atire is filled with air at 10 ∘c to a gauge pressure of 250 kpa. part a if the tire reaches a temperature of 45 ∘c, what fraction of the original air must be removed if the original pressure of 250 kpa is to be maintained?

The air fraction to be removed is 0.11

Given:

Initial temperature, T = = 283 K

Pressure, P = 250 kPa

Finally its temperature increases, T' = = 318 K

Solution:

Using the ideal gas equation:

PV = mRT

where

P = Pressure

V = Volume

m = no. of moles of gas

R = Rydberg's Constant

T = Temperature

Now,

Considering the eqn at constant volume and pressure, we get:

mT = m'T'

Thus

                     (1)

Now, the fraction of the air to be removed for the maintenance of pressure at 250 kPa:

From eqn (1):

= initial pressure

= final pressure = 2

= initial volume

= final volume = /4

= initial temperature = 20 c = 20 + 273 = 293 k

= final temperature = ?

using the equation