Question

A tire 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?

Answer 1

Answer:

The air fraction to be removed is 0.11

Given:

Initial temperature, T = $10^{\circ}$ = 283 K

Pressure, P = 250 kPa

Finally its temperature increases, T' = $45^{\circ}$ = 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

$\frac{m}{m'} = \frac{T'}{T}$                      (1)

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

$y = \frac{m - m'}{m} = 1 - \frac{m'}{m}$

From eqn (1):

$y = 1 - \frac{T}{T'}$

$y = 1 - \frac{283}{318} = 0.11$