Answer:
The air fraction to be removed is 0.11
Given:
Initial temperature, T = [tex]10^{\circ}[/tex] = 283 K
Pressure, P = 250 kPa
Finally its temperature increases, T' = [tex]45^{\circ}[/tex] = 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
[tex]\frac{m}{m'} = \frac{T'}{T}[/tex] (1)
Now, the fraction of the air to be removed for the maintenance of pressure at 250 kPa:
[tex]y = \frac{m - m'}{m} = 1 - \frac{m'}{m}[/tex]
From eqn (1):
[tex]y = 1 - \frac{T}{T'}[/tex]
[tex]y = 1 - \frac{283}{318} = 0.11[/tex]
