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):

answer: cesare beccaria developed the classical school of criminology with from philosopher jeremy bentham

explanation:

gpe = mass x gravity constant x height or… gpe = mgh

where… mass: measured in kilograms (kg)

gravity: gravity constant on earth = 10.00 m/s2

height: measured in meters (m)

gpe: measured in joules (j) or kilojoules (kj)

how do you solve gpe?

gravitational potential energy (gpe) is the energy an object has because of its position above the. ground.

gpe = mass x gravity constant x height or… gpe = mgh.

mass:

note: to earn full marks when solving science word problems, you must show your work:

formula.

example:

given values:

mass = 0.3 kg.