Question

Imagine a car tire that contains 5.1 moles of air when at a gauge pressure of 2.1×10^5N/m2 (the pressure above atmospheric pressure) and a temperature of 27 degrees C. The temperature increases to 37 degrees C, the volume decreases to 0.8 times the original volume, and the gauge pressure decreases to 1.6×10^5N/m2.
How many moles of air are left in the tire?

Answer 1

To solve this problem it is necessary to use the concepts given through the ideal gas equation.

For this it is defined that

$PV = nRT$

Where,

P = Pressure

V = Volume

R = Gas ideal constant

T = Temperature

n = number of moles.

In this problem we have two states in which the previous equation can be applied, so

$1) P_1V_1 = n_1RT_1$

$2) P_2V_2 = n_2RT_2$

From the first state we can calculate the Volume

$V_1 = \frac{n_1RT_1}{P_1}$

Replacing

$V_1 = \frac{5.1*8.314*300.15}{3.1*10^5}$

$V_1 = 0.041m^3$

From the state two we can calculate now the number of the moles, considering that there is a change of 0.8 from Volume 1, then

$n_2 = \frac{P_2(0.8*V_2)}{RT_2}$

$n_2 = \frac{2.6*10^5(0.8*0.041)}{8.314*310.15}$

$n_2 = 3.3moles$

Therefore there are 3.3moles of air left in the tire.