Question

The air pressure in a tire sitting outside is 30 psi on a nice fall day when the temperature is about 80 F. (Assuming that the density of the air inside the tire does not change with temperature and that the air may be treated as an ideal gas), what would the tire pressure be:

A) At 32 F?

B) At 100 F?

If the recommended tire pressure is between 28 and 32 psi, do you need to do anything in either case above? If so, which and what?

Answer 1

Answer:

Let us go to the basics first.

We know that Ideal gas equation provides us following relation:

PV = nRT..................Eqn.1

[Where, P = Pressure; V = volume; n = number of moles; R = universal gas constant; T = temperature]

Also, Density, D = m/V where, m = mass

=>D = nM / V ..........Eqn.2 where, m = nM; M = molar mass of gas

Also, Specific gas constant R_g = R / M....................Eqn.3

Using Eqn.3 & Eqn.1, we have:

PV = nMR_gT

Using Eqn.2, above equation becomes:

PV = DVR_gT

=>P = DR_gT.....................Eqn.4

(A) Now, for temperatures T₁ & T₂, eqn.4 will yield:

P₁ = DR_gT₁

& P₂ = DR_gT₂  (Given that density of air is constant)

Dividing these equations:

P₁ / P₂ = T₁ / T₂..........Eqn.5

[Using P₁ = 30psi; T₁ = 80 F = 299.817 K; T₂ = 32 F = 273.15 K]

=>30 / P₂ = 299.817 / 273.15

=>30 / P₂ = 1.0976

=>P₂ = 27.33 psi (Answer A)

(B)Again using Eqn.5:

P₁ / P₂ = T₁ / T₂

[Using P₁ = 30psi; T₁ = 80 F = 299.817 K; T₂ = 100 F = 310.928 K]

=>30 / P₂ = 299.817 / 310.928

=>30 / P₂ = 0.964

=>P₂ = 31.12 psi (Answer B)

As it is given that the recommended tire pressure is between 28 and 32 psi, so we need to increase pressure in tire in case A.

Thanks!!!