In the rubber chamber of a car wheel in a warm garage, where the air temperature was t1 = + 24 ° C

In the rubber chamber of a car wheel in a warm garage, where the air temperature was t1 = + 24 ° C, air was pumped to a pressure of 2.75 * 10 ^ 5 Pa. Then the wheel was rolled out into the street, where the air temperature was t2 = -3 ° С. What was the pressure in the chamber after thermal equilibrium was established?

Let the initial air temperature in the rubber chamber of a car wheel in a warm garage be T₁ = + 24 ° C = (24 + 273) K = 297 K, and the pressure p₁ = 2.75 ∙ 10 ^ 5 Pa. Then the wheel was rolled out onto the street, where, after establishing thermal equilibrium, the air temperature in the wheel became Т₂ = – 3 ° С = (- 3 + 273) К = 270 K. Since the process is isochoric (V = const), the steady pressure in the chamber р₂ we find according to Charles’s law:

р₁ / Т₁ = р₂ / Т₂ or р₂ = (Т₂ ∙ р₁) / Т₁.

Substitute the values ​​of physical quantities in the calculation formula:

p₂ = (270 K ∙ 2.75 ∙ 10 ^ 5 Pa) / 297 K;

p₂ = 2.5 ∙ 10 ^ 5 Pa.

Answer: the pressure in the chamber became 2.5 ∙ 10 ^ 5 Pa after the establishment of thermal equilibrium.