Question

A rigid container of O has a pressure of 340 kPa at a temperature of 713 K. What is the pressure at 273 K?

Answer 1

Answer:

P₂ = 130.18 kPa

Explanation:

In this case, we need to apply the Gay-Lussack's law assuming that the volume of the container remains constant. If that's the case, then:

P₁/T₁ = P₂/T₂   (1)

From here, we can solve for the Pressure at 273 K:

P₂ = P₁ * T₂ / T₁   (2)

Now, all we need to do is replace the given data and solve for P₂:

P₂ = 340 * 273 / 713

P₂ = 130.18 kPa

Hope this helps