Answer to Question #116425 in Chemistry for GILEANNA

Question

What would be the pressure of a gas, at 101.3 kPa, if the temperature is lowered from 368 K to 358 K?

Answer 1

According to Gay-Lussac's Law:

p₁/T₁ = p₂/T₂

where p₁ - initial pressure, T₁ - initial temperature, p₂ - final pressure, T₂ - final temperature.

From here:

p₂ = p₁ × T₂/T₁ = 101.3 kPa × 358 K / 368 K = 98.5 kPa

Answer: 98.5 kPa