Question

A gas at a fixed volume had a temperature change from 87.00 K to 65.25 K. At 65.25 K, the pressure was recorded as 18 Pa. What was the initial pressure?

Answer 1

Answer:

The initial pressure was 25.2 Pa.

Explanation:

According to the Gay-Lussac's Law, the pressure of given amount of gas is directly proportional to the temperature in kelvin while volume is kept constant.

Formula:

P₁/T₁ =P₂/T₂

P₁ = Initial pressure

T₁ = Initial temperature

P₂ = Final pressure

T₂ = Final temperature

Given data:

T₁ = 87 K

T₂ = 62.25 K

P₂ = 18 Pa

P₁ = ?

Now we will put the values in formula,

P₁/T₁ = P₂/T₂

P₁  = P₂/T₂×T₁

P₁  = (18 Pa / 62.25 K) × 87 K

P₁  = 0.2892 Pa/k × 87 K

P₁  = 25.2 Pa

The initial pressure was 25.2 Pa.