Question

A sample of chlorine gas is held at a pressure of 1023.6 kPa. When the pressure is decreased to 8114 kPa the

temperature is 36°C, What was the original temperature? Gas Law used:

Answer 1

T₁ = 39 K

Step-by-step explanation:

Given data:

Initial pressure = 1023.6 kpa

Final pressure = 8114 kpa

Final temperature = 36°C (36+ 273= 309K)

Initial temperature = ?

Solution:

P₁/T₁ = P₂/T₂

T₁ = P₁×T₂ /P₂

T₁ = 1023.6 kpa × 309 K /8114 kpa

T₁ = 316292.4 K. Kpa /8114 kpa

T₁ = 39 K

Thus original pressure was 39 k.