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:

T₁ = 39 K

Explanation:

Given dа ta:

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.