To find the original temperature of the gas when the pressure was 1 ATM, we can use the combined gas law, which states:
(P₁ × V₁) / T₁ = (P₂ × V₂) / T₂
Where: P₁ and T₁ are the initial pressure and temperature, P₂ and T₂ are the final pressure and temperature, and V₁ and V₂ are the initial and final volumes (assuming the volume remains constant).
In this case, we are given: P₁ = 1 ATM (the initial pressure) P₂ = 1.5 ATM (the final pressure) T₂ = 350 K (the final temperature)
Assuming the volume remains constant, V₁ = V₂, so we can simplify the equation to:
P₁ / T₁ = P₂ / T₂
Now we can plug in the known values:
(1 ATM) / T₁ = (1.5 ATM) / (350 K)
To solve for T₁, we can rearrange the equation:
T₁ = (1 ATM) / [(1.5 ATM) / (350 K)]
Simplifying further:
T₁ = (1 ATM) × (350 K) / (1.5 ATM)
T₁ = 233.33 K
Therefore, the original temperature of the gas was approximately 233.33 K when the pressure was 1 ATM.