What would be the original temperature of a gas that has a volume of 2.0 L and a pressure of 2.0 ATM and an unknown temperature
that the volume increased to 3.5 L in its pressure decreased to 1.0 ATM if the final temperature is measured to be 11°C
1 answer:
Answer:
= 51.57 °C
Explanation:
The combined gas equation shows that P₁V₁/T₁=P₂V₂/T₂ where P represents pressure, V represents volume and T absolute temperature
From the information provided in the question,
Since we are finding the initial or original temperature, we can make T₁ the subject of the formula.
T₁=P₁V₁T₂/P₂V₂
P₁=2.0 ATM
P₂= 1.0 ATM
V₁=2.0 L
V₂= 3.5 L
T₁=?
T₂= (11+273) K=284 K
Using these values in the formula:
T₁= (2.0 ATM × 2.0 L× 284 K)/(1.0 ATM × 3.5 L)
=324.57 K
324.57-273= 51.57 °C
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