Question

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

Answer 1

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