A sample of He gas (2.35 mol) occupies 57.9 L at 300.0 K and 1.00 atm. The volume of this sample is ________ L at 521 K and 1.00 atm.

Answer 1

To find the final volume of the sample at 521 K and 1.00 atm, we need to use the combined gas law formula:

(P₁V₁) / (T₁) = (P₂V₂) / (T₂)

Where:
P₁ = initial pressure (1.00 atm)
V₁ = initial volume (57.9 L)
T₁ = initial temperature (300.0 K)
P₂ = final pressure (1.00 atm)
V₂ = final volume (unknown)
T₂ = final temperature (521 K)

Plug in the values:

(1.00 atm * 57.9 L) / (300.0 K) = (1.00 atm * V₂) / (521 K)

Solve for V₂:

V₂ = (1.00 atm * 57.9 L * 521 K) / (300.0 K)

V₂ = 100.67 L

Therefore, the final volume of the sample at 521 K and 1.00 atm is approximately 100.67 L.