Answer:
V = 0.23 Lt
Explanation:
Data: O₂ 4.50 g = 0.14 mol (considering a molecular mass of 32 g for O₂)
V₁ = 15.0 Lt , T₁ = 203 K , P₁ = 5.0 atm
V₂ = ? , T₂ = 303 K , P₂ = 15.0 atm
If we consider that this gas beahves as an ideal gas, then:
PV = nRT
V = nRT/P
V₂ = nRT₂/P₂ , where R gas constant = 0.082 Lt*atm/K*mol
V₂ = (0.14x0.082x303)/15.0 ⇒ V₂ = 0.23 Lt
Which is consistent, considering that the gas is subjected to a high pressure (from 5 atm to 15 atm), its volume should decrease (from 15 Lt to 0.23 Lt)
