Asample of 4.50 g of diatomic oxygen (o2) occupies a volume of 15.0 l at a temperature of 203 k and a pressure of 5.0 atm. what will be the volume of the same sample if the pressure increases to 15.0 atm and the temperature is raised to 303 k?

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)

odds are decent that your handwriting right now looks a little bit different than it did five years ago, and will look different again in five more. when i look back through my notebooks from middle school to now, i see a long, slow progression from meticulous and teeny letters to bigger, sloppier ones, plus a few side detours: sometime in high school, for example, i decided those loopy cursive lowercase f’s were just the coolest, adopting them as the lone rep for cursive in my otherwise all-print handwriting.

explanation: