Question

A 0.477 mol sample of O2 ​ gas has a volume of 12.1 L at a certain temperature and pressure. If all this O2 ​ were converted to ozone (O3 ​) at the same temperature and pressure, what is the ozone volume (in liters)? 3O2 ​ (g) → 2O3 ​ (g)

Answer 1

If we are assuming ideal behavior, then the Ideal Gas Equation,
`PV = nRT`, is very helpful. The number of moles
`n` changes here by a factor of
`(2 \ mol \ O_3 )/(3 \ mol \ O_2)`, so you have
`0.318 \ mol \ O_3`.

So,
`V_1 = (n_1RT_1)/(P_1)` and
`V_2 = (n_2RT_2)/(P_2)`. Looking at the given information, P and T are constant (as is R, always). So, we can write
`(n_1)/(V_1) = (n_2)/(V_2)`. Solving for
`V_2`, we get
`(V_1n_2)/(n_1)`. Then we plug in 12.1 L for
`V_1`,
`0.318 \ mol` for
`n_2` and
`0.477 \ mol` for
`n_1`. Thus
`V_2 = 8.07 \ L \ O_3`.