Question

what is the volume of a 4.000 l balloon of oxygen when the pressure is increased from 2.000 atm to 6.000 atm at a constant temperature of 25.00 °c? this is a change in conditions problem, we need to define the variables: v₁ = 4.000 l t₁ = 298.15 k p₁ = 2.000 atm v₂ =? l t₂ = 298.15 k p₂ = 6.000 atm the problem doesnt mention the moles, n, thus the amount of moles was not changed and n₁ = 1. constant temperature means t₁ = 2 the equation simplifies to the following using symbols p₂ v₂ p₁ v₁ p₂ v₂ p₁ v₁ p₁ 3 ------- = ------- ------- = ------- solving for v₂ = ----------- n₂ t₂ n₁ t₁ 1 1 4 5 atm 6 l putting in the numbers v₂= ----------------- = 7 l 8 atm

Explanation:

Step1: Analyze the given gas law equation

$$\frac{P_{2}V_{2}}{n_{2}T_{2}}=\frac{P_{1}V_{1}}{n_{1}T_{1}}$$
Since \(n_{1}=n_{2}\) (moles not changed) and \(T_{1} = T_{2}\) (constant temperature), the equation simplifies to \(\frac{P_{2}V_{2}}{1}=\frac{P_{1}V_{1}}{1}\), and solving for \(V_{2}\) gives \(V_{2}=\frac{P_{1}V_{1}}{P_{2}}\)

Step2: Substitute the values

Given \(P_{1} = 2.000\) atm, \(V_{1}=4.000\) L, \(P_{2}=6.000\) atm.
Substitute into \(V_{2}=\frac{P_{1}V_{1}}{P_{2}}\), we have \(V_{2}=\frac{2.000\times4.000}{6.000}\)

Step3: Calculate the result

\(V_{2}=\frac{8.000}{6.000}=\frac{4}{3}\approx1.333\) L

Answer:

\(1.333\) L