Question

how many moles of gas were added to a balloon that started with 2.3 moles of gas and a volume of 1.4 l given that the final volume was7.2 l ?
12 mol
0.085 mol
9.5 mol
4.4 mol

Explanation:

Step1: Apply Avogadro's Law

Avogadro's Law states that \(\frac{V_1}{n_1}=\frac{V_2}{n_2}\) (at constant temperature and pressure), where \(V_1 = 1.4\space L\), \(n_1 = 2.3\space mol\), \(V_2 = 7.2\space L\). We solve for \(n_2\):
\(n_2=\frac{V_2\times n_1}{V_1}=\frac{7.2\times2.3}{1.4}\)
Calculate numerator: \(7.2\times2.3 = 16.56\)
Then \(n_2=\frac{16.56}{1.4}\approx11.83\space mol\) (approx 12 mol, but wait, no—wait, we need moles added, not final moles. Wait, no: Wait, the question is moles added. Wait, final moles \(n_2\), initial moles \(n_1 = 2.3\). So moles added is \(n_2 - n_1\). Wait, I made a mistake. Let's recalculate.

Wait, Avogadro's Law: \(\frac{V_1}{n_1}=\frac{V_2}{n_2}\) → \(n_2=\frac{V_2\times n_1}{V_1}=\frac{7.2\times2.3}{1.4}\). Let's compute that: \(7.2\times2.3 = 16.56\); \(16.56\div1.4\approx11.83\space mol\) (final moles). Then moles added is \(11.83 - 2.3\approx9.5\space mol\). Ah, that's the error. So step 1: Find final moles \(n_2\) via Avogadro's Law. Step 2: Subtract initial moles to get moles added.

Step1: Calculate final moles (\(n_2\))

Using \(\frac{V_1}{n_1}=\frac{V_2}{n_2}\) → \(n_2=\frac{V_2\times n_1}{V_1}\)
\(V_1 = 1.4\space L\), \(n_1 = 2.3\space mol\), \(V_2 = 7.2\space L\)
\(n_2=\frac{7.2\times2.3}{1.4}=\frac{16.56}{1.4}\approx11.83\space mol\)

Step2: Calculate moles added

Moles added = \(n_2 - n_1 = 11.83 - 2.3\approx9.5\space mol\)

Answer:

9.5 mol