Question

avogadro’s law states that the volume, v, of a gas is directly related to the number of moles, n, of the gas when temperature and pressure are constant (figure 1).

part a
a cylinder, with a piston pressing down with a constant pressure, is filled with 2.20 moles of a gas (n₁), and the volume is 46.0 l (v₁). if 0.00 moles of gas leak out, and the pressure and temperature remain the same, what is the final volume of the gas inside the cylinder?
express your answer with the appropriate units.
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v₂ =
value
units

Explanation:

Step1: Recall Avogadro's Law

Avogadro's Law states that \(\frac{V_1}{n_1}=\frac{V_2}{n_2}\) when temperature and pressure are constant. Here, \(V_1 = 46.0\space L\), \(n_1 = 2.20\space mol\), \(n_2 = 0.00\space mol\)? Wait, no, maybe a typo? Wait, probably \(n_2\) is the final moles? Wait, no, the problem says "if 0.00 moles of gas leak out"? Wait, no, maybe the initial moles \(n_1 = 2.20\space mol\), initial volume \(V_1 = 46.0\space L\), and final moles \(n_2=2.20 - 0.00\)? No, that can't be. Wait, maybe the problem is: A cylinder with a piston pressing down with constant pressure, filled with 2.20 moles of gas (\(n_1\)), volume \(V_1 = 46.0\space L\). If 0.00 moles leak out? No, that would mean \(n_2 = 2.20\space mol\), so \(V_2 = V_1\). But that doesn't make sense. Wait, maybe the leak is, say, some moles? Wait, maybe the user made a typo, but assuming the correct values: Let's say initial \(n_1 = 2.20\space mol\), \(V_1 = 46.0\space L\), final \(n_2\) (after leak) – wait, maybe the problem is that 0.00 moles leak out, so \(n_2 = 2.20\space mol\), so \(V_2 = 46.0\space L\). But that's odd. Wait, maybe the leak is 0.00 moles (no leak), so volume remains same. Or maybe the problem is different. Wait, let's re-express.

Wait, the problem: "A cylinder, with a piston pressing down with a constant pressure, is filled with 2.20 moles of a gas (\(n_1\)), and the volume is 46.0 L (\(V_1\)). If 0.00 moles of gas leak out, and the pressure and temperature remain the same, what is the final volume of the gas inside the cylinder?"

Using Avogadro's Law: \(\frac{V_1}{n_1}=\frac{V_2}{n_2}\)

Here, \(n_2 = n_1 - 0.00 = 2.20\space mol\)

So \(\frac{46.0\space L}{2.20\space mol}=\frac{V_2}{2.20\space mol}\)

Thus, \(V_2 = 46.0\space L\)

But that's trivial. Maybe the leak is, say, 1.10 moles? Wait, maybe the problem was mistyped. But assuming the given values, let's proceed.

Step2: Apply Avogadro's Law

\(\frac{V_1}{n_1}=\frac{V_2}{n_2}\)

Given \(V_1 = 46.0\space L\), \(n_1 = 2.20\space mol\), \(n_2 = 2.20 - 0.00 = 2.20\space mol\)

So \(V_2=\frac{V_1\times n_2}{n_1}=\frac{46.0\space L\times2.20\space mol}{2.20\space mol}=46.0\space L\)

Answer:

\(46.0\space L\)