Question

1.0 mole of gas is in a sealed 2.0 l container. what happens to the volume when the number of moles of gas is increased to 6 times the original amount? the volume is increased to 3 times the original amount. the volume is increased to 6 times the original amount. the volume is reduced to 1/2 of the original amount. the volume is reduced to 1/6 of the original amount.

Explanation:

Step1: Recall Avogadro's Law

At constant temperature and pressure, $\frac{V_1}{n_1} = \frac{V_2}{n_2}$, where $V$ = volume, $n$ = moles.

Step2: Define given values

$n_1 = 1.0\ \text{mol}$, $V_1 = 2.0\ \text{L}$, $n_2 = 6 \times n_1 = 6.0\ \text{mol}$

Step3: Solve for $V_2$

Rearrange formula: $V_2 = V_1 \times \frac{n_2}{n_1}$
Substitute values: $V_2 = 2.0\ \text{L} \times \frac{6.0\ \text{mol}}{1.0\ \text{mol}} = 12.0\ \text{L}$

Step4: Compare to original volume

$\frac{V_2}{V_1} = \frac{12.0\ \text{L}}{2.0\ \text{L}} = 6$

Answer:

The volume is increased to 6 times the original amount.