Given a gas is at 16 Land 1 mol, what is the volume of the gas when the number of moles changes to 3 mol? Round to one deeimal place. square

To solve this problem, we can use the ideal gas law, which states that the volume of a gas is directly proportional to the number of moles of the gas, given that the temperature and pressure remain constant.<br /><br />The ideal gas law can be expressed as:<br /><br />\[ V_1 / n_1 = V_2 / n_2 \]<br /><br />where:<br />- \( V_1 \) is the initial volume of the gas,<br />- \( n_1 \) is the initial number of moles of the gas,<br />- \( V_2 \) is the final volume of the gas,<br />- \( n_2 \) is the final number of moles of the gas.<br /><br />Given:<br />- \( V_1 = 16 \, \text{L} \)<br />- \( n_1 = 1 \, \text{mol} \)<br />- \( n_2 = 3 \, \text{mol} \)<br /><br />We need to find \( V_2 \).<br /><br />Using the ideal gas law:<br /><br />\[ V_2 = V_1 \times \frac{n_2}{n_1} \]<br /><br />Substitute the given values:<br /><br />\[ V_2 = 16 \, \text{L} \times \frac{3 \, \text{mol}}{1 \, \text{mol}} \]<br /><br />\[ V_2 = 16 \, \text{L} \times 3 \]<br /><br />\[ V_2 = 48 \, \text{L} \]<br /><br />Therefore, the volume of the gas when the number of moles changes to 3 mol is \( 48 \, \text{L} \).