To determine the new volume of the container when the number of moles of air is doubled, we can use the principles of the ideal gas law, specifically the direct relationship between the volume of a gas and the number of moles when temperature and pressure are held constant.
Given data:
- Initial volume ([tex]\( V_1 \)[/tex]) = 13.5 L
- Initial number of moles ([tex]\( n_1 \)[/tex]) = 1.50 moles
- New number of moles ([tex]\( n_2 \)[/tex]) = 3.00 moles
Since temperature and pressure are constant, the relationship between volume and number of moles is directly proportional. This can be represented mathematically as:
[tex]\[ \frac{V_1}{n_1} = \frac{V_2}{n_2} \][/tex]
We need to find the new volume ([tex]\( V_2 \)[/tex]). Rearranging the equation to solve for [tex]\( V_2 \)[/tex]:
[tex]\[ V_2 = V_1 \cdot \frac{n_2}{n_1} \][/tex]
Substitute the given values into the equation:
[tex]\[ V_2 = 13.5 \, \text{L} \cdot \frac{3.00 \, \text{moles}}{1.50 \, \text{moles}} \][/tex]
Calculate the fraction:
[tex]\[ \frac{3.00 \, \text{moles}}{1.50 \, \text{moles}} = 2 \][/tex]
Now multiply by the initial volume:
[tex]\[ V_2 = 13.5 \, \text{L} \cdot 2 = 27.0 \, \text{L} \][/tex]
Therefore, the new volume of the container when the number of moles of air is doubled to 3.00 moles is [tex]\( 27.0 \)[/tex] liters.
