Answer:
397 L
Explanation:
Recall the ideal gas law:
[tex]\displaystyle PV = nRT[/tex]
If temperature and pressure stays constant, we can rearrange all constant variables onto one side of the equation:
[tex]\displaystyle \frac{P}{RT} = \frac{n}{V}[/tex]
The left-hand side is simply some constant. Hence, we can write that:
[tex]\displaystyle \frac{n_1}{V_1} = \frac{n_2}{V_2}[/tex]
Substitute in known values:
[tex]\displaystyle \frac{(3.31 \text{ mol})}{(100 \text{ L})} = \frac{(13.15\text{ mol })}{V_2}[/tex]
Solving for V₂ yields:
[tex]\displaystyle V_2 = \frac{(100 \text{ L})(13.15)}{3.31} = 397 \text{ L}[/tex]
In conclusion, 13.15 moles of argon will occupy 397* L under the same temperature and pressure.
(Assuming 100 L has three significant figures.)
