Question

If 3.31 moles of argon gas occupies a volume of 100 L what volume does 13.15 moles of argon occupy under the same temperature and pressure

Answer 1

Answer:

397 L

Explanation:

Recall the ideal gas law:

$\displaystyle PV = nRT$

If temperature and pressure stays constant, we can rearrange all constant variables onto one side of the equation:

$\displaystyle \frac{P}{RT} = \frac{n}{V}$

The left-hand side is simply some constant. Hence, we can write that:

$\displaystyle \frac{n_1}{V_1} = \frac{n_2}{V_2}$

Substitute in known values:

$\displaystyle \frac{(3.31 \text{ mol})}{(100 \text{ L})} = \frac{(13.15\text{ mol })}{V_2}$

Solving for V₂ yields:

$\displaystyle V_2 = \frac{(100 \text{ L})(13.15)}{3.31} = 397 \text{ L}$

In conclusion, 13.15 moles of argon will occupy 397* L under the same temperature and pressure.

(Assuming 100 L has three significant figures.)