Answer
The final volume is equal to 14.6 L
Work Step by Step
1. Using Avogadro's law, solve for "$V_2$":
$\frac{V_1}{n_1} = \frac{V_2}{n_2}$
- Multiply both sides by "$n_2$"
$\frac{V_1}{n_1} \times n_2 = {V_2}$
2. Calculate the number of moles in 25.0 g of $Ne$:
Molar mass (Ne) : 20.18 g/mole
$25.0 \space g \space (Ne) \times \frac{1 \space moles \space (Ne)}{20.18 \space g \space (Ne)} = 1.24 \space moles \space (Ne)$
3. Since 1.24 moles of Ne atoms were added, the final number of moles is equal to "$n_1 + 1.24 \space moles$".
$n_2 = n_1 + 1.24 \space moles$
$n_2 = 1.50 \space moles + 1.24 \space moles = 2.74 \space moles$
4. Calculate the value of $V_2$:
$\frac{8.00 \space L}{1.50 \space moles} \times 2.74 \space moles = 14.6 \space L$
