Sure, let's solve this problem step-by-step.
Given:
- Initial moles of gas, [tex]\( n_1 = 0.50 \)[/tex] moles
- Initial volume, [tex]\( V_1 = 10.5 \)[/tex] liters
- Final moles of gas, [tex]\( n_2 = 2.0 \)[/tex] moles
We are asked to find the final volume [tex]\( V_2 \)[/tex].
We can use the relationship given by:
[tex]\[
\frac{n_1}{V_1} = \frac{n_2}{V_2}
\][/tex]
First, rearrange the equation to solve for [tex]\( V_2 \)[/tex]:
[tex]\[
V_2 = \frac{n_2 \cdot V_1}{n_1}
\][/tex]
Now, plug in the given values:
[tex]\[
V_2 = \frac{2.0 \, \text{moles} \cdot 10.5 \, \text{L}}{0.50 \, \text{moles}}
\][/tex]
Next, perform the calculation:
[tex]\[
V_2 = \frac{21.0 \, \text{L} \cdot \text{moles}}{0.50 \, \text{moles}}
\][/tex]
[tex]\[
V_2 = 42.0 \, \text{L}
\][/tex]
Therefore, the volume required to hold 2.0 moles of the gas under the same conditions is [tex]\( 42.0 \)[/tex] liters.
