To find the volume required to hold 2.0 moles of gas under the same conditions of pressure and temperature, we can use the concept of direct proportionality between volume and the number of moles in the context of the ideal gas law (PV = nRT). When the pressure (P) and temperature (T) are constant, the volume (V) of the gas is directly proportional to the number of moles (n).
Given:
- Initial moles of gas, [tex]\( n_1 = 5.0 \)[/tex] moles
- Initial volume of gas, [tex]\( V_1 = 15.0 \)[/tex] liters
- Final moles of gas, [tex]\( n_2 = 2.0 \)[/tex] moles
We can use the relationship:
[tex]\[
\frac{V_1}{n_1} = \frac{V_2}{n_2}
\][/tex]
Rearranging this to solve for the unknown volume [tex]\( V_2 \)[/tex], we get:
[tex]\[
V_2 = V_1 \times \frac{n_2}{n_1}
\][/tex]
Substituting in the given values:
[tex]\[
V_2 = 15.0 \, \text{L} \times \frac{2.0 \, \text{moles}}{5.0 \, \text{moles}}
\][/tex]
Simplifying the expression:
[tex]\[
V_2 = 15.0 \times 0.4
\][/tex]
[tex]\[
V_2 = 6.0 \, \text{L}
\][/tex]
Therefore, the volume required to hold 2.0 moles of the gas under the same conditions is:
[tex]\[
V_2 = 6.0 \, \text{L}
\][/tex]
