Answer :
Certainly! To solve this problem, we apply Boyle's Law, which states that for a gas at constant temperature, the pressure of the gas is inversely proportional to its volume. Mathematically, this relationship is expressed as:
[tex]\[ P_1 \times V_1 = P_2 \times V_2 \][/tex]
Where:
- [tex]\( P_1 \)[/tex] is the initial pressure,
- [tex]\( V_1 \)[/tex] is the initial volume,
- [tex]\( P_2 \)[/tex] is the final pressure, and
- [tex]\( V_2 \)[/tex] is the final volume.
Given:
- Initial pressure, [tex]\( P_1 = 5.00 \, \text{atm} \)[/tex]
- Initial volume, [tex]\( V_1 = 5.00 \, \text{L} \)[/tex]
- Final volume, [tex]\( V_2 = 10.00 \, \text{L} \)[/tex]
We need to find the new pressure, [tex]\( P_2 \)[/tex].
Using the relationship from Boyle's Law, rearrange the formula to solve for [tex]\( P_2 \)[/tex]:
[tex]\[ P_2 = \frac{P_1 \times V_1}{V_2} \][/tex]
Substitute the known values into the equation:
[tex]\[ P_2 = \frac{5.00 \, \text{atm} \times 5.00 \, \text{L}}{10.00 \, \text{L}} \][/tex]
[tex]\[ P_2 = \frac{25.00 \, \text{atm \, L}}{10.00 \, \text{L}} \][/tex]
[tex]\[ P_2 = 2.50 \, \text{atm} \][/tex]
So, the new pressure of the gas after expansion is [tex]\( 2.50 \, \text{atm} \)[/tex].
Therefore, the correct answer is option d: [tex]\( 2.50 \, \text{atm} \)[/tex].
[tex]\[ P_1 \times V_1 = P_2 \times V_2 \][/tex]
Where:
- [tex]\( P_1 \)[/tex] is the initial pressure,
- [tex]\( V_1 \)[/tex] is the initial volume,
- [tex]\( P_2 \)[/tex] is the final pressure, and
- [tex]\( V_2 \)[/tex] is the final volume.
Given:
- Initial pressure, [tex]\( P_1 = 5.00 \, \text{atm} \)[/tex]
- Initial volume, [tex]\( V_1 = 5.00 \, \text{L} \)[/tex]
- Final volume, [tex]\( V_2 = 10.00 \, \text{L} \)[/tex]
We need to find the new pressure, [tex]\( P_2 \)[/tex].
Using the relationship from Boyle's Law, rearrange the formula to solve for [tex]\( P_2 \)[/tex]:
[tex]\[ P_2 = \frac{P_1 \times V_1}{V_2} \][/tex]
Substitute the known values into the equation:
[tex]\[ P_2 = \frac{5.00 \, \text{atm} \times 5.00 \, \text{L}}{10.00 \, \text{L}} \][/tex]
[tex]\[ P_2 = \frac{25.00 \, \text{atm \, L}}{10.00 \, \text{L}} \][/tex]
[tex]\[ P_2 = 2.50 \, \text{atm} \][/tex]
So, the new pressure of the gas after expansion is [tex]\( 2.50 \, \text{atm} \)[/tex].
Therefore, the correct answer is option d: [tex]\( 2.50 \, \text{atm} \)[/tex].
