Answer :
To determine the final volume of the gas in the container, we'll use Boyle's Law, which states that the pressure and volume of a gas are inversely proportional when the temperature remains constant. Mathematically, Boyle's Law 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,
- [tex]\( V_2 \)[/tex] is the final volume.
We are given the following:
- Initial volume [tex]\( V_1 = 3.0 \text{ L} \)[/tex]
- Initial pressure [tex]\( P_1 = 150 \text{ kPa} \)[/tex]
- Final pressure [tex]\( P_2 = 2 \text{ atm} \)[/tex]
However, to use the formula properly, the pressures must be in the same units. We convert the initial pressure from kilopascals (kPa) to atmospheres (atm). The conversion factor is:
[tex]\[ 1 \text{ atm} = 101.325 \text{ kPa} \][/tex]
Thus,
[tex]\[ P_1 = \frac{150 \text{ kPa}}{101.325 \text{ kPa/atm}} \approx 1.480 \text{ atm} \][/tex]
Now we can use Boyle's Law:
[tex]\[ P_1 \times V_1 = P_2 \times V_2 \][/tex]
Substitute the known values:
[tex]\[ 1.480 \text{ atm} \times 3.0 \text{ L} = 2 \text{ atm} \times V_2 \][/tex]
Solving for [tex]\( V_2 \)[/tex]:
[tex]\[ V_2 = \frac{1.480 \text{ atm} \times 3.0 \text{ L}}{2 \text{ atm}} \][/tex]
[tex]\[ V_2 \approx \frac{4.440 \text{ atm} \cdot \text{L}}{2 \text{ atm}} \][/tex]
[tex]\[ V_2 \approx 2.220 \text{ L} \][/tex]
Therefore, the final volume of the gas when the pressure increases to [tex]\( 2 \text{ atm} \)[/tex] is approximately [tex]\( 2.22 \text{ L} \)[/tex]. Out of the given choices, the closest one is:
[tex]\[ 2.25 \text{L} \][/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,
- [tex]\( V_2 \)[/tex] is the final volume.
We are given the following:
- Initial volume [tex]\( V_1 = 3.0 \text{ L} \)[/tex]
- Initial pressure [tex]\( P_1 = 150 \text{ kPa} \)[/tex]
- Final pressure [tex]\( P_2 = 2 \text{ atm} \)[/tex]
However, to use the formula properly, the pressures must be in the same units. We convert the initial pressure from kilopascals (kPa) to atmospheres (atm). The conversion factor is:
[tex]\[ 1 \text{ atm} = 101.325 \text{ kPa} \][/tex]
Thus,
[tex]\[ P_1 = \frac{150 \text{ kPa}}{101.325 \text{ kPa/atm}} \approx 1.480 \text{ atm} \][/tex]
Now we can use Boyle's Law:
[tex]\[ P_1 \times V_1 = P_2 \times V_2 \][/tex]
Substitute the known values:
[tex]\[ 1.480 \text{ atm} \times 3.0 \text{ L} = 2 \text{ atm} \times V_2 \][/tex]
Solving for [tex]\( V_2 \)[/tex]:
[tex]\[ V_2 = \frac{1.480 \text{ atm} \times 3.0 \text{ L}}{2 \text{ atm}} \][/tex]
[tex]\[ V_2 \approx \frac{4.440 \text{ atm} \cdot \text{L}}{2 \text{ atm}} \][/tex]
[tex]\[ V_2 \approx 2.220 \text{ L} \][/tex]
Therefore, the final volume of the gas when the pressure increases to [tex]\( 2 \text{ atm} \)[/tex] is approximately [tex]\( 2.22 \text{ L} \)[/tex]. Out of the given choices, the closest one is:
[tex]\[ 2.25 \text{L} \][/tex]
