Sure, let's determine the pressure when the volume of the gas is decreased.
We start with Boyle's Law, which states that for a given mass of gas at constant temperature, the pressure of the gas varies inversely with its volume. Mathematically, Boyle's Law can be 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.
Given:
- [tex]\( P_1 = 76.5 \)[/tex] kPa,
- [tex]\( V_1 = 4.31 \)[/tex] L,
- [tex]\( V_2 = 2.6 \)[/tex] L.
We need to find [tex]\( P_2 \)[/tex].
Rearrange the equation to solve for [tex]\( P_2 \)[/tex]:
[tex]\[ P_2 = \frac{P_1 \times V_1}{V_2} \][/tex]
Substitute the given values into the equation:
[tex]\[ P_2 = \frac{76.5 \, \text{kPa} \times 4.31 \, \text{L}}{2.6 \, \text{L}} \][/tex]
Calculate the numerator:
[tex]\[ 76.5 \, \text{kPa} \times 4.31 \, \text{L} = 329.715 \, \text{kPa} \cdot \text{L} \][/tex]
Now divide by the final volume [tex]\( V_2 \)[/tex]:
[tex]\[ P_2 = \frac{329.715 \, \text{kPa}}{2.6 \, \text{L}} \][/tex]
Simplify the division:
[tex]\[ P_2 = 126.81346153846152 \, \text{kPa} \][/tex]
Rounded to a reasonable number of significant figures based on the given data:
[tex]\[ P_2 \approx 126.81 \, \text{kPa} \][/tex]
Therefore, the final pressure of the gas when the volume is decreased to 2.6 L is approximately 126.81 kPa.
