To solve this problem, we can use Boyle’s Law, which states that the pressure of a gas times its volume is constant when the temperature and the number of moles of gas remain unchanged. 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.
Given:
- [tex]\(P_1 = 6.20 \,\text{atm}\)[/tex]
- [tex]\(P_2 = 9.150 \,\text{atm}\)[/tex]
- [tex]\(V_2 = 322 \,\text{mL}\)[/tex]
- [tex]\(V_1 = ? \,\text{mL}\)[/tex]
We need to find the initial volume [tex]\(V_1\)[/tex].
1. Write down Boyle’s Law equation:
[tex]\[
P_1 \times V_1 = P_2 \times V_2
\][/tex]
2. Substitute the given values into the equation:
[tex]\[
6.20 \times V_1 = 9.150 \times 322
\][/tex]
3. Calculate the right side of the equation:
[tex]\[
9.150 \times 322 = 2943.30
\][/tex]
4. Now the equation looks like this:
[tex]\[
6.20 \times V_1 = 2943.30
\][/tex]
5. Solve for [tex]\(V_1\)[/tex] by dividing both sides of the equation by 6.20:
[tex]\[
V_1 = \frac{2943.30}{6.20}
\][/tex]
6. Perform the division to find [tex]\(V_1\)[/tex]:
[tex]\[
V_1 \approx 475.21 \,\text{mL}
\][/tex]
Therefore, the initial volume [tex]\(V_1\)[/tex] of the gas is approximately [tex]\(475.21 \,\text{mL}\)[/tex].
