To find the final volume of an oxygen sample when the pressure changes, we can use Boyle's Law, which states that for a gas at constant temperature, the product of pressure and volume is constant. Mathematically, this can be expressed as:
\[ P_1 V_1 = P_2 V_2 \]
Where:
- \( P_1 \) = initial pressure
- \( V_1 \) = initial volume
- \( P_2 \) = final pressure
- \( V_2 \) = final volume
First, we need to ensure that the pressures are in the same units. We have:
- \( P_1 = 575 , \text{mm Hg} \)
- \( P_2 = 1.25 , \text{atm} \)
Now, we will convert \( P_2 \) into mm Hg. We know that:
1 atm = 760 mm Hg
So, let's convert \( P_2 \):
\[ P_2 = 1.25 , \text{atm} \times 760 , \text{mm Hg/atm} = 950 , \text{mm Hg} \]
Now we have:
- \( P_1 = 575 , \text{mm Hg} \)
- \( P_2 = 950 , \text{mm Hg} \)
Now, let's apply Boyle's law. If we let \( V_1 \) be the initial volume and \( V_2 \) be the final volume, we can express the relationship as:
\[ 575 , \text{mm Hg} \cdot V_1 = 950 , \text{mm Hg} \cdot V_2 \]
To find \( V_2 \) in terms of \( V_1 \):
\[ V_2 = \frac{575 , \text{mm Hg} \cdot V_1}{950 , \text{mm Hg}} \]
Now, simplifying:
\[ V_2 = \frac{575}{950} \cdot V_1 \]
Calculating the fraction:
\[ V_2 = 0.6053 \cdot V_1 \]
This means the final volume \( V_2 \) is approximately 60.53% of the initial volume \( V_1 \) when the pressure is increased from 575 mm Hg to 950 mm Hg (which is equivalent to 1.25 atm).
You can plug in the value of \( V_1 \) (the initial volume) to find \( V_2 \). If you have a specific initial volume, please provide it for a numeric answer.
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