Question

final volume of oxygen sample when pressure is changed from 575mm Hg to 1.25atm

Answer 1

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.