300 mL of O2 are collected at a pressure of 645 mm of mercury (mm Hg). What volume will this have at one atmosphere pressure?
Use P1V1 = P2V2
P1 = 645 mm Hg
P1 = 760 mm Hg
V1 = 300 mL
V2 = ?
Post your work if you get stuck.
To find the volume of 300 mL of O2 at one atmosphere pressure, we need to use Boyle's Law, which states that the volume of a gas is inversely proportional to its pressure, assuming a constant temperature.
Boyle's Law equation is:
P1 x V1 = P2 x V2
Where:
P1 = Initial pressure = 645 mm Hg
V1 = Initial volume = 300 mL
P2 = Final pressure = 1 atm
V2 = Final volume (to be determined)
Since we want to find the final volume at one atmosphere pressure, we substitute the known values into the equation:
645 mm Hg x 300 mL = 1 atm x V2
Now, we can rearrange the equation to solve for V2:
V2 = (645 mm Hg x 300 mL) / 1 atm
First, let's convert the initial pressure of 645 mm Hg to atm:
1 atm = 760 mm Hg
So, we can rewrite the equation as:
V2 = (645 mm Hg x 300 mL) / 760 mm Hg
Now, we can calculate the final volume:
V2 = (645 mm Hg x 300 mL) / 760 mm Hg
V2 = 80625 / 760 mL
V2 ≈ 106.2 mL
Therefore, the volume of 300 mL of O2 at 1 atm pressure is approximately 106.2 mL.
To find the volume of the given oxygen at one atmosphere pressure, we can use Boyle's Law, which states that the product of pressure and volume is constant as long as temperature remains constant.
Boyle's Law formula:
P1 * V1 = P2 * V2
Where:
P1 = initial pressure (645 mm Hg)
V1 = initial volume (300 mL)
P2 = final pressure (one atmosphere pressure = 760 mm Hg)
V2 = final volume (to be determined)
Substituting the given values into the formula, we have:
(645 mm Hg) * (300 mL) = (760 mm Hg) * (V2)
To solve for V2, we can rearrange the equation as follows:
V2 = (645 mm Hg * 300 mL) / 760 mm Hg
V2 ≈ 255 mL
Therefore, the volume of the collected oxygen at one atmosphere pressure is approximately 255 mL.