A sample of oxygen that occupies 1.00x10^6 mL at 575 mm Hg is subjected to a pressure of 1.36 atm. What will the final volume of the sample be if the temperature is held constant?
How would I set up this equation?
P1V1 = P2V2
Note that you MUST use the same units; i.e., one P can't be in atm and the other in mm Hg.
To set up this equation, we can use the combined gas law equation, which relates the initial and final conditions of pressure and volume while keeping the temperature constant. The formula is as follows:
(P1 * V1) / T1 = (P2 * V2) / T2
Where:
P1 and P2 are the initial and final pressures, respectively.
V1 and V2 are the initial and final volumes, respectively.
T1 and T2 are the initial and final temperatures, respectively.
Since the temperature is held constant in this scenario, T1 and T2 will be the same, which allows us to simplify the equation to:
(P1 * V1) = (P2 * V2)
Now let's substitute the given values into the equation:
P1 = 575 mmHg (which needs to be converted to atm to match the units of P2)
V1 = 1.00x10^6 mL (which needs to be converted to L to match the units of V2)
P2 = 1.36 atm
V2 = ?
First, we need to convert mmHg to atm. Since 1 atm = 760 mmHg, we can use this conversion factor:
575 mmHg * (1 atm / 760 mmHg) = 0.757 atm
Next, we need to convert mL to L. Since 1 L = 1000 mL, we can use this conversion factor:
1.00x10^6 mL * (1 L / 1000 mL) = 1000 L
Now we can substitute the values into the equation:
(0.757 atm * 1000 L) = (1.36 atm * V2)
Simplifying:
757 atm * L = 1.36 atm * V2
Now, solve for V2:
V2 = (757 atm * L) / 1.36 atm
Calculating the final volume, we get:
V2 ≈ 557.8 L
Therefore, the final volume of the sample, when subjected to a pressure of 1.36 atm while keeping the temperature constant, will be approximately 557.8 L.