A liter(1000cubed)of gas is under a pressure of 80.00cm of mercury . What pressure is needed in order to reduce the original volume of the gas to 600.0cm cubed?
P1V1 = P2V2
Substitute the numbers and solve for the one unknown.
To solve this problem, we can use Boyle's Law, which states that the pressure and volume of a gas are inversely proportional when the temperature and amount of gas remain constant.
We can set up the following equation using Boyle's Law:
P1 * V1 = P2 * V2
Where:
P1 = initial pressure
V1 = initial volume
P2 = final pressure (what we need to find)
V2 = final volume
Given the following values:
P1 = 80.00 cm Hg
V1 = 1000 cm³
V2 = 600.0 cm³
We can rearrange the equation to solve for P2:
P2 = (P1 * V1) / V2
Substituting the values:
P2 = (80.00 cm Hg * 1000 cm³) / 600.0 cm³
P2 = 133.33 cm Hg
Therefore, the pressure needed to reduce the original volume of the gas to 600.0 cm³ is approximately 133.33 cm Hg.
To solve this problem, we can use Boyle's Law, which states that the pressure and volume of a gas are inversely proportional when temperature is held constant.
Boyle's Law equation:
P1 * V1 = P2 * V2
Where:
P1 = initial pressure
V1 = initial volume
P2 = final pressure
V2 = final volume
Given:
P1 = 80.00 cm Hg
V1 = 1000 cm³
V2 = 600.0 cm³ (final volume)
We need to find P2, the pressure needed to reduce the original volume to 600.0 cm³.
Using Boyle's Law equation, we can rearrange it to solve for P2:
P2 = (P1 * V1) / V2
Substituting the given values:
P2 = (80.00 cm Hg * 1000 cm³) / 600.0 cm³
Simplifying the equation:
P2 = 80000 cm Hg / 600.0 cm³
Calculating the pressure:
P2 ≈ 133.33 cm Hg
Therefore, the pressure needed to reduce the original volume of the gas to 600.0 cm³ is approximately 133.33 cm Hg.