A gas that occupies a volume of 6.75L at 89.0atm will occupy what volume at 68.55mm Hg if the temperature remains constant?
To solve this problem using Boyle's Law, we can use the formula:
P₁V₁ = P₂V₂
Where:
P₁ = initial pressure
V₁ = initial volume
P₂ = final pressure
V₂ = final volume
Given:
P₁ = 89.0 atm
V₁ = 6.75 L
P₂ = 68.55 mm Hg
Note: We need to convert the pressure units to use the same units throughout the equation.
Converting mm Hg to atm:
1 atm = 760 mm Hg
P₂ = 68.55 mm Hg / 760 mm Hg/atm
P₂ = 0.090 ATM (approx.)
Now, we can substitute the given values into the formula and solve for V₂:
P₁V₁ = P₂V₂
(89.0 atm)(6.75 L) = (0.090 atm)(V₂)
V₂ = (89.0 atm)(6.75 L) / (0.090 atm)
V₂ ≈ 6693.75 L
Therefore, the gas would occupy approximately 6693.75L at 68.55 mm Hg if the temperature remains constant.
To determine the final volume of the gas at a different pressure, we can use the combined gas law equation:
(P1 * V1) / T1 = (P2 * V2) / T2
Where:
P1 = initial pressure
V1 = initial volume
T1 = initial temperature
P2 = final pressure
V2 = final volume
T2 = final temperature
In this case, we are assuming that the temperature remains constant, meaning T1 = T2.
Given values:
P1 = 89.0 atm
V1 = 6.75 L
P2 = 68.55 mm Hg
First, we need to convert the pressure from mm Hg to atm:
1 atm = 760 mm Hg
P2 = 68.55 mm Hg / 760 mm Hg/atm
P2 = 0.0902 atm
Now we can substitute the known values into the combined gas law equation:
(89.0 atm * 6.75 L) / T1 = (0.0902 atm * V2) / T1
We cancel out T1 from both sides of the equation, assuming constant temperature:
89.0 atm * 6.75 L = 0.0902 atm * V2
Now we can solve for V2:
V2 = (89.0 atm * 6.75 L) / 0.0902 atm
V2 ≈ 6639.6 L
Therefore, the gas will occupy approximately 6639.6 liters at a pressure of 68.55 mm Hg, assuming the temperature remains constant.