A helium-filled balloon contains 143 mL of
gas at a pressure of 0.99 atm. What volume
will the gas occupy at standard pressure?
To find the volume the gas will occupy at standard pressure, we can use Boyle's Law, which states that the volume of a gas is inversely proportional to its pressure when the temperature is constant.
The formula for Boyle's Law is: P₁V₁ = P₂V₂
P₁ and V₁ represent the initial pressure and volume, while P₂ and V₂ represent the final pressure and volume.
Given:
P₁ = 0.99 atm
V₁ = 143 mL
P₂ = Standard pressure (1 atm)
V₂ = ?
Now we can plug the values into the formula and solve for V₂:
0.99 atm * 143 mL = 1 atm * V₂
Simplifying the equation:
143 mL = V₂
Therefore, the volume the gas will occupy at standard pressure is 143 mL.
To find the volume of gas at standard pressure, we can use the combined gas law, which states that the ratio of initial pressure to the initial volume is equal to the ratio of final pressure to the final volume.
The combined gas law formula is:
(P1 * V1) / T1 = (P2 * V2) / T2
In this case, we are given the initial pressure (P1) as 0.99 atm and the initial volume (V1) as 143 mL. We need to find the final volume (V2) at standard pressure, which is 1 atm.
The standard temperature is usually 273.15 K, but for simplicity, we will assume the temperature remains constant throughout.
Now, we can rearrange the formula to solve for V2:
V2 = (P1 * V1 * T2) / (P2 * T1)
Substituting the known values into the equation:
V2 = (0.99 atm * 143 mL * 273.15 K) / (1 atm * 273.15 K)
Note that the temperature unit (Kelvin) cancels out.
V2 = (0.99 * 143) mL
Calculating this, we find:
V2 = 141.57 mL
Therefore, the volume of the gas at standard pressure will be 141.57 mL.