The volume of a gas is 250 mL at 340.0 kPa pressure. What will the volume be when the pressure is reduced to 50.0 kPa, assuming the temperature remains constant?
(P1V1) = (P2V2)
To find the volume of the gas when the pressure is reduced, we can use Boyle's Law. Boyle's Law states that the pressure and volume of a gas are inversely proportional, assuming the temperature remains constant.
Boyle's Law can be mathematically represented as:
P₁V₁ = P₂V₂
Where:
P₁ and P₂ are the initial and final pressures, respectively.
V₁ and V₂ are the initial and final volumes, respectively.
In this case, the initial volume (V₁) is given as 250 mL, the initial pressure (P₁) is 340.0 kPa, and the final pressure (P₂) is 50.0 kPa. We need to find the final volume (V₂).
Using Boyle's Law equation, we can rearrange it to solve for V₂:
V₂ = (P₁ * V₁) / P₂
Substituting the values we have:
V₂ = (340.0 kPa * 250 mL) / 50.0 kPa
First, let's convert mL to L to ensure consistency:
V₂ = (340.0 kPa * 0.250 L) / 50.0 kPa
Cancelling units:
V₂ = (340.0 * 0.250) / 50.0
V₂ = 1.7 L
Therefore, when the pressure is reduced to 50.0 kPa, assuming the temperature remains constant, the volume of the gas will be 1.7 L.