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?
Boyle's law is perfect for this:
P1*V1=P2*V2
To find the volume of the gas when the pressure is reduced to 50.0 kPa, 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 expressed as:
P₁V₁ = P₂V₂
Where:
P₁ is the initial pressure (340.0 kPa)
V₁ is the initial volume (250 mL)
P₂ is the final pressure (50.0 kPa)
V₂ is the final volume (unknown)
We can rearrange the equation to solve for V₂:
V₂ = (P₁V₁) / P₂
Now, let's substitute the given values into the equation to find the final volume:
V₂ = (340.0 kPa * 250 mL) / 50.0 kPa
First, let's convert the volume from milliliters (mL) to liters (L) since the pressure is given in kilopascals (kPa). To convert from mL to L, divide the volume by 1000:
V₂ = (340.0 kPa * 0.250 L) / 50.0 kPa
Now, we can simplify the equation:
V₂ = 8.5 L / 50.0 kPa
Dividing L by kPa, we get the final volume in the unit of liters per kilopascal:
V₂ ≈ 0.17 L/kPa
Therefore, the volume of the gas when the pressure is reduced to 50.0 kPa, assuming constant temperature, is approximately 0.17 L.