A sample of a gas has a volume of 250. mL at 2.10 atm and 227 oC. What volume will it have at standard temperature and pressure (STP)?
(P1V1/T1) = (P2V2/T2)
To find the volume of the gas at standard temperature and pressure (STP), you can use the combined gas law equation:
(P₁ x V₁) / (T₁) = (P₂ x V₂) / (T₂)
Where:
P₁ = initial pressure of the gas (2.10 atm)
V₁ = initial volume of the gas (250 mL)
T₁ = initial temperature of the gas (227 °C + 273.15 = 500.15 K) (Converted to Kelvin)
P₂ = final pressure at STP (1 atm)
V₂ = final volume at STP (We need to find this)
T₂ = final temperature at STP (273.15 K)
Let's plug in these values and solve for V₂:
(2.10 atm x 250 mL) / (500.15 K) = (1 atm x V₂) / (273.15 K)
Now, let's rearrange the equation to solve for V₂:
(2.10 atm x 250 mL x 273.15 K) / (500.15 K) = 1 atm x V₂
Now, we can calculate V₂:
V₂ = (2.10 atm x 250 mL x 273.15 K) / (500.15 K)
V₂ ≈ 289 mL
Therefore, the volume of the gas at standard temperature and pressure (STP) is approximately 289 mL.