A simple of oxygen gas at room temperature and 1.03atm occupies a volume of 22dm³. Find the volume of the oxygen gas at STP?
(P1V1/T1) = (P2V2/T2)
Remember T must be in kelvin. Also you must decide what room T is.
To find the volume of the oxygen gas at STP (Standard Temperature and Pressure), we need to apply the ideal gas law. The ideal gas law equation is:
PV = nRT
Where:
P = pressure (in atmospheres)
V = volume (in liters)
n = number of moles
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature (in Kelvin)
Given:
P = 1.03 atm
V = 22 dm³ (since 1 dm³ = 1 L)
T = ?
n = ?
To find the number of moles, we can use the ideal gas law equation:
n = PV / RT
Let's assume the temperature of the gas is the same at both conditions, which is room temperature (approximately 298 K). Substituting the given values into the equation:
n = (1.03 atm)(22 L) / (0.0821 L·atm/mol·K)(298 K)
n ≈ 0.937 moles
Now that we know the number of moles, we can find the volume at STP using the ideal gas law:
PV = nRT
Considering STP conditions (standard temperature 0°C or 273 K, and standard pressure 1 atm), we can rewrite the equation:
P₁V₁ / T₁ = P₂V₂ / T₂
P₁ = 1.03 atm
V₁ = 22 L
T₁ = 298 K
P₂ = 1 atm (STP)
V₂ = ? (to be determined)
T₂ = 273 K (STP)
Substituting the values:
(1.03 atm)(22 L) / (298 K) = (1 atm)(V₂) / (273 K)
Simplifying the equation:
(22 L)(1.03 atm)(273 K) = (1 atm)(V₂)(298 K)
Solving for V₂:
V₂ = (22 L)(1.03 atm)(273 K) / (1 atm)(298 K)
V₂ ≈ 27.5 L
Therefore, the volume of the oxygen gas at STP is approximately 27.5 liters.