WHAT MASS IN GRAMS OF OXYGEN GAS IS CONTAINED IN C 8.5L TANK AT 32 DEGREE CELSIUS AND 3.23 ATMOSPHERE
17.5989
To determine the mass of oxygen gas in grams contained in a tank, we need to use the Ideal Gas Law equation:
PV = nRT
Where:
P = pressure (in atm)
V = volume (in liters)
n = number of moles
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature (in Kelvin)
First, let's convert the temperature from Celsius to Kelvin:
T(K) = T(°C) + 273.15
T(K) = 32 + 273.15
T(K) = 305.15 K
Now, we can substitute the given values into the equation:
PV = nRT
(3.23 atm) * (8.5 L) = n * (0.0821 L·atm/mol·K) * (305.15 K)
(27.455 atm·L) = n * (25.010515 L·atm/mol)
To solve for moles (n), we divide both sides by (25.010515 L·atm/mol):
n = (27.455 atm·L) / (25.010515 L·atm/mol)
n = 1.0974 mol
Now, we can calculate the mass using the molar mass of oxygen gas:
molar mass of oxygen (O2) = 32 g/mol
mass = n * molar mass
mass = (1.0974 mol) * (32 g/mol)
mass = 35.11 g
Therefore, the mass of oxygen gas in grams contained in the 8.5 L tank at 32 degrees Celsius and 3.23 atmospheres is approximately 35.11 grams.