a 16 gram sample of oxygen has a volume of 11.2L at standard pressure . what is the temp of the oxygen ?
Use PV = nRT
n = grams/molar mass O2
Solve for T (in kelvin)
so when i get my answer
i add 273 to it?
Maybe yes, maybe no, but definitely you don't add 273.
If you want T in kelvin that is your answer. If you want it in degrees C, you subtract 273 from the calculation.
K = 273 + C
To find the temperature of the oxygen, we can 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, we need to convert the given mass of oxygen into moles. To do this, we'll use the molar mass of oxygen, which is 32 g/mol.
Given mass of oxygen = 16 grams
Molar mass of oxygen = 32 g/mol
Number of moles of oxygen = (Given mass of oxygen) / (Molar mass of oxygen)
Number of moles of oxygen = 16 g / 32 g/mol
Number of moles of oxygen = 0.5 mol
Now, let's rearrange the ideal gas law equation to solve for temperature:
T = (PV) / (nR)
Substituting the given values into the equation:
P = standard pressure = 1 atm
V = volume = 11.2 L
n = number of moles = 0.5 mol
R = ideal gas constant = 0.0821 L·atm/(mol·K)
T = (1 atm * 11.2 L) / (0.5 mol * 0.0821 L·atm/(mol·K))
Simplifying the equation:
T = 22.4 / 0.04105
T ≈ 545.3 K
Therefore, the temperature of the oxygen is approximately 545.3 Kelvin.