Which equation is set up correctly to determine the volume of a 3.2 mole sample of oxygen gas at 50°C and 101.325 kPa?
V=(3.2 mol)(8.314 L⋅kPa/K⋅mol)(323 K)101.325 kPa
V=(3.2 mol)(8.314 L⋅kPa/K⋅mol)(50∘C)1 atm
V=(8.314 L⋅kPa/K⋅mol)(50∘C)(3.2 mol)(101.325 kPa)
V=(3.2 mol)(8.314 L⋅kPa/K⋅mol)(1 atm)(323 K)
V=(8.314 L⋅kPa/K⋅mol)(50∘C)(3.2 mol)(101.325 kPa)
The correct equation to determine the volume of a 3.2 mole sample of oxygen gas at 50°C and 101.325 kPa is:
V=(3.2 mol)(8.314 L⋅kPa/K⋅mol)(323 K)101.325 kPa
To determine the volume of a gas sample, we can use the ideal gas law equation, which states:
PV = nRT
Where:
P is the pressure of the gas
V is the volume of the gas
n is the number of moles of the gas
R is the ideal gas constant
T is the temperature of the gas in Kelvin
In the given question, we are given the following information:
Number of moles (n) = 3.2 moles
Temperature (T) = 50°C (convert to Kelvin by adding 273.15) = 323 K
Pressure (P) = 101.325 kPa
Now we need to determine the correct equation that uses the given information and the ideal gas law equation:
V = (3.2 mol)(8.314 L·kPa/K·mol)(323 K) / 101.325 kPa
This equation correctly sets up the calculation to determine the volume of the 3.2-mole sample of oxygen gas at 50°C and 101.325 kPa.
Therefore, the correct equation is:
V = (3.2 mol)(8.314 L·kPa/K·mol)(323 K) / 101.325 kPa