What volume (in liters) is occupied by 0.01 moles of carbon monoxide, CO, at 9 degrees celcius and 0.973 atm?
Use the Ideal Gas Law:
PV = nRT
Substitute:
P = 0.973 atm
V = Unknown. Solve for this variable.
n = 0.01mol CO
R = 0.0821 L.atm/K.mol [a constant]
T = 9 + 273 = 282 K
To determine the volume occupied by 0.01 moles of carbon monoxide (CO) at a specific temperature and pressure, 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, let's convert the given temperature from Celsius to Kelvin by adding 273.15:
9 degrees Celsius + 273.15 = 282.15 Kelvin
Now we can plug in the values into the ideal gas law equation:
PV = nRT
0.973 atm * V = 0.01 moles * 0.0821 L·atm/mol·K * 282.15 K
0.973 V = 0.008442315 L·atm
To isolate the V (volume), divide both sides of the equation by 0.973:
V = 0.008442315 L·atm / 0.973
V ≈ 0.0087 liters
Therefore, approximately 0.01 moles of carbon monoxide (CO) at 9 degrees Celsius and 0.973 atm occupy a volume of 0.0087 liters.