What volume will 12.0 g of oxygen gas occupy at 25 degrees celcius and a pressure of 52.7 kPa?
Use PV = nRT. n = number of mols = grams/molar mass. Remember T must be in kelvin and note the correct spelling of celsius. I would convert kPa to atm but that isn't absolutely necessary if you use the correct R as a constant.
To solve this problem, we need to use the ideal gas law equation:
PV = nRT
where:
P = pressure (in Pa)
V = volume (in m³)
n = number of moles of gas
R = ideal gas constant (8.314 J/(K*mol))
T = temperature (in K)
However, before we can use this equation, we need to convert the given values to suitable units.
Given:
Pressure (P) = 52.7 kPa = 52,700 Pa
Temperature (T) = 25 degrees Celsius = 25 + 273.15 = 298.15 K
The molar mass of oxygen (O₂) is approximately 32.00 g/mol.
To find the number of moles (n) of oxygen gas, we can use the formula:
n = mass / molar mass
n = 12.0 g / 32.00 g/mol
n = 0.375 mol
Now, we can substitute the known values into the ideal gas law equation and solve for the volume (V):
PV = nRT
V = (nRT) / P
V = (0.375 mol * 8.314 J/(K*mol) * 298.15 K) / 52,700 Pa
V ≈ 0.004 m³ (rounded to three decimal places)
Therefore, 12.0 g of oxygen gas will occupy approximately 0.004 m³ at 25 degrees Celsius and a pressure of 52.7 kPa.
To find the volume of the oxygen gas, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure (in kPa)
V = volume (in liters)
n = moles of gas
R = ideal gas constant (0.0821 L·atm/(mol·K) or 8.31 J/(mol·K))
T = temperature (in Kelvin)
First, we need to convert the given values into the appropriate units.
Pressure: 52.7 kPa
Temperature: 25 degrees Celsius
To convert the temperature from Celsius to Kelvin, we add 273.15:
T in Kelvin = 25 + 273.15 = 298.15 K
Next, we need to calculate the number of moles of oxygen gas. To do this, we use the formula:
n = m/M
Where:
m = mass of gas (in grams)
M = molar mass of gas (in g/mol)
The molar mass of oxygen (O2) is approximately 32 g/mol.
n = 12.0 g / 32 g/mol ≈ 0.375 mol
Now we have all the values needed to solve for the volume of the gas. Rearranging the ideal gas law equation to solve for V:
V = (nRT) / P
Substituting the known values:
V = (0.375 mol * 0.0821 L·atm/(mol·K) * 298.15 K) / 52.7 kPa
Now we need to convert kPa to atm, since the ideal gas constant is in atm. 1 atm is approximately equal to 101.325 kPa:
V = (0.375 mol * 0.0821 L·atm/(mol·K) * 298.15 K) / (52.7 kPa * (1 atm / 101.325 kPa))
Now we can calculate the volume:
V ≈ 8.24 L
Therefore, 12.0 g of oxygen gas will occupy approximately 8.24 liters at 25 degrees Celsius and a pressure of 52.7 kPa.