During an experiment, 17.5 mL of oxygen gas were collected over water at room temperature (25 C) and 100.2 kPa of atmospheric pressure. The vapor pressure of water at this temperature is 2.6 kPa.
a) What is the pressure of the "dry" oxygen gas?
b) How many moles of oxygen gas were produced?
c) What is the molar volume of oxygen gas at the conditions in the laboratory?
1. Ptotal = pO2 + pH2O
100.2 = pO2 + 2.6
Solve for pO2.
2. Use PV = nRT and solve for n.
3. Convert mols you have to what volume would be 1 mol.
Post your work if you get stuck.
a) Well, it seems like the oxygen gas needs a towel to dry off! Since the vapor pressure of water is 2.6 kPa, we need to subtract that from the total pressure of 100.2 kPa. So, the pressure of the "dry" oxygen gas is 100.2 kPa - 2.6 kPa = 97.6 kPa.
b) Alright, let's calculate the number of moles of oxygen gas. We can use the ideal gas law here: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin.
Since we have the pressure (97.6 kPa), the volume (17.5 mL), and the temperature (25 C or 298 K), we can rearrange the formula to solve for moles: n = (PV) / (RT).
Plugging in the values, we get n = (97.6 kPa * 17.5 mL) / (8.314 J/(mol·K) * 298 K).
Now, we need to remember that 1 L = 1000 mL, so we have to convert the volume to liters: 17.5 mL = 0.0175 L.
Calculating further, n = (97.6 kPa * 0.0175 L) / (8.314 J/(mol·K) * 298 K).
By canceling out units, we find n ≈ 0.00914 moles of oxygen gas.
c) To determine the molar volume of oxygen gas at these conditions, we can divide the volume (0.0175 L) by the number of moles (0.00914 moles). So, the molar volume of oxygen gas is approximately 1.91 L/mol.
Keep in mind that my calculations are all for educational purposes, and humorously so. If you need precise results, do the math carefully!
a) To find the pressure of the "dry" oxygen gas, we need to subtract the vapor pressure of water from the total pressure.
Pressure of "dry" oxygen gas = Total pressure - Vapor pressure of water
Pressure of "dry" oxygen gas = 100.2 kPa - 2.6 kPa
Pressure of "dry" oxygen gas = 97.6 kPa
b) To find the number of moles of oxygen gas produced, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure (in Pa)
V = volume (in m^3)
n = number of moles
R = ideal gas constant (8.314 J/(mol·K))
T = temperature (in Kelvin)
First, we need to convert the volume of oxygen gas collected to m^3:
Volume of oxygen gas = 17.5 mL = 0.0175 L = 0.0175 dm^3 = 0.0175 m^3
Now we can rearrange the ideal gas law equation to solve for the number of moles:
n = PV / RT
n = (Pressure of "dry" oxygen gas) * (Volume of oxygen gas) / (R * Temperature)
Since the temperature is given in Celsius, we need to convert it to Kelvin:
Temperature in Kelvin = 25 Celsius + 273.15
Temperature in Kelvin = 298.15 K
n = (97.6 kPa) * (0.0175 m^3) / ((8.314 J/(mol·K)) * 298.15 K)
Now we can calculate the number of moles of oxygen gas.
c) The molar volume of a gas is the volume occupied by one mole of the gas at a particular temperature and pressure. To calculate the molar volume, we can use the relationship between the number of moles, volume, and molar volume:
Molar volume = Volume of oxygen gas / Number of moles
Molar volume = (0.0175 m^3) / (number of moles from part b)
Now we can calculate the molar volume of oxygen gas.
To find the answers to these questions, we can use the ideal gas law equation:
PV = nRT
Where:
P = Pressure (in Pa)
V = Volume (in m³)
n = Number of moles
R = Ideal gas constant (8.314 J/mol·K)
T = Temperature (in Kelvin)
Now let's solve each question step by step:
a) What is the pressure of the "dry" oxygen gas?
To find the pressure of the dry oxygen gas, we need to subtract the vapor pressure of water from the total pressure. The formula for this is:
P(dry) = P(total) - P(water)
Given values:
P(total) = 100.2 kPa
P(water) = 2.6 kPa
First, let's convert the given values to Pa:
P(total) = 100.2 kPa * 1000 Pa/kPa = 100,200 Pa
P(water) = 2.6 kPa * 1000 Pa/kPa = 2,600 Pa
Now we can substitute the values into the formula:
P(dry) = 100,200 Pa - 2,600 Pa
P(dry) = 97,600 Pa
Therefore, the pressure of the "dry" oxygen gas is 97,600 Pa.
b) How many moles of oxygen gas were produced?
To find the number of moles, we need to rearrange the ideal gas law equation to solve for n:
n = PV / RT
Given values:
P = P(dry) = 97,600 Pa
V = 17.5 mL = 17.5 cm³ = 17.5 * 10^(-6) m³ (converted to m³)
R = 8.314 J/mol·K
T = 25 °C = 25 + 273.15 K (converted to Kelvin)
Now let's substitute the values into the formula:
n = (97,600 Pa * 17.5 * 10^(-6) m³) / (8.314 J/mol·K * (25 + 273.15) K)
n = 0.000077527 moles
Therefore, approximately 0.0000775 moles of oxygen gas were produced.
c) What is the molar volume of oxygen gas at the conditions in the laboratory?
The molar volume of a gas is the volume occupied by one mole of the gas at a given temperature and pressure. It can be calculated using the formula:
Molar volume = V / n
Given values:
V = 17.5 mL = 17.5 cm³ = 17.5 * 10^(-6) m³ (converted to m³)
n = 0.0000775 moles (approximated value from the previous question)
Now let's substitute the values into the formula:
Molar volume = (17.5 * 10^(-6) m³) / 0.0000775 moles
Molar volume = 0.226774 m³/mol
Therefore, the molar volume of oxygen gas at the conditions in the laboratory is approximately 0.2268 m³/mol.