7.40 g of liquid methanol, CH3OH, are placed in an empty 20.0 L container at 45.0°C. What will be the pressure in the container when equilibrium is established? Assume the temperature remains constant and the volume of any liquid is negligible. The vapor pressure of methanol at 45.0°C is 205 mm Hg.
Won't the pressure be the vapor pressure of the methanol (assuming there is enough liquid methanol to establish equilibrium between liquid and vapor)?
You can calculate how much of the methanol will evaporate by
n = PV/RT. There will be some liquid left at equilibrium.
To find the pressure in the container when equilibrium is established, we need to consider the vapor pressure of methanol at the given temperature and the amount of methanol present in the container.
First, let's convert the given vapor pressure from millimeters of mercury (mm Hg) to atmospheres (atm).
1 atm = 760 mm Hg
So, the vapor pressure of methanol at 45.0°C is equivalent to 205 mm Hg / 760 mm Hg/atm ≈ 0.27 atm.
Next, we need to determine the number of moles of methanol present in the container. To do this, we'll use the molecular weight of methanol.
The molecular weight of methanol (CH3OH) can be calculated as follows:
C = 12.01 g/mol
H = 1.01 g/mol
O = 16.00 g/mol
Adding up the atomic weights: 12.01 + 3(1.01) + 16.00 = 32.04 g/mol
Now we can calculate the number of moles of methanol:
moles = mass / molecular weight
moles = 7.40 g / 32.04 g/mol ≈ 0.231 moles
Since methanol is a volatile liquid, it will partially vaporize and exert a partial pressure in the container. According to Dalton's Law of Partial Pressures, the total pressure in the container will be the sum of the partial pressure of methanol vapor and the partial pressure of air.
P_total = P_methanol + P_air
Given that the volume of the liquid is negligible, we can assume that the number of moles of air in the container remains constant. So, the partial pressure of air can be considered constant as well.
Since the temperature remains constant, the partial pressure of methanol vapor will reach equilibrium when it reaches the vapor pressure of methanol at 45.0°C, which is 0.27 atm.
Therefore:
P_total = P_methanol + P_air
P_total = 0.27 atm + P_air
To find the total pressure in the container, we need to determine the partial pressure of air, or P_air. However, this information is not provided in the question, so we cannot calculate the exact pressure in the container.
In summary, the pressure in the container when equilibrium is established will be the sum of the vapor pressure of methanol at 45.0°C (0.27 atm) and the partial pressure of air in the container, which is unknown.