A 20.0L bulb is thermostated to 298.0K. After being evacuated 1.00g H2O is injected into the bulb. What will be pressure in the bulb and how many grams of H2O will remain as a liquid? (Vapour pressure of H2O at 298.0 is 23.8 torr.)

To determine the pressure in the bulb after injecting 1.00 g of H2O and to find out how much of it will remain as a liquid, we can use the ideal gas law and the concept of vapor pressure. Let's break this down step by step.

Step 1: Calculate the number of moles of H2O injected

First, we need to convert the mass of water injected into moles. The molar mass of water (H2O) is approximately 18.02 g/mol. Using this, we can find the number of moles:

Number of moles (n) = mass (g) / molar mass (g/mol)

n = 1.00 g / 18.02 g/mol ≈ 0.0555 moles

Step 2: Calculate the pressure exerted by the water vapor

Next, we need to consider the vapor pressure of water at 298.0 K, which is given as 23.8 torr. This means that at this temperature, the maximum pressure that water vapor can exert is 23.8 torr. Since we are injecting water into a vacuum, the pressure in the bulb will initially be equal to the vapor pressure of the water.

Step 3: Determine the total pressure in the bulb

According to the ideal gas law, the pressure exerted by the gas can be calculated using the formula:

P = (nRT) / V

Where:

• P = pressure in atm
• n = number of moles of gas
• R = ideal gas constant (0.0821 L·atm/(K·mol))
• T = temperature in Kelvin (298.0 K)
• V = volume in liters (20.0 L)

Substituting the values:

P = (0.0555 moles × 0.0821 L·atm/(K·mol) × 298.0 K) / 20.0 L

P ≈ 0.0678 atm

Step 4: Convert pressure to torr

To convert this pressure into torr (since 1 atm = 760 torr):

P ≈ 0.0678 atm × 760 torr/atm ≈ 51.6 torr

Step 5: Determine the total pressure in the bulb

The total pressure in the bulb will be the sum of the vapor pressure of the water and the pressure exerted by the gas:

Total Pressure = Vapor Pressure + Pressure from injected water

Total Pressure = 23.8 torr + 51.6 torr = 75.4 torr

Step 6: Calculate the amount of liquid water remaining

Now, we need to find out how much of the injected water remains as a liquid. Since the vapor pressure of water at 298.0 K is 23.8 torr, we can use this to determine how much water can exist as vapor at this pressure.

Using the ideal gas law again, we can find the number of moles of water vapor that corresponds to the vapor pressure:

P = (nRT) / V

Rearranging gives us:

n = PV / RT

Substituting the vapor pressure (23.8 torr converted to atm: 23.8 torr / 760 torr/atm ≈ 0.0314 atm):

n = (0.0314 atm × 20.0 L) / (0.0821 L·atm/(K·mol) × 298.0 K)

n ≈ 0.0255 moles

Step 7: Convert moles of vapor back to grams

Now, converting moles of water vapor back to grams:

Mass = n × molar mass

Mass = 0.0255 moles × 18.02 g/mol ≈ 0.459 g

Final Calculation: Determine liquid water remaining

Initially, we injected 1.00 g of water. The amount that has evaporated is approximately 0.459 g. Therefore, the remaining liquid water is:

Remaining liquid = Initial mass - Mass evaporated

Remaining liquid = 1.00 g - 0.459 g ≈ 0.541 g

Summary of Results

In summary, after injecting 1.00 g of H2O into the bulb:

• The total pressure in the bulb will be approximately 75.4 torr.
• Approximately 0.541 g of H2O will remain as a liquid.