A 0.75 L bottle is cleaned, dried, and closed in a room where the air is 20°C and 42% relative humidity (that is, the water vapor in the air is 0.42 of the equilibrium vapor pressure at 20°C). The bottle is brought outside and stored at 0.0°C. (See Table 5.2)
(a) What mass of water condenses inside the bottle?
We don't have table 5.2. Water at 20C has a vapor pressure of 17.4 mm Hg. R.H. = vapor pressure/17.4 so you can use that to calculate pressure and use PV = nRT to determine the moles H2O. I assume table 5.2 will give you the vapor pressure of water at 0C. Use that pressure and PV = nRT to determine moles at OC. Convert each value for moles to grams and subtract to find the amount of water condensed. Check my thinking.
To find the mass of water that condenses inside the bottle, we need to calculate the difference between the amount of moisture inside the bottle when it was closed in the room and the amount of moisture at the lower temperature outside.
To calculate the mass of water that initially exists in the bottle, we can use the relative humidity and the equilibrium vapor pressure at 20°C. The equilibrium vapor pressure at 20°C can be found in Table 5.2. Let's assume it is P1.
So, the initial amount of water in the bottle can be calculated as:
Initial water vapor pressure = Relative humidity at 20°C * Equilibrium vapor pressure at 20°C
= (0.42) * P1
The final amount of water vapor in the bottle can be estimated using the vapor pressure at 0.0°C. Let's assume it is P2.
Now, the difference in the amount of water vapor can be determined by subtracting the initial amount from the final amount:
Difference in water vapor = Final water vapor pressure - Initial water vapor pressure
= P2 - (0.42) * P1
Next, we can use the ideal gas law to calculate the number of moles of water vapor that condenses, assuming the volume of the bottle remains constant. The ideal gas law equation is given by:
PV = nRT
Where:
P is the pressure of the gas (P2 in our case)
V is the volume of the bottle (0.75 L in this case)
n is the number of moles of gas
R is the ideal gas constant
T is the temperature in Kelvin (0.0°C = 273.15 K)
We can rearrange the equation to solve for n:
n = PV / RT
Now that we have the number of moles of water vapor, we can calculate the mass of water that condenses by multiplying the number of moles by the molar mass of water (18.015 g/mol).
So, the final step is to calculate the mass of water condensation:
Mass of water condensation = Number of moles of water vapor * Molar mass of water
By following these steps, you should be able to determine the mass of water that condenses inside the bottle.
To calculate the mass of water that condenses inside the bottle, we need to determine the change in vapor pressure when the bottle is moved from the room temperature to the outside temperature. We will use the equation:
ΔP = P_final - P_initial
Where:
ΔP = Change in vapor pressure
P_final = Vapor pressure at 0.0°C
P_initial = Vapor pressure at 20°C
First, we need to find the equilibrium vapor pressure at both temperatures using the relative humidity values given in the question. We can use Table 5.2 for this information.
At 20°C:
Relative Humidity = 0.42
Equilibrium Vapor Pressure at 20°C = Relative Humidity * Vapor Pressure at 20°C
= 0.42 * (Vapor Pressure at 20°C from Table 5.2)
At 0.0°C:
Vapor Pressure at 0.0°C = Vapor Pressure at 0.0°C from Table 5.2
Now, we can calculate the change in vapor pressure:
ΔP = Vapor Pressure at 0.0°C - Vapor Pressure at 20°C
Finally, we can use the ideal gas law to find the mass of water that condenses:
PV = nRT
Where:
P = Pressure (ΔP)
V = Volume of the bottle (0.75 L)
n = Number of moles (unknown)
R = Ideal Gas Constant (0.0821 L·atm/(K·mol))
T = Temperature Change (20°C - 0.0°C)
We can rearrange the equation to solve for the number of moles (n):
n = (PV) / (RT)
Now, we can solve for the mass of water condensing:
Mass = Number of moles * Molar mass of water
The molar mass of water is approximately 18.015 g/mol.
Let's calculate the mass of water condensing using the step-by-step process outlined above.