When determining the vapor pressure of water, the bubble volume is 1.2mL at 5 degrees celsius.
-How many moles of dry air were in the bubble? (assume 1 atm)
-If the bubble volume is 8.6mL at 75 degrees celsius, what is the partial pressure of the dry air at this temp?
To determine the moles of dry air in the bubble, we can use the ideal gas law equation:
PV = nRT
Where:
P = Pressure (in atm)
V = Volume (in liters)
n = Moles of gas
R = Ideal gas constant (0.0821 L·atm/(mol·K))
T = Temperature (in Kelvin)
Let's solve each part of the question separately:
1. Determining the moles of dry air at 5 degrees Celsius and a bubble volume of 1.2 mL:
First, we need to convert the temperature to Kelvin by adding 273.15:
T = 5 degrees Celsius + 273.15 = 278.15 K
Next, convert the volume from mL to L:
V = 1.2 mL = 1.2 * 10^(-3) L
Since the pressure is assumed to be 1 atm, we can substitute these values into the ideal gas law and solve for n:
1 * (1.2 * 10^(-3)) = n * (0.0821) * 278.15
n = (1 * (1.2 * 10^(-3))) / ((0.0821) * 278.15)
Calculate the value of n to get the moles of dry air in the bubble at this temperature.
2. Calculating the partial pressure of dry air at 75 degrees Celsius and a bubble volume of 8.6 mL:
Again, we need to convert the temperature to Kelvin:
T = 75 degrees Celsius + 273.15 = 348.15 K
Next, convert the volume from mL to L:
V = 8.6 mL = 8.6 * 10^(-3) L
Since we are asked for the partial pressure of dry air, we need to subtract the vapor pressure of water at this temperature from the total pressure.
First, find the vapor pressure of water at 75 degrees Celsius using a reference table or online resource. We'll assume it is 3170 Pa.
Now, convert the vapor pressure to atm:
Vapor pressure = 3170 Pa * (1 atm/101325 Pa) = 0.03132 atm
To find the partial pressure of dry air, subtract the vapor pressure from the total pressure (which is assumed to be 1 atm):
Partial pressure of dry air = 1 atm - 0.03132 atm
Calculate the final partial pressure of dry air at this temperature.