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 at 5 degrees Celsius, we can use the ideal gas law. The ideal gas law equation is given as:

PV = nRT

Where:
P = pressure
V = volume
n = number of moles
R = ideal gas constant
T = temperature

Given that the bubble volume is 1.2 mL at 5 degrees Celsius, we need to convert the volume to liters and the temperature to Kelvin.

1 mL = 0.001 L
5 degrees Celsius = 5 + 273.15 Kelvin

Substituting the values into the ideal gas law equation:

(1 atm) * (0.001 L) = n * (0.0821 L·atm/(K·mol)) * (278.15 K)

Simplifying the equation:

0.001 atm·L = 0.0821 * n * 278.15 K

To solve for "n" (number of moles), divide both sides of the equation by (0.0821 × 278.15):

n = (0.001 atm·L) / (0.0821 L·atm/(K·mol) × 278.15 K)

n ≈ 0.00004799 moles

Therefore, there are approximately 0.00004799 moles of dry air in the bubble at 5 degrees Celsius.

Now, to determine the partial pressure of the dry air at a higher temperature of 75 degrees Celsius and a bubble volume of 8.6 mL. We can use the same ideal gas law equation and procedure.

Convert 8.6 mL to liters and 75 degrees Celsius to Kelvin:

8.6 mL = 0.0086 L
75 degrees Celsius = 75 + 273.15 Kelvin

Substituting the values into the ideal gas law equation:

P * (0.0086 L) = n * (0.0821 L·atm/(K·mol)) * (348.15 K)

Simplifying the equation:

0.0086 atm·L = 0.0821 * n * 348.15 K

To solve for "P" (partial pressure), divide both sides of the equation by (0.0086 × 0.0821 × 348.15):

P = (0.0086 atm·L) / (0.0821 L·atm/(K·mol) × 348.15 K)

P ≈ 0.0292 atm

Therefore, the partial pressure of the dry air at 75 degrees Celsius and a bubble volume of 8.6 mL is approximately 0.0292 atm.