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 number of moles of dry air in a bubble, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure of the gas
V = volume of the gas
n = number of moles of the gas
R = ideal gas constant
T = temperature in Kelvin
First, we need to convert the temperature from Celsius to Kelvin. To do that, we add 273.15 to the temperature in Celsius.
1. Converting 5 degrees Celsius to Kelvin:
T1 = 5 + 273.15 = 278.15 K
Now, let's calculate the number of moles of dry air using the ideal gas law equation:
P1 × V1 = n × R × T1
Given:
P1 = 1 atm (assuming 1 atm)
V1 = 1.2 mL = 0.0012 L (converting mL to L)
R = 0.0821 L·atm/(mol·K) (ideal gas constant)
Substituting these values into the equation, we can solve for n:
(1 atm) × (0.0012 L) = n × (0.0821 L·atm/(mol·K)) × (278.15 K)
n = (1 atm × 0.0012 L) / (0.0821 L·atm/(mol·K) × 278.15 K)
n ≈ 5.82 × 10^-5 moles
Therefore, there are approximately 5.82 × 10^-5 moles of dry air in the bubble at 5 degrees Celsius.
Moving on to the second question:
To determine the partial pressure of dry air at 75 degrees Celsius, we can again use the ideal gas law equation:
P × V = n × R × T
Converting 75 degrees Celsius to Kelvin:
T2 = 75 + 273.15 = 348.15 K
Given:
V2 = 8.6 mL = 0.0086 L (converting mL to L)
Now, we need to find the new pressure, denoted as P2. We already know that the number of moles n remains the same, and the temperature has changed.
P2 × V2 = n × R × T2
Substituting the known values:
P2 × 0.0086 L = (5.82 × 10^-5 moles) × (0.0821 L·atm/(mol·K)) × (348.15 K)
Solving for P2:
P2 = [(5.82 × 10^-5 moles) × (0.0821 L·atm/(mol·K)) × (348.15 K)] / 0.0086 L
P2 ≈ 1.76 atm
Therefore, the partial pressure of dry air in the bubble at 75 degrees Celsius is approximately 1.76 atm.