If 0.500L of dry N2 was bubbles into an inverted flask through water to saturate it at 25.0C, what would be the total volume at the lab atmospheric pressure. Our pressure was : 757.5 mmHg
find the water vapor pressure from the temperature, and subtract that pressure from your observed pressure. At 25C, water vapor pressure is 23.8 mmHg (check that)
so your observed pressure then was 757.5mmHg-23.8mmHg.
Now, you have pressure and volume at that pressure, convert the volume to lab pressure.
V2=.5L*(755.5-23.8)/755.5
To find the total volume at the lab atmospheric pressure, we need to consider the ideal gas law equation, which states:
PV = nRT
Where:
P is the pressure
V is the volume
n is the number of moles of the gas
R is the ideal gas constant (0.0821 L·atm/(mol·K))
T is the temperature in Kelvin
First, we need to convert the lab atmospheric pressure from mmHg to atm by dividing it by 760 (1 atm = 760 mmHg). So, the lab atmospheric pressure would be 757.5 mmHg / 760 mmHg = 0.996 atm.
Given the following information:
Initial volume of dry N2: 0.500 L
Lab atmospheric pressure: 0.996 atm
Temperature: 25.0°C
To calculate the total volume at the lab atmospheric pressure, we need to find the number of moles of N2 using the ideal gas law equation:
n = PV / RT
First, we need to convert the temperature to Kelvin by adding 273.15 (25.0°C + 273.15 = 298.15 K).
Now, we can calculate the number of moles of N2:
n = (0.996 atm) * (0.500 L) / [(0.0821 L·atm/(mol·K)) * (298.15 K)]
Simplifying this equation, we get:
n = 0.0201 moles
Since gases behave ideally, we can use the ideal gas law to find the total volume at the lab atmospheric pressure:
V = nRT / P
V = (0.0201 moles) * (0.0821 L·atm/(mol·K)) * (298.15 K) / (0.996 atm)
Simplifying this equation, we get:
V ≈ 0.497 L
Therefore, the total volume of the gas at the lab atmospheric pressure would be approximately 0.497 L.