Answer:
0.422 L
Step-by-step explanation:
To find the volume of gas at standard conditions when dry, we need to apply the concept of Dalton's Law of Partial Pressures and the ideal gas law.
Step 1: Convert the given pressure to atm units.
Given pressure: 767.4 mmHg
1 atm = 760 mmHg (by definition)
Pressure in atm = 767.4 mmHg / 760 mmHg/atm = 1.011 atm (rounded to three decimal places)
Step 2: Convert the given volume to liters.
Given volume: 400.0 mL
1 L = 1000 mL (by definition)
Volume in liters = 400.0 mL / 1000 mL/L = 0.400 L (rounded to three decimal places)
Step 3: Apply Dalton's Law of Partial Pressures.
Dalton's Law of Partial Pressures states that the total pressure of a gas mixture is the sum of the partial pressures of each gas component. In this case, we have two gases: the gas of interest and water vapor.
The partial pressure of water vapor at 24.0 °C is 23.76 mmHg (at 100% relative humidity). We need to subtract this from the total pressure to get the partial pressure of the gas of interest.
Partial pressure of gas of interest = Total pressure - Partial pressure of water vapor
Partial pressure of gas of interest = 1.011 atm - 23.76 mmHg / 760 mmHg/atm = 0.979 atm (rounded to three decimal places)
Step 4: Apply the ideal gas law to find the volume at standard conditions.
The ideal gas law states that PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.
At standard conditions, the pressure is 1 atm, and the temperature is 0 °C or 273.15 K.
R = 0.0821 L atm / (mol K) (ideal gas constant)
Rounded to three decimal places, the equation becomes:
(0.979 atm)(0.400 L) / (1 atm) = (n)(0.0821 L atm / (mol K))(273.15 K)
Solving for n (number of moles):
n = [(0.979 atm)(0.400 L)] / [(0.0821 L atm / (mol K))(273.15 K)] = 0.0186 mol (rounded to four decimal places)
Step 5: Find the volume of the gas at standard conditions using the molar volume of gases.
At standard conditions (0 °C or 273.15 K, 1 atm), the molar volume of an ideal gas is 22.71 L/mol.
Volume at standard conditions = n (molar volume of gas)
Volume at standard conditions = 0.0186 mol × 22.71 L/mol = 0.422 L (rounded to three decimal places)
So, the volume of the gas at standard conditions when dry is 0.422 L.