Assume that 0.0200 mole of a gas is collected in the lab by water displacement. If the
temperature of the gas is 26.0oC and the total pressure in the gas tube is 745 mm Hg (including water
vapor), what is the volume in liters of the dry gas? (Use the ideal gas law.)
Ptotal = pH2O + pgas
745 mm = pH2O(look up vapor pressure H2O @ 26 C) + pgas. Solve for pgas which is the pressure of the DRY gas.
Then PV = nRT. You know P (use pressure of dry gas), n, R, and T; solve for V in liters.
To find the volume of the dry gas, we can use the ideal gas law equation, which is:
PV = nRT
Where:
P = Pressure (in atm)
V = Volume (in liters)
n = Number of moles
R = Ideal Gas Constant (0.0821 L.atm/mol.K)
T = Temperature (in Kelvin)
First, we need to convert the pressure from mmHg to atm and the temperature from degrees Celsius to Kelvin.
To convert mmHg to atm, we divide the pressure by 760mmHg/atm:
745 mmHg / 760 mmHg/atm = 0.9789 atm
To convert Celsius to Kelvin, we add 273.15 to the temperature:
26.0oC + 273.15 = 299.15 K
Now we can substitute the values into the ideal gas law equation and solve for the volume:
PV = nRT
V = (nRT) / P
V = (0.0200 mol * 0.0821 L.atm/mol.K * 299.15 K) / 0.9789 atm
V ≈ 6.42 L
Therefore, the volume of the dry gas is approximately 6.42 liters.