A sample of propane gas, C3H8, was collected over water at 25 degree Celsius and 745 torr. The volume of the wet gas is 1.25 L. What will be the volume of the dry propane at standard pressure?
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To find the volume of the dry propane at standard pressure, we need to use the ideal gas law equation: PV = nRT.
First, we need to find the number of moles of propane gas (n) in the sample. We can use the ideal gas law equation to find this:
PV = nRT
Where:
P = pressure (in atm or Pa)
V = volume (in liters or m^3)
n = number of moles
R = ideal gas constant (0.0821 L.atm/(mol.K) for pressure in atmospheres)
T = temperature (in Kelvin)
Given:
Pressure = 745 torr (1 atm = 760 torr, so 745 torr is approximately 0.9789 atm)
Volume = 1.25 L
Temperature = 25 degrees Celsius = 25 + 273.15 Kelvin = 298.15 K
Plugging in the values:
(0.9789 atm)(1.25 L) = n(0.0821 L.atm/(mol.K))(298.15 K)
Solving for n:
n = (0.9789 atm)(1.25 L) / (0.0821 L.atm/(mol.K))(298.15 K)
n ≈ 0.0616 mol
Now we know the number of moles of propane, we can assume it is an ideal gas and convert it to standard pressure.
Standard pressure is defined as 1 atm. Since the number of moles (n) of propane gas does not change, we can use the formula PV = nRT to find the new volume at standard pressure.
Plugging in the values:
(1 atm)(V) = (0.0616 mol)(0.0821 L.atm/(mol.K))(298.15 K)
Solving for V:
V ≈ (0.0616 mol)(0.0821 L.atm/(mol.K))(298.15 K) / 1 atm
V ≈ 1.80 L
Therefore, the volume of the dry propane at standard pressure will be approximately 1.80 liters.