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?
where would i get N?
To find the volume of the dry propane gas at standard pressure, we need to use the concept of the ideal gas law. The ideal gas law equation is given as:
PV = nRT
Where:
P is the pressure of the gas (in this case, standard pressure),
V is the volume of the gas at the given temperature and pressure,
n is the number of moles of the gas,
R is the ideal gas constant (0.0821 L·atm/(mol·K)), and
T is the temperature of the gas in Kelvin.
First, let's convert the temperature from Celsius to Kelvin:
T(K) = T(°C) + 273.15
T(K) = 25 + 273.15
T(K) = 298.15 K
Next, let's find the number of moles of propane gas using the ideal gas law:
PV = nRT
Since the volume of the wet gas is given as 1.25 L and the pressure is given as 745 torr, we can plug in these values along with the ideal gas constant and the temperature (in Kelvin) to solve for the number of moles (n):
(745 torr) * (0.00131579 atm/1 torr) * (1.25 L) = n * (0.0821 L·atm/(mol·K)) * (298.15 K)
Simplifying the equation:
0.9775043625 atm·L = n * 24.493752015
n = 0.0399483709 mol
Now that we have the number of moles, we can use it to find the volume of the dry propane gas at standard pressure. At standard pressure, the pressure (P) is 1 atm:
PV = nRT
(1 atm) * V = (0.0399483709 mol) * (0.0821 L·atm/(mol·K)) * (298.15 K)
Simplifying the equation:
V = (0.0399483709 mol) * (24.493752015 L)
V = 0.9775043625 L
Therefore, the volume of the dry propane gas at standard pressure is approximately 0.9775 L.
Use PV = nRT and solve for V.
For P, use 745 torr - vapor pressure H2O @ 25C.