A mixture of nitrogen and water vapours is admitted to a flask at 760 ...
To solve this problem, we need to use the ideal gas law equation, PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature. We can rearrange this equation to solve for V:
V = (nRT) / P
First, we need to calculate the number of moles of nitrogen and water vapors present in the flask. To do this, we can use the partial pressure of nitrogen and the total pressure of the mixture.
Partial pressure of nitrogen (P_nitrogen) = Total pressure - Partial pressure of water vapors
P_nitrogen = 760 torr - 722 torr = 38 torr
Next, we need to convert the pressure from torr to atmospheres since the gas constant (R) is given in atm:
P_nitrogen = 38 torr * (1 atm / 760 torr) = 0.05 atm
Now, we can calculate the number of moles of nitrogen using the ideal gas law equation:
n_nitrogen = (P_nitrogen * V) / (R * T)
Since the flask is completely dry, the moles of nitrogen before and after the experiment will be the same.
n_nitrogen(before) = n_nitrogen(after)
(P_nitrogen * V_before) / (R * T_before) = (P_nitrogen * V_after) / (R * T_after)
Since the temperature and gas constant are constant, we can simplify the equation:
V_before = V_after
Now, we can calculate the volume of the flask using the equation:
V = (n_nitrogen * R * T) / P_nitrogen
V = (n_nitrogen * R * T) / P_nitrogen
Using the given values:
n_nitrogen = n_nitrogen(after) = (0.05 atm * V_after) / (R * T)
V = ((0.05 atm * V_after) / (R * T)) * R * T / (0.05 atm)
V = V_after
Substituting the given values:
V = (0.9 g / (28 g/mol)) * (0.0821 L * atm / (mol * K)) * (300 K) / (0.05 atm)
V = 0.9 g * 0.0821 L / 28 g * 300 K / 0.05
V = 24.63 L
Therefore, the volume of the flask is approximately 24.63 L.
A mixture of nitrogen and water vapours is admitted to a flask at 760 ...
PH2O=PTotal−PN2
Given : PTotal = 760 torr, PN2 = 722 torr, T=300K
PH2O=760−722=38 torr
PH2O=38/760 atm
nH2O=0.9/18(mass/mol. mass)
Now,
PV=nRT
V=nRT/P
V=0.9×0.821×300×760/18×38
V=24.63 L