Water has a van der Waals constants of a = 5.464 L·atm/mol2 and b = 0.03049 L/mol. What is the pressure of 5.20 moles of water vapor in a 1.10 L container at 100.0°C?
I Don't understand this problem!
http://en.wikipedia.org/wiki/Van_der_Waals_equation_of_state
Use the modern form of the equation. You are given a,b,n,R, and T.
To solve this problem, we can use the van der Waals equation, which is an equation of state that corrects for the non-ideal behavior of gases. The equation is:
(P + (an^2/V^2))(V - nb) = 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.
In this case, we are given:
a = 5.464 L·atm/mol^2
b = 0.03049 L/mol
n = 5.20 mol
V = 1.10 L
T = 100.0°C = 373.15 K (converting Celsius to Kelvin by adding 273.15)
Now we can plug in the given values into the van der Waals equation and solve for P:
(P + (a(n^2)/(V^2)))(V - nb) = nRT
(P + ((5.464 L·atm/mol^2) * (5.20 mol)^2) / (1.10 L)^2))(1.10 L - (0.03049 L/mol) * 5.20 mol) = (5.20 mol)(0.0821 L·atm/(mol·K))(373.15 K)
Simplifying the equation:
(P + (56.4512 L^2·atm/(mol^2))(1.10 L - 0.158828 L)) = (5.20 mol)(0.0821 L·atm/(mol·K))(373.15 K)
(P + (56.4512 L^2·atm/(mol^2))(0.941172 L)) = (5.20 mol)(0.0821 L·atm/(mol·K))(373.15 K)
(P + 53.2035 L^3·atm/(mol^2))(0.941172 L) = (5.20 mol)(0.0821 L·atm/(mol·K))(373.15 K)
(P + 50.0702 L^3·atm/(mol^2)) = (5.20 mol)(0.0821 L·atm/(mol·K))(373.15 K) / (0.941172 L)
(P + 50.0702 L^3·atm/(mol^2)) = 34.349 L·atm
P + 50.0702 L^3·atm/(mol^2) = 34.349 L·atm
P = 34.349 L·atm - 50.0702 L^3·atm/(mol^2)
Now substitute in the given values and solve for P:
P = (34.349 L·atm) - (50.0702 L^3·atm/(mol^2))
P = -15.7212 L·atm
Therefore, the pressure of 5.20 moles of water vapor in a 1.10 L container at 100.0°C is approximately -15.7212 L·atm.