Question

For chlorine gas, the van der Waals constants are: a = 0.658 J·m3/mol2, and b = 5.62 × 10-5 m3/mol. A 3.0-L tank contains 10.0 moles of chlorine gas at a temperature of 625 K. What is
the pressure in the container?
A) 15.0 MPa
B) 17.3 MPa
C) 13.5 MPa
D) 18.2 MPa
E) 11.0 MPa

Answer 1

18.2

Answer 2

The pressure in the container can be calculated using the van der Waals equation:

(P + a n2/V2) (V - n b) = n R T

where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, T is the temperature, and a and b are the van der Waals constants.

First, we need to calculate the volume of the gas in the container since it is given as 3.0 L. The volume occupied by the gas molecules themselves (i.e. the "dead space" which is not available for molecule motion) is given by nb, where n is the number of moles and b is the van der Waals constant. Thus, the actual volume of the gas is V = 3.0 L + n b = 3.0 L + 10.0 mol x 5.62 x 10^-5 m^3/mol = 3.00056 L.

Now we can substitute the given values into the van der Waals equation and solve for P:

(P + a n^2/V^2) (V - n b) = n R T

(P + 0.658 x 10^2/ (3.00056 x 10^-3)^2) (3.00056 x 10^-3 - 10.0 x 5.62 x 10^-5) = 10.0 x 8.314 x 625

(P + 2.550) (0.0299944) = 52018.5

P = (52018.5/0.0299944) - 2.550

P = 1730798.2 Pa

Converting to MPa: P = 17.3 MPa

Therefore, the pressure in the container is option (B) 17.3 MPa.