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Use the Van der Waals equation of state to calculate the pressure of 2.80 mol of H2S at 481 k in a 4.20 l vessel. Van der Waals constant for H2S is 4.481 atm*l^2 / mol^2, and 0.0434g/mol.
Accepted Answer
The Van der Waals equation of state is: (P + a(n/V)^2)(V-nb) = nRT. Where:
P = pressure
V = volume
T = temperature
n = number of moles
R = ideal gas constant = 0.08206 L⋅atm/(mol⋅K)
a = Van der Waals constant for attractive forces
b = Van der Waals constant for repulsive forces
Therefore, P = (nRT/(V-nb)) - a(n/V)^2.
P = ((2.80 mol
0.08206 L⋅atm/(mol⋅K)
481 K) / (4.20 L - (0.0434 g/mol
2.80 mol))) - (4.481 atm
L^2 / mol^2
(2.80 mol / 4.20 L)^2)
P = 12.9 atm.
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