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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. r= 0.08206 L*atm / mol*k
Accepted Answer
The Van der Waals equation of state is: (P + a(n/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 (0.08206 L
atm/mol
K)
T is the temperature
a is the Van der Waals constant for intermolecular forces
b is the Van der Waals constant for the volume of the molecules
To solve for pressure (P), rearrange the equation:
P = (nRT / (V - nb)) - a(n/V)^2
Plug in the given values:
P = ((2.80 mol)(0.08206 L
atm/mol
K)(481 K) / (4.20 L - (0.0434 L/mol)(2.80 mol))) - (4.481 atm
L^2/mol^2)(2.80 mol / 4.20 L)^2
P ≈ 12.8 atm
Therefore, the pressure of 2.80 mol of H2S at 481 K in a 4.20 L vessel is approximately 12.8 atm.
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