Use the van der Waals equation of state to calculate the pressure of 3.40 mol of

Question

Use the van der Waals equation of state to calculate the pressure of 3.40 mol of H2S at 475 K in a 5.60 L vessel. Van der Waals constants can be found in this table. Number P 22.07 atm Use the ideal gas equation to calculate the pressure under the same conditions. Number P 23.93 atm In a 15.5 L vessel, the pressure of 3.40 mol of H2S at 475 K is 8.55 atm when calculated using the ideal gas equation and 8.42 atm when calculated using the van der Waals equation of state. Why is the percent difference in the pressures calculated using the two different equations greater when the gas is in the 5.60 L vessel compared to the 15.5L vessel? The attractive forces between molecules become less of a factor at the higher pressure in the 5.60 L vessel. The molecular volume is a smaller part of the total volume of the 560 L vessel The molecular volume is a larger part of the total volume of the 5.60 L vessel. The attractive forces between molecules become a greater factor at the higher pressure in the 5.60 L vessel.

Explanation / Answer

Part a

van der Waal's equation of state for a real gas

(P + n2a / V2)(V- nb) = nRT

a = 4.544 bar L2/mol2

b = 0.04339 L/mol

(P + 3.42*4.544/5.62) (5.6 - 3.4*0.04339) = 3.4*8.314*10^-2*475

(P + 1.675)(5.452474) = 134.2711

P = 22.95 bar x 0.987atm/bar

P = 22.652 atm

Part b

From ideal gas equation

P = nRT/V

= 3.4 x 0.0821 x 475 / 5.6

= 23.677 atm

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