Question

The amount n = 2.00 mol of a van der Waals gas with a = 0.245...

The amount n = 2.00 mol of a van der Waals gas with a = 0.245 m6 Pa mol-2 occupies a volume of 0.840 L if the gas is at a temperature of 85.0 K and at a pressure of 2850 kPa. From this information, calculate the van der Waals constant b and the pressure p of this gas sample when it occupies a volume of 1.680 dm3 at T = 255 K.

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Answer #1

[(2850000 Pa) + 0.245(2.00)2 ] * [(0.840 m3 - (2.00)(b)] = (2.00) (8.314472 Pa.m3/(mol.K) (85.0 K)

(0.840 m3)2

[(2850000) + (1.167)] * [0.84 - 2b] = 1383.8

2850001.67 (0.84 - 2b) = 1383.8

2394001.4028 - 5700003.34b = 1383.8

2392617. 60 = 5700003.34b

b = 0.419 m3/mol

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part B

[P + (0.245 m6Pa mol-2) (2.00 mole)2] *  [(0.001680 m3) - (2.00 moles) (0.419) = (2.00) (8.314472 Pa.m3/(mol.K) (255 K)

(0.001680 m3)2

[P + 291.7] * [0.001680 - 0.838] = 4240.14

[P + 291.7] * (- 0.83632) = 4240.14

-0.83632 P - 243.954 = 4240.14

4484.09 = 0.83632 P

P = 5361.69 Pa

P = 5.362 KPa

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