Question

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.

Answer #1

[(2850000 Pa) + 0.245(2.00)^{2} ] *
[(0.840 m^{3} - (2.00)(b)] = (2.00) (*8.314472*
*Pa.m ^{3}/(mol.K)* (85.0 K)

*(0.840 m ^{3})^{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 m^{3}/mol

---------------------------------------------------------------

part B

[P + (0.245
m^{6}Pa mol^{-2}) (2.00 mole)^{2}]
* [(0.001680 m^{3}) - (2.00 moles) (0.419) =
(2.00) (*8.314472* *Pa.m ^{3}/(mol.K)* (255
K)

(0.001680 m^{3})^{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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