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Unformatted text preview: K and Vm,1 = 0.00600 m .mol , the compression factor has the value Z1 given by: Z1 = (P1 Vm,1 / R T1) = 1 + [bT1 – a]/(RT1Vm,1)  3
Z1 = 1 + (3.93 10 x 200. – 1.00)/(8.3145 x 200. X 0.00600) = 0.979 With three significant figures, we obtain: Z1 = 0.979 Since the compression factor is less than unity, it suggests that the molar volume of the gas is lower than the molar volume of the gas calculated using the ideal gas law, hence, that intermolecular attractions are predominantly attractive. 5 Part b. (5 points) Can the above equation of state account for the critical behavior of this gas? Justify your answer. 2
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It cannot because the solution to the conditions (∂P/∂Vm)T = 0 and (∂ P/∂Vm )T = 0 (see Question 3, part c) is not physically meaningful. The first and second partial derivatives of P vs Vm are obtained from the equation of state as: 2
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(∂P/∂Vm)T =  RT/Vm – 2 [bT – a]/Vm and (∂ P/∂Vm )T = 2RT/Vm + 6 [bT – a]/Vm 3
Equate these two partial differentials to 0 and multiply the first partial by Vm and the second by 4
Vm , you will get: RTVm + 2 [bT – a] = 0 and 2RTVm + 6 [bT – a] = 0 From which you conclude that Tc = a/b and Vmç = 0. Using these values in the expression for P leads to Pc = ∞. Obviously, this solution is physically meaningless! Hence, the equation of state considered in this question cannot describe the critical behavior of a gas. Question 12: (12 points) Exactl...
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This note was uploaded on 01/26/2014 for the course CHEM 3615 taught by Professor Aresker during the Spring '07 term at Virginia Tech.
 Spring '07
 AREsker
 Physical chemistry, pH

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