3 the work of expansion is always negative since the

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Unformatted text preview: .mol , intermolecular interactions in this gas are predominantly attractive or repulsive. Justify your answer without relying on whether T1 is greater or lower than TB. The Boyle’s temperature, TB, is obtained when the second Virial coefficient B (or B’) is equal to 0 in the Virial Equation of States. So, we need to rewrite the above equation of state in the form of the Virial equation of state. This can be easily achieved if we remember that the Virial series are obtained as polynomial functions of P or of Vm to represent the compression factor Z = PVm/RT We have the expression of P as a function of Vm and T, so by replacing P by its expression in Z, we obtain: 2 Z = (RT/Vm + [bT – a]/Vm ) x (Vm/RT) = 1 + [bT – a]/(RTVm) So, we conclude that the second virial coefficient of the Virial Series as a function of powers of Vm is given by: B = [bT – a]/(RT) - 3 B is equal to 0 when bT – a = 0 or when TB = a/b = 1/3.33x10 = 300. K 3 - 1 At the state defined by T1 = 200. K and Vm,1 = 0.00500 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 + (3.33 10 x 200. – 1.00)/(8.3145 x 200. X 0.00500) = 0.9598 With three significant figures, we obtain: Z1 = 0.960 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. Part b. (5 points) Can the above equation of state account for the critical behavior of this gas? Justify your answer. 2 2 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 3 2 2 3 4 (∂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 5 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) Exactly two moles of an ideal gas are placed in a cylinder, which is sealed from the surroundings by a freely moving, massless adiabatic piston. The cylinder has diathermic walls and is placed in an oven o kept at 28 C, where the atmospheric pressure is Patm = 1.00 atm. A mass M of dry ice is placed on top of the piston. Once, the gas in the cylinder equilibrates under the combined actions of the mass M of dry ice and the atmospheric pressure, the gas pressure reads P = 1.54 atm. Now consider the following two isothermal experiments: In experiment (A), the block of dry ice is removed very quickly (say in one millisecond) from the top of the piston and the gas is allowed to expand. In experiment (B), the block of dry ice is allowed to undergo very slow sublimation (say over a period of a few hours) and...
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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.

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