Test01Sol-docB

# Hence for each experiment u 0 for ideal gases the

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: = 1/3 and PT = 0.900 bar Hence, PA = 0.300 bar Question 11: (10 points) A non- ideal gas is described using the following equation of state: RT bT − a P= + 2 Vm Vm - 3 where a = 1.00 S.I. unit and b = 3.93 10 S.I. unit are specific parameters proposed for this gas. Part a. (5 points) Determine the Boyle’s temperature TB predicted by this equation of state for this gas and indicate 3 - 1 whether at the state defined by T1 = 200. K and Vm,1 = 0.00600 m .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 = 1/3.93x10 = 254 K 3 - 1 At the state defined by T1 = 200....
View Full Document

## 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.

Ask a homework question - tutors are online