HW4 Solution - CHEM 3615 Problem Set #4 Real Gases Due...

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Page 1 of 12 CHEM 3615 Problem Set #4 –Real Gases Due September 18, 2007 Finish reading Chapter 2. Problems 1 through 7 are due at the start of class and will be graded. 1. In the last problem set, you plotted Z values for methane as a function of temperature expressed in terms of nT B for the two-term virial expansion for a van der Waal's gas: RT p RT a b 1 RT V p Z + = = Plot the derivative of the compression factor with respect to pressure at constant temperature as a function of temperature (100K < T < 6000K) and make a second plot where the temperature varies between (100K < T < 1x10 6 K) using a logarithmic scale. In both cases, pick your y axis in such a way as to clearly see any maximum in temperature. At what temperature (T max ) does the slope maximize in terms of the Boyle temperature? Prove this mathematically. Finally, what is the value of the maximum slope in terms of the van der Waal's constants a and b? {Units for T max are K, the maximum slope is in units of Pa -1 } Solution: RT p ) RT a b ( 1 Z + = , and 2 2 T T R a RT b p Z = (this is the slope of the Z - p plot). where 2 6 mol m Pa 227 . 0 a = and 1 3 5 mol m 10 28 . 4 b × = So we can get: 2 6 T T 00328 . 0 T 10 15 . 5 p Z × = The derivative of the compression factor with respect to pressure at constant temperature as a function of temperature is plotted as following: the first of which has the temperature of 100K~6000K, while the second has 100 K ~ 1 × 10 6 K. 3 2 2 T T R a 2 RT b p Z T + = and 4 2 3 T 2 2 T R a 6 RT b 2 p Z T = then, at T=T max , where 0 T R a 2 RT b p Z T 3 max 2 2 max T = + = so, B max T 2 Rb / a 2 T = = . To show it is a maximum, at T = T max = 2 T B 0 a b R 8 1 ) Rb / a 2 ( R a 6 ) Rb / a 2 ( R b 2 T R a 6 RT b 2 p Z T 3 4 2 4 2 3 4 2 3 T 2 2 < = = =
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Page 2 of 12 The maximum slope is a 4 b a 4 b a 2 b T R a RT b 2 2 2 2 max 2 max = = For methane, 2 6 mol m Pa 227 . 0 a = and 1 3 5 mol m 10 28 . 4 b × = So maximum slope is 2.02 × 10 -9 Pa -1 . This only refers to the initial slope. This means big gas molecules with weak attractions have large positive maximum slopes while small molecules with strong attractions have small positive maximum slopes. Slope vs Temperature (100K~6000K) 2.5 2.0 1.5 1.0 0.5 0.0 -0.5 -1.0 ( Z/ p) T x 10 9 /Pa -1 6000 4000 2000 0 T /K Figure 1A T B T = 2T B
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Page 3 of 12 Slope vs Temperature (100K~1 × 10 6 K) 2. The law of corresponding states predicts two gases have identical states when they have identical reduced variables. Using the critical constants in the data section of your textbook, tell me when nitrogen has the same compression factor as helium at 10.0 K, 60.0 mL/mol, and 1.50 atm.
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This homework help was uploaded on 04/21/2008 for the course CHEM 3615 taught by Professor Aresker during the Fall '07 term at Virginia Tech.

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HW4 Solution - CHEM 3615 Problem Set #4 Real Gases Due...

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