HW2 Solutions - 0.0166x 2 0.0754x 1 R 2 = 1 1 1.1 1.2 1.3...

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Unit 3 Homework 2 Solutions 1. A plot of lnP vs 1/T will yield the slope R H vap ο - = -5291.5 ο H vap = 44.0 kJ/mol y = -5291.5x + 20.924 R 2 = 1 2.70 2.90 3.10 3.30 3.50 0.00328 0.0033 0.00332 0.00334 0.00336 0.00338 0.0034 0.00342 1/T ln P 2. Start from the Clapeyron Equation since this involves the solid-liquid coexistence curve: V T H dT dP trs trs = Since we want the dependence of temperature on pressure, use the equation: H V T dP dT trs trs = - = - - - - 1 1 3 1 1 3 3 08206 . 0 / 314 . 8 1 10 / 5 . 6009 ) / 63 . 1 )( 2 . 273 ( mol K atm dm mol K J dm cm mol J mol cm dP dT = P T -0.00751 K atm -1 atm atm K K atm K T P 133 00751 . 0 1 00751 . 0 = - - = - = P f – 1 = 133 atm P f = 134 atm You can also integrate the equation to get ( 29 1 2 1 2 P P H V T T ln - = and solve for pressure.
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3. (a) (b) y = 0.0002x
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Unformatted text preview: + 0.0166x 2 + 0.0754x + 1 R 2 = 1 1 1.1 1.2 1.3 1.4 1.5 1.6 1 2 3 4 5 6 7 molality (mol/kg) osmotic coefficient 4. The plot of φ vs. m is shown on the right. 0754 . 0166 . 004 . 0002 . 1 2 3 + +-=-m m m m φ Then integrate from m = 0 to m = 2 molal. 2 2 3 4 2 2 3 0754 . 2 0166 . 3 004 . 4 0002 . 0754 . 0166 . 004 . 0002 . 1 m m m m m m m dm m + +-= ∫ + +-= -1741 . 1508 . 0332 . 0107 . 0008 . = + +-= Now solve dm m m ∫-+-= 1 1 ln γ to get 3625 . 1741 . ) 1508 . 0664 . 032 . 0032 . ( ln = + + +-= 24-50 24.53...
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