hw5_solutions_F - CH353M Homework assignment 5 (due Friday...

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CH353M Homework assignment 5 ( due Friday Nov. 11 ) 1(8) . n 1 moles of compound 1 form an ideal mixture with n 2 moles of compound 2. The process is performed at constant P and T . Let 0 1 V and 0 2 V be the molar volumes of the pure compounds. (a) What is the total volume of the mixture? Is it larger, smaller, or the same as the total volume of the pure compounds? G = 00 11 2 2 1 1 1 2 (l n ) n ) n n n R Tx n R Tx µµ µ += + + + where x i = n i /(n 1 +n 2 ). V = 12 0 0 121 1 2 2 ,, Tn n G n nn V n V PP P  ∂∂ =+= +   is independent of the mole fractions and is the same as before mixing (b) What is the change of the Gibbs free energy, G , upon mixing? G = G - 1 1 2 2 ln ln n n n R Txn R −= + (c) What is A ? Α = G – PV since both P and V (see Part a) remain unchanged, we have A= G = ln ln n R R + (d) What is U ? () 0 UH P V ∆= ∆− = (using results from part a and from the lecture)
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2. For the purposes of this problem, a soda drink in a can consists of water with CO 2 dissolved in it. The drink is in equilibrium with pure gaseous CO 2 . As we know from our experience, the overall pressure P inside the can is higher than the atmospheric pressure. Assume that CO 2 and water form an ideal solution [FYI: This assumption is actually incorrect; The correct treatment of the problem would lead to qualitatively the same answer]. A (3). Soda cans made by different manufactures generally have different pressures inside. Consider two cans of soda; The pressure inside can 1 is P 1 = 1.5 atm and the pressure inside can 2 is P 2 = 2 atm. The mole fraction of CO 2 in the drink in the first can is x 1 . What is the corresponding mole fraction x 2 in the second can? According to Raoult’s law, the amount of CO 2 dissolved in water is proportional to the partial pressure 2 CO P of CO 2 in gas phase; since we neglect the pressure of H 2 O vapor 2 CO P is the total pressure in the gas phase 22 2 * CO CO CO Px P = , ( 1 ) where 2 * CO P is the pressure at which gaseous carbon dioxide is in equilibrium with pure liquid CO 2 . From this equation, 21 21 // 4 / 3 PP x x == or x 2 = 4x 1 /3 B (3) . Explain, using appropriate equations, why you see the formation of bubbles when you open a can of soda. When you open the can, the pressure of the gas decreases. According to Eq. (1), the equilibrium concentration of dissolved CO 2 must decrease; The excess CO 2 leaves the solution in the form of bubbles.
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C (3) . Will the temperature T freeze at which the soda in the can freezes increase of decrease as the CO 2 pressure in the can is increased? Compare your result with the effect of pressure on the melting point of pure water. According to what we have learned in the class, dissolving CO 2 will result in a depression of the freezing point (you can establish that by putting a soda can in a freezer). Neglecting the pressure dependence of the freezing point of pure water, the new freezing point is: 2 2 2 pure CO freeze H O melting RT x TT H =− However the freezing point of pure water itself depends on the pressure as well, according to the Clapeyron equation: dP/dT = - (-) (-) melting ls H HH TV V =
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This note was uploaded on 09/06/2009 for the course CH 353M taught by Professor Lim during the Spring '08 term at University of Texas at Austin.

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hw5_solutions_F - CH353M Homework assignment 5 (due Friday...

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