Quiz3key_Sp10

Quiz3key_Sp10 - (T=298 K, P = 1 atm). A quantity of liquid...

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CEE 320 Quiz #3 Environmental Engineering F 12 February 2010 PLEASE WRITE YOUR NAME ON THE BACK Quiz #3 solution key Potentially useful data: PV = nRT where R = 82.05 x 10 -3 mol -1 L atm K -1 Density of water: 1 kg L -1 = 1,000 kg m -3 1,000 L = 1 m 3 1,000 mL = 1 L Investigating MTBE Methyl tert-butyl ether (MTBE) is a gasoline additive and common groundwater contaminant. Some of its properties are given below: Chemical formula C 5 H 12 O Molecular mass 88 g mole -1 Saturation vapor pressure (25 o C) 0.33 atm Henry’s law constant 1.7 M atm -1 (a) Consider a sealed container (volume = 1 L) that contains 0.1 L of pure liquid MTBE at equilibrium. The remainder of the space is gas. What is the mass concentration of MTBE in the gas phase? Assume that the temperature is 25 o C. P MTBE V air = n MTBE RT n MTBE = P MTBE V air RT n MTBE = 0.33 atm ( ) 0.9 L ( ) 82.05 × 10 -3 L atm mol K ( ) 298 K ( ) = 0.012 mol C MTBE = n MTBE V air = 0.012 mol 0.9 L 88 g mol = 1.2 g L (b) Consider another sealed container (volume = 3 L) that contains 1 L of water and 2 L of air
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Unformatted text preview: (T=298 K, P = 1 atm). A quantity of liquid MTBE equal to 10-3 mole is injected into the water. What is the equilibrium partial pressure of MTBE in the gas phase? Mole balance eqns: Henry s law: = = n total = n air + n water n air = P MTBE V air RT n water = C MTBE V water C MTBE = K H P MTBE CEE 320 Quiz #3 Environmental Engineering F 12 February 2010 Substitute Henry s Law equation into n water : Substitute n air and n water into n total eqn and solve for P MTBE : n water = K H P MTBE V water n total = P MTBE V air RT + K H P MTBE V water n total = P MTBE V air RT + K H V water P MTBE = n total V air RT + K H V water = [ ] mol L L atm mol K K + mol L atm L = [ ] atm P MTBE = 10 3 mol 2 L 82.05 10-3 L atm mol K ( ) (298 K) + 1.7 mol L atm ( ) 1 L ( ) = 5.6 10 4 atm...
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Quiz3key_Sp10 - (T=298 K, P = 1 atm). A quantity of liquid...

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