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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 ʼ 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 ʼ 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 Λ Ν Μ Ξ Π Ο P MTBE = n total V air RT + K H V water Λ Ν Μ Ξ Π Ο = [ ] mol L L ⋅ atm mol ⋅ K ⋅ K + mol L ⋅ atm ⋅ L Λ Ν Μ Μ Ξ Π Ο Ο = [ ] 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 ( ) Ρ Σ Τ Φ Υ Υ = 5.6 × 10 − 4 atm...
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This note was uploaded on 05/04/2010 for the course CEE 320 taught by Professor Mcmahon during the Spring '10 term at University of Wisconsin.
- Spring '10