Hence 2 1 167 kpa 2 185 kpa

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Unformatted text preview: mol ⎞ π 1 = (1) ⎜ 10 ⎟ ( 0.75 ) ⎜ 0.082 ⎟ ( 298 K ) ⎜ ⎝ ⎝ ⎝ 1000 mmol ⎟ ⎠ L⎠ mol-K ⎠ π 1 = 0.183 atm Since 101.3 kPa is equal to 1 atm, the osmotic pressure is approximately 18.5 kPa. We also know the osmotic pressure gradient across the membrane, 167 kPa. Hence: Δπ = π 2 − π 1 167 kPa = π 2 − 18.5 kPa π 2 = 185.4 kPa Now that we know the osmotic pressure of the second, more concentrated chamber, we can solve for the concentration of mannose in this chamber as follows: π 2 = gCσ RT 1 atm L-atm ⎞ ⎛ ⎞ ⎛ ⎛ 1 mol ⎞ 185.4 kPa ⎜ ⎟ = (1)C ( 0.75 ) ⎜ 0.082 ⎟ ( 298 K ) ⎜ ⎝ 101.325 kPa ⎠ ⎝ ⎝ 1000 mmol ⎟ ⎠ mol-K ⎠ 1 atm ⎛ ⎞ 185.4 kPa ⎜ ⎝ 101.325 kPa ⎟ ⎠ C= L-atm ⎞ ( 0.75 ) ⎛ 0.082 ⎜ ⎟ ( 298 K ) ⎝ mol-K ⎠ C = 0.0999 mol L Since there is one 1L in each compartment, there’s 0.09...
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This note was uploaded on 01/22/2014 for the course BIOE 110 taught by Professor Sanjaykumar during the Spring '13 term at University of California, Berkeley.

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