PMonSep29

PMonSep29 - 1 Thermodynamics 2 Irreversible changes Water...

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Unformatted text preview: 1 Thermodynamics 2 Irreversible changes Water can be cooled below its freezing point and if it is not disturbed it will not freeze. This phenomenon is known as super cooling. Calculate the change in entropy when 1 mole of liquid water at -5 °C and 1 bar changes to ice at the same temperature and pressure. The system is not reversible so it has to be viewed as a set of reversible problems. VERY IMPORTANT 3 Reversible steps of supercooled water freezing Consider the entropy changes when the liquid is heated to its freezing point, allowed to freeze reversibly and cooled back down to -5 °C. Step 1 Heat the Water dT C dH P = ( 29 dT T C S P ∫ = ∆ / Heat capacity of ice and water are different … Water C P 75.3 JK-1 mol-1 Ice C P 36.8 JK-1 mol-1 4 Step 1 Heat the Water Step 2 Freeze the Water ∆ fus H for ice = 6.02 kJ mol-1 1- 1- 1 2 mol K J 39 . 1 268 273 ln 3 . 75 ln = × = = ∆ T T C S P 1- 1- mol K J 05 . 22 273 6020- =- = ∆ = ∆ T H S Sign is negative because heat is being removed. 5 Step 3 Cool the Ice Overall ∆ S = 1.39 - 22.05 – 0.68 = -21.34 JK-1 mol-1 The sign is negative which implies there has been a decrease in entropy. This is intuitively good because ice is more ordered than water. However somewhere in the universe there has been a larger + ve change in entropy to make this happen. 1- 1- 1 2 mol K J 68 . 273 268 ln 8 . 36 ln- = × = = ∆ T T C S P 6 Supercooling and Superheating Just as a liquid may be supercooled, it may also be superheated. Superheating occurs when the liquid is heated above its boiling point without boiling. http://www.youtube.com/watch?v=YGqealb3FBk&mode=related&search= http://www.youtube.com/watch?v=fSPzMva9_CE 7 Entropy as a function of volume and temperature Consider an ideal gas that undergoes a change in volume and temperature. Calculate the entropy change associated with this change. PdV dq dw dq dU- = + = PdV dU dq + = dV V nRT dT nC dq V + = 8 dV V nRT dT nC dq V + = dV V nR T dT nC T dq dS V + = = + = ∆ 1 2 1 2 ln ln V V nR T T nC S V Integrate 9 Example Calculate the correction factor needed to convert entropy values listed at 25 °C and 1 atm. to values at 25 °C and 1 bar. The temperature is constant so the first term is 0. V 2 /V 1 = P 1 /P 2 . Therefore we have: + = ∆ 1 2 1 2 ln ln V V nR T T nC S V 1- 1- 5 2 1 mol K J 109 . 10 101324 ln 314 . 8 1 ln = × × = = ∆ P P nR S 10 Entropy as a function of pressure and temperature + = ∆ 1 2 1 2 ln ln V V nR T T C S V + + = ∆ = + = ∆ 2 1 1 2 1 2 2 1 1 2 1 2 ln ln ln ln ln P P...
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This note was uploaded on 03/17/2010 for the course CH 3530 taught by Professor Consors during the Fall '10 term at WPI.

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PMonSep29 - 1 Thermodynamics 2 Irreversible changes Water...

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