l07 - 20.110J / 2.772J / 5.601J Thermodynamics of...

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20 .110/5.60 Fall 2005 Lecture #7 page 1 Entropy, Reversible and Irreversible Processes and Disorder Examples of spontaneous processes T 1 T 2 Connect two metal blocks thermally in an isolated system ( U = 0) Initially 12 T ( ) () += đđ 21 1 1 1 2 TT đ 2 q qq dS −= = = > 0 for spontaneous process if đ đ 1 1 0 in f both cases heat flows 0 rom hot to cold as expected >> << ⇒⇒ gas V vac. V Joule expansion with an ideal gas 1 mol gas ( V,T ) adiabatic = 1 mol gas (2 V , ) = 0 = 0 w = 0 Need a reversible path to compute S from q! Close the cycle and go back to the initial state reversibly and isothermally backwards SS =−∆ 20.110J / 2.772J / 5.601J Thermodynamics of Biomolecular Systems Instructors: Linda G. Griffith, Kimberly Hamad-Schifferli, Moungi G. Bawendi, Robert W. Field
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20 .110/5.60 Fall 2005 Lecture #7 page 2 0 rev q 1 m o l g a s ( 2 V,T ) = 1 mol gas ( V , T ) đ đ backwards 2 1 ln 2 wRd SR TT ∆== = = ∫∫ ln2 0 ∆= > spontaneous IMPORTANT!! To calculate S for the irreversible process, we needed to find a reversible path so we could determine đ đ and . Mixing of ideal gases at constant and p n A A (g, A , ) + B A (g, B , ) = (A + B) (g, , ) n V A A n V B B nn VV = + A A n V B B spontaneous mixing To calculate , we need to find a reversible path between the two states. mix A B A + B constant T piston permeable to A only piston to B only back to initial state demix function of state 20.110J / 2.772J / 5.601J
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This note was uploaded on 11/11/2011 for the course BIO 20.010j taught by Professor Lindagriffith during the Spring '06 term at MIT.

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l07 - 20.110J / 2.772J / 5.601J Thermodynamics of...

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