20_562ln08

20_562ln08 - MIT OpenCourseWare http:/ocw.mit.edu 5.62...

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MIT OpenCourseWare http://ocw.mit.edu 5.62 Physical Chemistry II Spring 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms .
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5.62 Lecture #20: Virial Equation of State Goal: Derive Virial Eqn. of State p = V A = kT ln V Q N , T N , T pressure ( 2 π mkT ) 3 N /2 Z N , V , T ) = ( 2 π mkT ) 3 N /2 V N exp N β N ⎞ ⎤ Q = N ! h 3 N ( N ! h 3 N 2 V ln Q = ln ( 2 π mkT ) 3 N /2 + N ln V + N N N ! h 3 N 2 V Plugging ln Q into equation for p … p = kT ( constants ) N lnV ( N 2 β /2V ) V + V + V N β N 2 NkT N 2 kT β = kT 0 + V 2V 2 = V 2V 2 Nk = nR, nN a = N N a n β ⎛ nRT pV N a β ⎛ RT RT + B 2 T V units Volume/mol . pV = nRT 2 V n pV = RT 2 V ( ) RT pV = RT + B 2 T V ( ) RT Virial Equation of State ( ) = 2 π N a 2nd VIRIAL COEFFICIENT B 2 T = N 2 a β 0 dr r 2 [e u(r)/kT 1] [The 1st VIRIAL COEFFICIENT, B 1 (T), is 1!]
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Lecture #20, Page 2 As T → ∞ , B 2 ( T ) 0 because [ e u ( r ) kT 1 ] 0 At finite high T, B 2 ( T ) > 0 At low T, B 2 ( T ) < 0 low T — attractive forces dominate pressure is lower than ideal high T — repulsive forces dominate p is higher than ideal (pV = RT) Typical Values of B 2 (T) … in cm
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20_562ln08 - MIT OpenCourseWare http:/ocw.mit.edu 5.62...

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