5.62 Physical Chemistry II Spring 2008

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1 MIT OpenCourseWare Physical Chemistry II Spring 2008 For information about citing these materials or our Terms of Use, visit:

2 5.62 Lecture #20: Virial Equation of State Goal: Derive Virial Eqn. of State p = A V N,T pressure = kt lnq V N,T (2πmkT ) N /2 Z N,V,T ) = ( 2πmkT ) N /2 V N exp Nβ N Q = ( N!h N N!h N 2 V lnq = ln ( 2πmkT ) N /2 + N lnv + Nβ N N!h N 2 V Plugging ln Q into equation for p p = kt (constants) N lnv (N 2 β / 2V ) + + V V V N βn 2 NkT N 2 ktβ = kt 0 + V 2V 2 = V 2V 2 Nk = nr, nn a = N N a nβ nrt pv N a β RT pv = nrt RT + B2 2 V n pv = RT 2 V pv = RT + B 2 T V ( ) RT Virial Equation of State ( ) RT T V units Volume/mol. B 2nd VIRIAL COEFFICIENT 2 T = N a β 2 dr r 2 [e u (r)/kt 1] 0 [The 1st VIRIAL COEFFICIENT, B 1 (T), is 1!] ( ) = 2πN a

3 5.62 Spring 2008 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 mol 1 500K 400K 00K 200K Ar C2H

4 5.62 Spring 2008 Lecture #20, Page For ρ = n = mol cm V T(K) p ideal (atm) p actual Ar % dif p actual C 2 H 6 % dif ε = 124K k [ε is well depth. We will see this later.] ε 200K k Trend is toward too low p at low T and too high p at high T. There is a difference between Ar and benzene in the sense that benzene seems always to have too low p. If we include more terms in Z RT RT pv = RT + B 2 ( T) V + B ( T) + = B n (T)RTV 1 n [B 1 (T) 1] V 2 n =1 2nd VIRIAL rd VIRIAL COEFF COEFF Calculate B 2 (T) for Hard Sphere Potential where sum of two atomic radii Hard sphere potential: u(r)= r< 0 r B 2 (T) = N 2 a β β = 4π 0 dr r 2 [e u(r ) kt 1] β = 4π dr r 2 ( e 1) + 4π 0 dr r 2 ( e 0 1)

5 5.62 Spring 2008 Lecture #20, Page 4 β = 4π r2 dr + 4π 0 0 β = 4π B 2 (T ) = 2 π Na INDEPENDENT OF TEMPERATURE What is physical significance of B 2 for hard sphere potential? IT IS THE EXCLUDED VOLUME Simple geometric argument independent of statistical mechanics: a volume 4 π N a 4π 2π 2 = N a each volume includes 2 atoms same as B 2 calculated from stat. mech. Hard-sphere equation of state, correct through B 2 (T), is RT RT RT pv = RT + B 2 RT + B 2 p because V V = p B p 2 V 2 p(v B 2 ) = RT p V N a 2π = RT excluded volume Compare to van der Waals eqn. of state: small

6 5.62 Spring 2008 Lecture #20, Page 5 a p + V 2 (V b) = RT excluded volume The true molar volume V is reduced by b. A volume V + b is required to give values of p, R, T that are consistent with the ideal gas law. So far we have considered only the repulsive part of the potential. Now include attractions: e.g., square well, Sutherland, or Lennard-Jones. r < ( ) = Square well potential: u r ε b r < λ 0 r > λ See Non-Lecture: Result is excluded volume + term of opposite sign. r < Sutherland potential: u ( r ) = 6 [goal is to express β in terms of, ε] ε r r ( ) kt β = 4π dr r e u r r 6 kt = 4π dr r 2 e 1 + 4π dr r 2 e ε ( ) expand 4 π 4πε 6 kt dr r 4 = 4 π ε kt 4π dr r 2 (1+ ε 6 r 6 kt 1) for modest (i.e.,not too small) kt (weak attraction) If T is too small, must keep more terms in the expansion. β = 4 π ε 1 β is T-dependent and can be positive at low-t and negative at high-t kt B 2 T ( ) = Na β T π N a ε kt ( ) = π N a hard sphere from attractive part of u(r) High T: T-independent, excluded volume repulsion dominates

7 5.62 Spring 2008 Lecture #20, Page 6 Low T: linear variation of B 2 (T) vs. 1/T. Use this to determine ε. Equation of state: RT pv = RT + B 2 V RT + B 2p p(v B 2 ) = RT 2π 2 p V N a + p π N a ε kt = RT 2π p V N a + 2 π 2 N a ε V = RT 2π Define b = N a, a = 2 π 2 N a ε insert the 2 terms of B 2 (T) p p nn = = a = N a kt pv N V V p(v b) + a / V ( p + a / V 2 )(V b) = RT (ab / V 2 0) van der Waals Eqn. of State! replace p/kt in second term Non-Lecture Square well potential: u r ( ) kt β = 4π dr r e u r r < ( ) = ε b r < λ 0 r > λ = 4π dr r 2 ( e λ 0 1) + 4π dr r 2 ( e ε b kt 1) + 4π dr r 2 λ (e 0 1) = 4 π + 4 (λ) (e ε b kt 1) + 0 π expand this ( ) for modest kt > ε b (weak attraction) 1 + ε b kt 1 β= 4 π + 4 π (λ 1)ε b kt i.e., if T is not too low B 2 T ( ) = Na 2 2 β T 2 excluded volume + term of opposite sign! ( ) = π N a π N a (λ 1)ε b kt

8 5.62 Spring 2008 Lecture #20, Page 7 RT pv = RT + B 2 V RT + B 2 p p(v B 2 ) = RT 2π p V N + p 2 π N a (λ a 1)ε b kt = RT p p nn = = a = N a kt PV N V V 2π p V N a + 2 π 2 N ( a λ 1)ε b V = RT 2π Define b = N a, a = 2 π 2 N ( a λ 1)ε b p(v b) + a / V ( p + a / V 2 )(V b) = RT (ab / V 2 0) van der Waals Eqn. of State!

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