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Solutions7

# Solutions7 - PHYS 333 Winter 2008 Assignment 7 1 The...

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PHYS 333 Winter 2008 Assignment 7 1. The Helmholtz function The Helmholtz function is defined as F = U - TS Starting from this definition, prove the following identities ∂F ∂T V,N = - S (1) ∂F ∂V T,N = - P (2) These require F to be written as a function of T , V and N . But since dF = dU - TdS - SdT , and (from the 2nd Law) dU = TdS - PdV , we have dF = TdS - PdV - TdS - SdT = - PdV - SdT . The required identities come from the mathematical statement that dS must also be dS = ∂F ∂T V,N dT + ∂F ∂V T,N dV . ∂F ∂N T,V = μ (3) The previous “proof” did not explicitly reference N . To include it, write the 2nd Law as TdS = dU + PdV - μdN , or dU = TdS - PdV + μdN . In turn this gives dF = - PdV - SdT + μdN , so that the final term gives the required identity. T 2 ( F/T ) ∂T V,N = - U (4) Start from the definition of F , and write it as U = F + TS But we also know (!) that S = - ∂F ∂T V,N , so that U = F - T ∂F ∂T V,N By inspection, therefore, U = - T 2 ( F/T ) ∂T V,N Alternatively, starting from the definition: F T = U T - S 1

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so that d F T = d U T - dS or d F T = 1 T dU - U T 2 dT - dS But, dU = TdS - PdV + μdN , so d F T = - P T dV - U T 2 dT + μdN and U T 2 = - ( F/T ) ∂T V,N as required. F = μN - PV (5) In class, (see also Schroeder, page 164), we saw that G = μN . Since F = G - PV , we have the required result. Explicitly verify these identities for the Ideal Gas , for which PV = Nk B T U = 3 2 Nk B T S = Nk B [ln [ A ( V/N )( U/N ) 3 / 2 ] + (5 / 2)] μ = - k B T ln [ A ( V/N )( U/N ) 3 / 2 ] First, write F = 3 2 Nk B T - Nk B T [ln [ A ( V/N )( U/N ) 3 / 2 ] + (5 / 2)] . Substituting for U , and keeping V and N constant, we obtain ∂F ∂T V,N = 3 2 Nk B - Nk B [ln [ A ( V/N )( U/N ) 3 / 2 ] + (5 / 2)] - T Nk B T = - S and, keeping T and N constant ∂F ∂V T,N = 0 - Nk B T 1 V = - P Finally, keeping T and V constant, we obtain ∂F ∂N T,V = (3 / 2) k B T - k B T ln [ A ( V/N )( U/N ) 3 / 2 ] - (5 / 2) k B T + k B T = μ The identity involving F/T can be demonstrated by writing F T = 3 2 Nk B - Nk B [ln [ A ( V/N )( U/N ) 3 / 2 ] + (5 / 2)] 2
Then, keeping V and N constant ( F/T ) ∂T V,N = 0 - Nk B × (3 / 2) 1 T = - U/T 2 The identity F = μN - PV is obtained directly, using the equation of state to substitute for PV : F = ( - k B T ln [ A ( V/N )( U/N ) 3 / 2 ]) N - Nk B T exactly as required. and for the VdW gas , for which ( P + a N 2 V 2 )( V - b N ) = Nk B T U = C V T - a N 2 V + U 0 S = C V ln T/T 0 + Nk B ln [( V - b N ) /N ] + S 0 μ = - k B T ln [( V - b N ) /N ] - ( C V /N ) T ln T/T 0 + k B Tb N/ ( V - b N ) - 2 a N/V + [ C V + Nk B - S 0 ] T/N + U 0 /N The parameters C V , U 0 and S 0 are not constants: they depend on N . However, the parameters A , a , b and T 0 are all constants. In order to show N explicitly, I have defined the parameters in the VdW equation to be a = a/N 2 and b = b/N in terms of the usual parameters a and b

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Solutions7 - PHYS 333 Winter 2008 Assignment 7 1 The...

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