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 ² V,N = - S (1) ± ∂F ∂V ² 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 = ³ ∂F ∂T ´ V,N dT + ³ ∂F ∂V ´ T,N dV . ± ∂F ∂N ² 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 ² V,N = - U (4) Start from the deﬁnition of F , and write it as U = F + TS But we also know (!) that S = - ³ ∂F ∂T ´ V,N , so that U = F - T ± ∂F ∂T ² V,N By inspection, therefore, U = - T 2 ± ( F/T ) ∂T ² V,N Alternatively, starting from the deﬁnition: F T = U T - S 1

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so that d ± F T ² = d ± U T ² - dS or d ± F T ² = 1 T dU - U T 2 dT - dS But, dU = TdS - PdV + μdN , so d ± F T ² = - P T dV - U T 2 dT + μdN and U T 2 = - ± ( F/T ) ∂T ² 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 ² 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 ² T,N = 0 - Nk B T 1 V = - P Finally, keeping T and V constant, we obtain ± ∂F ∂N ² 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 ² 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

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

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