Peeter Joot [email protected]

Thermodynamic identities Impressed with the clarity of Baez’s entropic force discussion on differential forms [1], let’s use that methodology to find all the possible identities that we can get from the thermodynamic identity (for now assuming N is fixed, ignoring the chemical potential.) This isn’t actually that much work to do, since a bit of editor regular expression magic can do most of the work. Our starting point is the thermodynamic identity ¯ + dW ¯ = TdS − PdV, dU = dQ

(1.1)

0 = dU − TdS + PdV.

(1.2)

or It’s quite likely that many of the identities that can be obtained will be useful, but this should at least provide a handy reference of possible conversions. Differentials in P, V

This first case illustrates the method.

0= dU −TdS + PdV        ∂U ∂U ∂S ∂S = dP + dV − T dP + dV + PdV ∂P V ∂V P ∂P ∂V P       V     ∂U ∂S ∂U ∂S = dP −T + dV −T +P . ∂P V ∂P V ∂V P ∂V P Taking wedge products with dV and dP respectively, we form two two forms      ∂U ∂S −T 0 = dP ∧ dV ∂P V ∂P V      ∂U ∂S 0 = dV ∧ dP −T +P . ∂V P ∂V P Since these must both be zero we find 

∂U ∂P



 =T V

1

∂S ∂P

(1.3)

(1.4a) (1.4b)

 (1.5a) V

 P=−

∂U ∂V





−T P

∂S ∂V

 .

(1.5b)

P

Differentials in P, T 0 = dU  −TdS + PdV            (1.6) ∂U ∂S ∂S ∂V ∂V ∂U dP + dT − T dP + dT + dP + dT, = ∂P T ∂T P ∂P T ∂T P ∂P T ∂T P or



     ∂U ∂S ∂V 0= −T + ∂P T ∂P T ∂P T       ∂U ∂S ∂V 0= −T + . ∂T P ∂T P ∂T P

(1.7a) (1.7b)

Differentials in P, S 0 = dU         −TdS + PdV ∂U ∂V ∂V ∂U dP + dS − TdS + P dP + dS , = ∂P S ∂S P ∂P S ∂S P or

   ∂V ∂U = −P ∂P S ∂P S     ∂U ∂V T= +P . ∂S P ∂S P

(1.8)



(1.9a) (1.9b)

Differentials in P, U 0 = dU − TdS     + PdV      ∂S ∂S ∂V ∂V = dU − T dP + dU + P dP + dU , ∂P U ∂U P ∂P U ∂U P or



∂S ∂U





∂V ∂U P     ∂V ∂S =P . T ∂P U ∂P U

0 = 1−T

2

(1.10)



+P

(1.11a) P

(1.11b)

Differentials in V, T 0 = dU  −TdS + PdV        ∂U ∂U ∂S ∂S = dV + dT − T dV + dT + PdV, ∂V T ∂T V ∂V T ∂T V or



 ∂S 0= −T +P ∂V T T     ∂U ∂S =T . ∂T V ∂T V ∂U ∂V



(1.12)



(1.13a) (1.13b)

Differentials in V, S 0 = dU   −TdS + PdV  ∂U ∂U dV + dS − TdS + PdV, = ∂V S ∂S V or

 ∂U P=− ∂V S   ∂U T= . ∂S V

(1.14)



(1.15a) (1.15b)

Differentials in V, U 0 = dU − TdS          + PdV ∂S ∂V ∂V ∂S dV + dU + P dV + dU = dU − T ∂V U ∂U V ∂V U ∂U V or

   ∂S ∂V 0 = 1−T +P ∂U V ∂U V     ∂S ∂V T =P . ∂V U ∂V U

(1.16)



(1.17a) (1.17b)

Differentials in S, T 0 = dU  − TdS  + PdV         ∂U ∂U ∂V ∂V = dS + dT − TdS + P dS + dT , ∂S T ∂T S ∂S T ∂T S or

   ∂U ∂V 0= −T+P ∂S T ∂S T     ∂U ∂V 0= +P . ∂T S ∂T S

(1.18)



3

(1.19a) (1.19b)

Differentials in S, U 0 = dU − TdS + PdV  = dU − TdS + P or

∂V ∂S



 dS + U

∂V ∂U



 dU

(1.20)

S

  ∂V 1 =− P ∂U S   ∂V T=P . ∂S U

(1.21a) (1.21b)

Differentials in T, U 0 = dU − TdS          + PdV ∂S ∂V ∂V ∂S dT + dU + P dT + dU , = dU − T ∂T U ∂U T ∂T U ∂U T or

(1.22)



   ∂S ∂V 0 = 1−T +P ∂U T ∂U T     ∂V ∂S =P . T ∂T U ∂T U

4

(1.23a) (1.23b)

Bibliography [1] John Baez. Entropic forces, 2012. URL http://johncarlosbaez.wordpress.com/2012/02/01/ entropic-forces/. [Online; accessed 07-March-2013]. 1

5

Peeter Joot [email protected] Thermodynamic ...

Impressed with the clarity of Baez's entropic force discussion on differential forms [1], let's use that methodology to find all the possible identities that we can get ...

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