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(2) The natural independent variables for U are (S,V), from dU = Tds - pdV. U = U (S,V) gives simple expressions for I and p as T = (35), and p = - (3). Suppose you use (V, T) instead. Show that U = U (V,T) leads to a much more complicated expression for p, namely au dT F = f (OV), II + f(V). S (2V)₁ T T² T where f (V) is an unknown function of V.

(2) The natural independent variables for U are (S,V), from dU = Tds - pdV. U = U (S,V) gives simple expressions for I and p as T = (35), and p = - (3). Suppose you use (V, T) instead. Show that U = U (V,T) leads to a much more complicated expression for p, namely au dT F = f (OV), II + f(V). S (2V)₁ T T² T where f (V) is an unknown function of V.

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(2) The natural independent variables for U are (S,V), from dU = TdS – pdV. U = U (S, V) = gives simple expressions for T and p as T instead. Show that U = (24) ² and P: == where f(V) is an unknown function of V. U (V, T) leads to a much more complicated expression for p, namely dT F = S (2 V ) , I/7/71 T² au (). Suppose you use (V,T) S + ƒ(V),
Transcribed Image Text:(2) The natural independent variables for U are (S,V), from dU = TdS – pdV. U = U (S, V) = gives simple expressions for T and p as T instead. Show that U = (24) ² and P: == where f(V) is an unknown function of V. U (V, T) leads to a much more complicated expression for p, namely dT F = S (2 V ) , I/7/71 T² au (). Suppose you use (V,T) S + ƒ(V),
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