P t t p we also have the identity dgpt comparing

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Unformatted text preview: T ⎠V ⎝ ∂V ⎠T Let us calculate the change of entropy in a process where both V and T are changed. ⎛ ∂S ⎞ ⎛ ∂S ⎞ ⎛ ∂P⎞ ⎛ ∂S ⎞ ⎟ dV + ⎜ ∂T ⎟ dT = ⎜ ∂T ⎟ dV + ⎜ ∂T ⎟ dT ⎝ ∂V ⎠ T ⎝ ⎠V ⎝ ⎠V ⎝ ⎠V We have dS = ⎜ (4) ⎛ ∂S ⎞ ⎟ yet. However this ⎝ ∂T ⎠V We do not have any convenient way of calculating or measuring ⎜ derivative can be easily expressed in terms of the heat capacity, which can be measured. Consider a process at a constant V. Then we have δq = n CV dT = T dS so that ⎛ ∂S ⎞ = CV / T = nCV / T ⎝ ∂T ⎟ V ⎠ ⎜ Eq. 4 finally becomes: CV ⎛ ∂P⎞ dS = ⎜ ⎟ dV + T dT ⎝ ∂T ⎠ V (5) Exercise: show that Eq. 5 reduces to S = Rn ln V + CV ln T for an ideal gas whose CV is temperature independent. How to express U in terms of measurable quantities. ⎛ ∂P ⎞ ⎟ dV . Combining these ⎝ ∂T ⎠V We have dU = TdS – P dV. For dS we obtained: dS = n C v dT/T + ⎜ two equations, we have: ⎛ ∂P⎞ - P] dV ⎝ ∂T ⎟ V ⎠ dU = C v n dT + [T ⎜ ⎛ ∂P ⎞ ⎟ = P so we get ⎝ ∂T ⎠V...
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This note was uploaded on 02/13/2014 for the course CH 353M taught by Professor Lim during the Spring '08 term at University of Texas at Austin.

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