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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.
 Spring '08
 LIM
 Physical chemistry, pH

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