**Integration of the Thermodynamic Identity for an Ideal Gas**From the thermodynamic identity at a constant number of particles, we have:\[d\sigma = \frac{dU}{\tau} + \frac{pdV}{\tau} = \frac{1}{\tau} \left( \frac{\partial U}{\partial \tau} \right)_V d\tau + \frac{1}{\tau} \left( \frac{\partial U}{\partial V} \right)_\tau dV + \frac{pdV}{\tau}\]Show by integration that for an ideal gas the entropy is:\[\sigma = C_v \log \tau + N \log V + \sigma_1\]where \( \sigma_1 \) is a constant independent of \( \tau \) and \( V \).

Answered: Image /qna-images/answer/21a555bc-7d16-4863-980d-a5a88d7519fc.jpg

2. (Kittel 6.5) Integration of the thermodynamic identity for an ideal gas. From the thermodynamic identity at a constant number of particles we have JU dU do = + T pdV T T dt+ Show by integration that for an ideal gas the entropy is o=C₂ log T+ Nlog V + 0₁ T where o, is a constant independent of 7 and V. T dV + pdV T

2. (Kittel 6.5) Integration of the thermodynamic identity for an ideal gas. From the thermodynamic identity at a constant number of particles we have JU dU do = + T pdV T T dt+ Show by integration that for an ideal gas the entropy is o=C₂ log T+ Nlog V + 0₁ T where o, is a constant independent of 7 and V. T dV + pdV T

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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