Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 18 - The Micro/Macro Connection - Exercises and Problems - Page 525: 60

Answer

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Work Step by Step

a) We know that the thermal energy of a monatomic gas is given by $$E_{th,1}=\frac{3}{2}n_1RT$$ and the thermal energy of a diatomic gas is given by $$E_{th,2}=\frac{5}{2}n_2RT$$ The total thermal energy of the mixture is now given by $$E_{th,tot}=E_{th,1}+E_{th,2}$$ $$E_{th,tot}=\frac{3}{2}n_1RT+\frac{5}{2}n_2RT$$ $$E_{th,tot}=\frac{RT}{2}(3n_1+5n_2 )$$ Recalling that $E_{th}=nC_{\rm V}T$, so for the mixture, $E_{th}=(n_1+n_2)C_{\rm V}T$ Hence, $$(n_1+n_2)C_{\rm V}\color{red}{\bf\not} T=\frac{R\color{red}{\bf\not} T}{2}(3n_1+5n_2 )$$ Thus, $$\boxed{C_{\rm V}=\dfrac{(3n_1+5n_2 )R}{2(n_1+n_2)}}$$ --- b) When $n_1=0$, which means that the only gas there is the diatomic. Hence, $$C_{\rm V}=\dfrac{(0+5n_2 )R}{2(0+n_2)}=\frac{5}{2}R$$ which is the expected value. Now let's test it when $n_2=0$ and the only gas there is the monatomic gas. $$C_{\rm V}=\dfrac{(3n_1+0)R}{2(n_1+0)}=\frac{3}{2}R$$ which is the expected value.
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