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 523: 36

Answer

a) $T_B\gt T_A$ b) $5200\;\rm J,\;7800\;J$

Work Step by Step

a) To find which gas has the higher initial temperature, we need to find the initial temperature of each gas. We know that the thermal energy of a monatomic gas is given by $$E_{th}=nC_{\rm V}T=\frac{3}{2}nRT\tag 1$$ Hence, $$T=\dfrac{2E_{th}}{nR}$$ Hence, the initial temperature of gas A is $$T_{iA}=\dfrac{2E_{(th,iA)}}{3n_AR}=\dfrac{2(5000)}{3(2)(8.31)}$$ Hence, $$T_{iA}=\bf 201\;\rm K$$ And the initial temperature of gas B is $$T_{iB}=\dfrac{2E_{(th,iB)}}{3n_BR}=\dfrac{2(8000)}{3(3)(8.31)}$$ Hence, $$T_{iB}=\bf 214\;\rm K$$ Therefore, $$\boxed{T_{iB}\gt T_{iA}}$$ --- b) We assume that the system (gas A+gasB) is an isolated system which means that there is no energy that comes in or out of the system. So the heat lost by one of them is gained by the other. $$Q_A+Q_B=0$$ And hence, $$Q_A=-Q_B\tag 2$$ Hence, after a long time, the system will reach an equilibrium point at which the two gases are having the same temperature. Now we need to find the final temperature of the system. We know that $$\Delta E_{th}=Q+W$$ and there is no work done on any of our two samples since their volumes remain constant. Hence, $\Delta E_{th,A}=Q_A$ and $\Delta E_{th,B}=Q_B$ and from (2), we got $$\Delta E_{th,A}=-\Delta E_{th,B}$$ So, $$ \color{red}{\bf\not} \frac{3}{2}n_A \color{red}{\bf\not} R(T_f-T_{iA})=- \color{red}{\bf\not} \frac{3}{2}n_B \color{red}{\bf\not} R(T_f-T_{iB})$$ $$ n_A T_f-n_A T_{iA}=- n_B T_f+n_BT_{iB}$$ $$ n_A T_f+ n_B T_f= n_BT_{iB}+n_A T_{iA}$$ $$ (n_A + n_B )T_f= n_BT_{iB}+n_A T_{iA}$$ $$T_f=\dfrac{n_BT_{iB}+n_A T_{iA}}{n_A + n_B}$$ Plugging the known; $$T_f=\dfrac{(3)(214)+(2)(201)}{2+ 3}=\bf 208.8\;\rm K$$ Now we can use (1) to find the final thermal energy of each gas: $$E_{(th,fA)}=\frac{3}{2}n_ART_f=\frac{3}{2}(2)(8.31)(208.8)$$ $$E_{(th,fA)}=\color{red}{\bf 5200}\;\rm J$$ $$E_{(th,fB)}=\frac{3}{2}n_BRT_f=\frac{3}{2}(3)(8.31)(208.8)$$ $$E_{(th,fB)}=\color{red}{\bf 7800}\;\rm J$$
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