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UBC MECH327 A4 Q3.tex
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\documentclass[10pt]{article} \usepackage{amssymb,amsmath} \usepackage[hmargin=1cm,vmargin=1cm]{geometry} \usepackage{cancel} \begin{document} {\large UBC MECH327 Assignment 4 Q3} \begin{align*} &m_1=0.5\text{ kg},\quad T_1=353\text{ K},\quad P_1=1\text{ bar}.\\ &m_2=1.0\text{ kg},\quad T_2=323\text{ K},\quad P_2=2\text{ bar}.\\ \\ &\text{After the valve is opened and the two connected tanks reach equilibrium,}\\ &\text{the air in it has mass $m$ (kg), temperature $T$ (K) and pressure $P$ (bar).}\\ \\ &\text{As the tanks are insulated, the sum of heat transfer of both tanks must be zero.}\quad Q=mc\Delta T.\\ &m_1c(T-T1)+m_2c(T-T2)=0,\quad m_1(T-T1)=-m_2(T-T2),\\ \\ &T=\frac{m_1T_1+m_2T_2}{m_1+m_2}=\frac{0.5\times 353+1.0\times 323}{0.5+1.0}=333\text{ K}=60^\circ\text{C}.\\ \\ &\text{Ideal gas assumed:}\quad PV=nRT,\quad PV=\frac{m}{M}RT,\quad\text{where M is average molar mass of air.}\\ &V=\frac{mT}{P}\cdot\frac{R}{M}.\\ \\ &V=V_1+V_2,\quad\frac{mT}{P}\cdot\frac{R}{M}=\frac{m_1T_1}{P_1}\cdot\frac{R}{M}+\frac{m_2T_2}{P_2}\cdot\frac{R}{M}.\\ &\frac{T}{P}=\frac{\frac{m_1T_1}{P_1}+\frac{m_2T_2}{P_2}}{m_1+m_2}=225.333.\\ \\ &P=\frac{T}{225.333}=\frac{333}{225.333}=1.47781\text{ bar}.\\ \end{align*} \end{document}