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Category > Applied Sciences Posted 15 Jun 2017 My Price 5.00

A gas analyzes 60% methane and 40% ethylene by volume

 

A gas analyzes 60% methane and 40% ethylene by volume. It is desired to store 12.3 kgof this gas mixture in a cylinder having a capacity of 5·14 ´ 10-2 m3 at a maximum temperatureof 45°C. Calculate the pressure inside the cylinder by(a) assuming that the mixture obeys the ideal gas laws;(b) using the compressibility factor determined by the pseudo critical point method.
 
 

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Alpha Geek
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Status NEW Posted 15 Jun 2017 10:06 AM My Price 5.00

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file 1497521748-Answer.docx preview (267 words )
A-----------nsw-----------er ----------- Pa-----------rt -----------a ------------ ga-----------s m-----------ixt-----------ure----------- by----------- vo-----------lum-----------e ------------100----------- li-----------ter-----------s m-----------eth-----------ane----------- = -----------60 -----------lit-----------ers----------- et-----------hyl-----------ene----------- = -----------40 -----------lit-----------ers----------- fo-----------r i-----------dea-----------l g-----------as -----------mix-----------tur-----------e ------------ vo-----------lum-----------e %----------- = -----------mol----------- fr-----------act-----------ion----------- % -----------CH4----------- ga-----------s =----------- 0.-----------6 ,----------- C2-----------H4 -----------= 0-----------.4 -----------Mas-----------s o-----------f g-----------as -----------mix-----------tur-----------e =----------- 12----------- .3----------- kg----------- = -----------12.-----------3*1-----------000----------- =1-----------230-----------0 g----------- V -----------=5.-----------14 -----------*10----------- ^------------2 m-----------3 =----------- 5.-----------14*-----------100-----------0 /-----------100----------- =5-----------1.4----------- li-----------ter-----------s ,----------- T -----------=45----------- +2-----------73.-----------15 -----------= 3-----------18.-----------15 -----------K A-----------vea-----------rag-----------e m-----------ole-----------cul-----------ar -----------wt -----------of -----------gas----------- mi-----------xtu-----------re -----------= 0-----------.6 -----------*16----------- + -----------0.4----------- * -----------( 1-----------2*2----------- +4-----------*1 -----------) =----------- 9.-----------6 +-----------11.-----------2 =-----------20.-----------8 g----------- /m-----------ol -----------no -----------of -----------mol-----------es -----------of -----------gas----------- mi-----------xtu-----------re -----------= 1-----------230-----------0 /-----------20.-----------8
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