1.

The density of steam at `100^(@)C` and `10^(5) Pa` pressure is `0.6 Kgm^(-3)`. Calculate the compresibility factor of steam.

Answer» We know `Z=(PV)/(nRT)` and `d=(PM)/(RT)`
`thereforeZ=(PV)/(nRT)=(PV)/((m//w)RT)=(MPV)/(wRT)`
Since `d=(w)/(V)`
`thereforeZ=(MP)/(dRT)=((18xx10^(-3))xx10^(5))/(0.6xx8.31xx373.15)=0.967`


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