The graph of compressibility factor (Z) vs. P for one mole of a real gas is shown in following diagram. The graph is plotted at constant temperature 273K. If the slope of graph at very high pressure `((dZ)/(dP))` is `((1)/(2.8))atm^(-1)` , then calculate volume of one mole of real gas molecules (in L/mol)
Given : `N_(A)=6xx10^(23)` and `R=(22.4)/(273)L atmK^(-1)mol^(-1)`

Answer

2

Explanation

`Z=1+(Pb)/(RT)" high pressure"`
`(dZ)/(dP)=(b)/(RT)=(1)/(2.8)`
`b=(RT)/(2.8)=(22.4)/(2.8)=4xx(N_(A)xx(4)/(3)piR^(3))`
`(N_(A)xx(4)/(3)piR^(3))="Volume of 1 mole gas"`
`=(5.6)/(2.8)=2`