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) = 4(N_(A) xx (4)/(3)pir^(3))`
`(N_(A) xx (4)/(3)pir^(3)) =` Volume of `1` mol gas molecules