When perssure is increase at constant temperature, volume of gas decreases, AB `to` gases,Bc `to` vapour +liquid ,CD `to` liquid critical point .At this all the physical properties of liquid phase as density of liquid = density if vapour.
`T_(e)` or critical temp : Temperature above which a gas can not be liquified
`V_(e)` or critical volume : Volume occupied by one mole of gas at `T_(e)` abd `P_(e).`
Critical constant Using van der Waals' Equations
`(P+(a)/(V_(m)^(2)))(V_(m)-b)=RT`
`implies(PV_(m)^(2)+a)(V_(m)-b)=RTV_(m)^(2)`
`PV_(m)^(3)+aV_(m)-PbV_(m)^(2)=ab-RTV_(m)^(2)=0`
`impliesV_(m)^(3)-V_(m)^(2)(b+(RT_(e))/(P))+(a)/(p)(V)/(m)-(ab)/(p)=0`
Since equation is cubic in `V_(m)` hence there will be roots of equation at any temperature and pressure.At critical point van der Waals' equation will be
`V_(m)^(3)-V_(m)^(2)(b+(RT_(c))/(P_(c)))+(a)/(p_(c))V_(m)-(ab)/(p_(c))=0" " ....(i)`
But at critical point all three roots of the equation should be equal, hence equation should be :
`V_(m)=V_(c)`
`(V_(m)-V_(c))^(3)=0`
`V_(m)-3V_(m)^(2)V_(c)+3V_(m)V_(c)^(2)-V_(c)^(3)=0`....(ii)
Comparing with equation (i)
`b+(RT_(c))/(p_(c))=3V_(c)` ...(iii)
`(a)/(p_(c))=3V_(e)^(2)`...(iv)
`(ab)/(p_(c))=V_(c)^(3)`....(v)
from eqns.(iv) and (v0, `V_(c)=3b`
from eq. (iv) `p_(c)=(a)/(3V_(c)^(2))` substituting `p_(c)=(a)/(3(3b)^(2))=(a)/(27b^(2))`
from e.q (iii) `(RT_(c))/(p_(c))=3V_(c)-b=9b-b=8b`
`impliesT_(c)=(8a)/(27Rb)`
At critical point, the slope of PV curve (slope of isotherm) will be zero at all other point slpoe will be negative zero is the maximum value of slope
`((delP)/(delV_(m)))_(Tc)=0`...(iv)
`(del)/(delV_(m))((delP)/(delV_(m)))_(Tc)=0`
[Mathematically such points are known as point of inflection (where first two derivatives becomes zero) Using the two `T_(c)P_(c)` and `V_(c)` can be calculated
A scientist proposed the following equation of state
`P=(RT)/(V_(m))-(B)/(V_(m)^(2))+(C)/(V_(m)^(2))` If this equation leads to the critical behaviour then critical temperature is
