When pressure is increases at constant temp volume of gas decreases `AB rarr gases, BC rarr vpour +liquid, CD rarr liquid`
critical point: At this point all the physical propeties of liquid phase will be same as the physical properties in vapour such as, density of liquid = density of vapour
`T_(c)` or critical temp: Temperature abive which a gas can not be liquified
`P_(c)` or critical pressure: minimum pressure which must be applied at critical temp to convert the gas into liqid.
`V_(c)` or critical volume: volume occupied by one mole of gas at `T_(c) & P_(c)`
CRITICAL CONSTANT USING VANDERWAAL EQUATIONS:
`{:((P+(a)/(V_(m)^(2)))(V_(m)-gb)=RT,rArr,(PV_(m)^(2)+a)(V_(m)-b)=RTV_(m)^(2)),(PV_(m)^(3)+aV_(m)-PbV_(m)^(2) -ab -RTV_(m)^(2) =0, rArr, V_(m)^(3)+V_(m)^(2) (b+(RT)/(P))+(a)/(P)(V)/(m)-(ab)/(P) = 0):}`
since equation is cubic in `V_(m)` hence there will be three roots of equation of any temperature and pressure. At critical point all three roots will coincide and will given single value of `V_(m) = V_(c)` at critical point. Vander Waal 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)^(3) - 3V_(m)^(2) V_(c) +3V_(m) V_(c)^(2) -V_(c)^(3) = 0 ..(2)`
comparing with equation (1)
`b +(RT_(c))/(P_(c)) = 3V_(c)` ...(i) `(a)/(P_(c)) = 3V_(c)^(2)` ...(ii) `(ab)/(P_(c)) = V_(c)^(3)` ..(iii)
From (ii) and (iii), `V_(c) = 3b`
From (ii) `P_(c) = (a)/(3V_(c)^(2))` substituting `P_(c) = (a)/(3(3b)^(2)) = (a)/(27b^(2))`
From (i) `(RT_(c))/(P_(c)) = 3V_(c) -b = 9b -b = 8b rArr T_(c) = (8a)/(27Rb)`
At critical point, the slope of `PV` curve (slope of isotherm) will be zero at all other point slope will be negative zero is the maximum value of slope.
`((delP)/(delV_(m)))_(TC) =0 ..(i) (del)/(delV_(m)) ((delP)/(delV_(m)))_(TC) = 0 ...(ii)`
{Mathematically such points an known as point of inflection (where first two derivatives becimes zero)}
using the two `T_(c)P_(c)` and `V_(c)` can be calculated by

If the critical constants for a hypotentical gas are `V_(c) = 150 cm^(3) mol^(-1). P_(c) = 50 atm` and `T_(c) = 300K`. then the radius of the molecule is: [Take `R = (1)/(12)Ltr atm mol^(-1)K^(-1)]`

Options

A. `((75)/(2piN_(A)))^(1//3)`
B. `((75)/(8piN_(A)))^(1//3)`
C. `((3)/(piN_(A)))^(1//3)`
D. `((3)/(256piN_(A)))^(1//3)`

Answer

D

Explanation

`P_(c), V_(c)` and `T_(c)` are given hence 'a' and 'b' should be calculated using `P_(c)` and `T_(c)` as it is more reliable.
`P_(c) = (a)/(27b^(2)), T_(c) = (8a)/(27Rb)`
`(P_(c))/(T_(c)) = (R )/(8b) rArr b = (300 xx 1//12)/(8 xx 50) = (1)/(16)`
`4 xx (4)/(3) pi r^(3). N_(A) = (1)/(16) rArr r = ((3)/(256piNa))^(1//2)`