The critical temperature and critical density of `CO^(2)` are `27^(@)C` and `0.45 gcm^(-3)` respectively The van der waals' constnat a is .

Options

A. `0.22 litre^(2) atm mo1^(-2)`
B. `1.23 litre^(2) atm mo1^(-2)`
C. `2.71 litre^(2) atm mo1^(-2)`
D. `3.15 litre^(2) atm mo1^(-2)`

Explanation

`V_(c) = (mo1.wt)/("critical density")`
`= (44)/(0.45) = 97.78cm^(3) mo1^(-1)`
Now `V_(c) = 3b`
`:. B = (97.78)/(3) = 32.59cm^(3) mo1^(-1)`
`= 32.59 xx 10^(-3) litre mo1^(-1)`
`:' (P_(c)V_(c))/(RT_(c)) = (3)/(8)`
`:. P_(c) = (3xx0.0821 xx300)/(8 xx97.78 xx 10^(-3)) =94.45 atm`
`:' P_(c) = (a)/(27b^(2))`
`:. a = 27b^(2)P_(c)`
` =27 xx (32.59 xx 10^(-3))^(2) xx 94.45`
`=2.708 litre^(2) atm mo1^(-2)` .