Calculate the pressure exerted by one mole of `CO_(2)` gas at `273 K` van der Waals constant `a=3.592 dm^(6) atm mol^(-2)`. Assume that the volume occupied by `CO_(2)` molecules is negligible.

Options

A. `0.99 atm`
B. `17.85 atm`
C. `34.96 atm`
D. `18.20 atm`

Answer

C

Explanation

`[P+(a)/(V^(2))][V-b]=RT`
or `V^(2) P -RTV + a =0`
`V = + RT root(+-)(R^(2)T^(2)-4aP)/(2P)`
Since `V` has only one value
`R^(2) T^(2) = 4aP`
or `P=(R^(2)T^(2))/(4a)`
`=((0.0821)^(2)xx273xx273)/(4 xx3.5.592)`
` =34.96` atm .