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The pressure exerted by `1 mol` of `CO_(2)` at `273 K` is `34.98 atm`. Assuming that volume occupied by `CO_(2)` molecules is negligible, the value of van der Waals constant for attraction of `CO_(2)` gas isA. `3.59 dm^(6) atm mol^(-2)`B. `2.59 dm^(6) atm mol^(-2)`C. `1.25 dm^(6) atm mol^(-2)`D. `1.59 dm^(6) atm mol^(-2)` |
Answer» `[P+(a)/(V^(2))][V-b]=RT` `:. [P+(a)/(V^(2))]V=RT` or `V^(2)P-RTV+a=0` `V=(+-RT+-sqrt(R^(2)T^(2)-4Pa))/(2P)` Since, `V` is constant at given `P` and `T`, `V` can have only one value or discriminant `=0` `:. R^(2)T^(2)=4Pa` or `a=(R^(2)T^(2))/(4P)` `=((0.821)^(2)xx(273)^(2))/(4xx34.98)` `=3.59 dm^(6) atm mol^(-2)` |
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