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The pressure exerted by 1 mole of  \small CO_2 at 273K is 34.98 atm assuming that the volume occupied by \small CO_2  molecules is negligible. The value of Van der Waals constant (in dm6 atm mole-2) for attraction of  \small CO_2  gas is:

Option: 1


Option: 2


Option: 3


Option: 4


Answers (1)


As the volume occupied by the molecules of  CO_2  is negligible, thus the value of b is negligible.
Thus the Van der Waals equation is given as follows:

\left [ P+\frac{a}{V^2} \right ]V=RT

V=\frac{RT\pm \sqrt{R^2T^2-4Pa}}{2P}

Since V can have only one value, thus discriminant must be zero.

\therefore R^2T^2=4Pa

a=\frac{R^2T^2}{4P}=\frac{(0.0821)^2\times (273)^2}{4 \times 34.98}=3.59dm^6\: atm\: mol^{-2}


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