Critical temperature for $C O_{2}$ and $C H_{4}$ are ${31.1}^{o}$ C and $- {81.9}^{o} C$ respectively. Which of these has stronger intermolecular forces and why?

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Solution

Critical Temperature is given as
$T_{C} = \frac{8 a}{27 R b}$
Where $a$ defines the intermolecular attraction between the molecule.
Thus, more the critical temperature, more will be the attraction and less will be the repulsion.
As $T_{C}$ for $C O_{2}$ is higher than $T_{C}$ for $C H_{4}$ , thus attraction or stronger molecular forces will be higher in $C O_{2}$ .

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C O_{2}

{31.1}^{0}

- {81.9}^{0}

Critical temperature for carbon dioxide and methane are 31.1 °C and –81.9 °C respectively. Which of these has stronger intermolecular forces and why?

{C O}_{2}

{C H}_{4}

31.1

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