The critical temperatures for carbon dioxide | vedclass

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(A) The critical temperature $(T_c)$ of a gas is a measure of the strength of intermolecular forces of attraction. Higher $T_c$ indicates stronger int

Vedclass · Accepted Answer

Carbon dioxide is easier to liquefy than methane.

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(A) The critical temperature $(T_c)$ of a gas is a measure of the strength of intermolecular forces of attraction. Higher $T_c$ indicates stronger intermolecular forces,making the gas easier to liquefy.For $CO_2$,$T_c = 31.1 \, ^\circ C$,and for $CH_4$,$T_c = -81.9 \, ^\circ C$.Since $31.1 \, ^\circ C > -81.9 \, ^\circ C$,the intermolecular forces in $CO_2$ are stronger than in $CH_4$.Therefore,$CO_2$ can be liquefied more easily than $CH_4$ at a given temperature.

The critical temperatures for carbon dioxide $(CO_2)$ and methane $(CH_4)$ are $31.1 \, ^\circ C$ and $-81.9 \, ^\circ C$ respectively. Which of the following statements is correct regarding their liquefaction?

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