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(A) The compressibility factor $Z$ is defined as $Z = \frac{PV}{nRT}$.For real gases,the deviation from ideal behavior is represented by the dip in th

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$D$

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(A) The compressibility factor $Z$ is defined as $Z = \frac{PV}{nRT}$.For real gases,the deviation from ideal behavior is represented by the dip in the $Z$ vs $P$ curve.$A$ greater dip indicates stronger intermolecular forces of attraction.Stronger intermolecular forces of attraction correspond to a higher critical temperature $(T_c)$,as it is easier to liquefy the gas.In the given graph,the curve for gas $D$ shows the maximum deviation (the deepest dip) from the ideal behavior $(Z=1)$.Therefore,gas $D$ has the strongest intermolecular forces and the highest critical temperature.

The $Z$ vs $P$ graph for four gases is given below at the same temperature.
Which of the gases will have the highest value of critical temperature $(T_c)$?

Similar Questions

Consider the following gases and their corresponding $ 'a'$ values given in parentheses: $CH_4 (2.25), N_2 (1.35), O_2 (1.36)$,and $CO (1.46)$. Which of these gases will be the most difficult to liquefy?

Which gas shows real behavior?

Define the compressibility factor $(Z)$ in terms of the molar volume of a gas.

At low pressure,the van der Waals equation becomes:

$A$ real gas obeying Van der Waals equation will resemble an ideal gas if the:

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The $Z$ vs $P$ graph for four gases is given below at the same temperature.Which of the gases will have the highest value of critical temperature $(T_c)$?