One of the assumptions of the kinetic theory of gases states that "there is no force of attraction between the molecules of a gas." How far is this statement correct? Is it possible to liquefy an ideal gas? Explain.
(N/A) This statement is correct only for an ideal gas. It is impossible to liquefy an ideal gas because there are no intermolecular forces of attraction between the molecules of an ideal gas. Liquefaction requires the presence of intermolecular forces to bring molecules closer together to form a liquid phase.
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For gases $W, X, Y$ and $Z$,the van der Waals constants are given below. Which gas has the highest critical temperature?
Gas $a$ (atm $L^2$ mol$^{-2}$) $b$ ($L$ mol$^{-1}$) $W$ $4.0$ $0.027$ $X$ $8.0$ $0.030$ $Y$ $6.0$ $0.032$ $Z$ $12.0$ $0.027$
| Gas | $a$ (atm $L^2$ mol$^{-2}$) | $b$ ($L$ mol$^{-1}$) |
|---|---|---|
| $W$ | $4.0$ | $0.027$ |
| $X$ | $8.0$ | $0.030$ |
| $Y$ | $6.0$ | $0.032$ |
| $Z$ | $12.0$ | $0.027$ |
Which among the following gases liquefy easily?
Gases possess characteristic critical temperature which depends upon the magnitude of intermolecular forces between the particles. Following are the critical temperatures of some gases.
$Gases$ $H_2$ $He$ $O_2$ $N_2$ $Critical \ temperature \ (K)$ $33.2$ $5.3$ $154.3$ $126$
From the above data,what would be the decreasing order of liquefaction of these gases?
| $Gases$ | $H_2$ | $He$ | $O_2$ | $N_2$ |
| $Critical \ temperature \ (K)$ | $33.2$ | $5.3$ | $154.3$ | $126$ |
From the above data,what would be the decreasing order of liquefaction of these gases?
Medium
View SolutionWhich intermolecular force is most responsible for allowing $Xe$ gas to liquefy?
DifficultJEE MAIN 2016
View SolutionUnder what conditions can an ideal gas,which follows the kinetic gas equation,be liquefied?
Medium
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