Critical temperatures for A- B- C and D gases are 26- C- 10- C- -80- C and 15- C respectively- Which gas will be liquefied more easily-

The correct option is A A Critical temperature of a gas #160; T_C = 8a27Rb⇒ a ∝ T_C where a is measure of vander Waal's forces #160;of attraction. ...

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Byju's Answer

Standard XII

Chemistry

Liquefaction of Gases and Critical Constants

Critical temp...

Question

Critical temperatures for A, B, C and D gases are $26^{\circ} C, 10^{\circ} C, - 80^{\circ} C$ and $15^{\circ} C$ respectively. Which gas will be liquefied more easily?

A

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B

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C

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D

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Solution

The correct option is A $A$
Critical temperature of a gas
$T_{C} = \frac{8 a}{27 R b} \Rightarrow a \propto T_{C}$

where a is measure of vander Waal's forces of attraction. Higher the critical temperature of a gas greater the intermolecular forces of attraction between the molecules of gas and easily the gas can be liquefied.

Hence gas A whose critical temperature $(25^{\circ} C)$ is highest among all the given options will be liquified more easily.

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Critical temperatures for A, B, C and D gases are 26∘C,10∘C,−80∘C and 15∘C respectively. Which gas will be liquefied more easily?

Critical temperatures for A, B, C and D gases are $26^{\circ} C, 10^{\circ} C, - 80^{\circ} C$ and $15^{\circ} C$ respectively. Which gas will be liquefied more easily?