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Liquefaction of Gases and Critical Constants

Question

$T_{C}$ and $P_{c}$ of a gas are 400 K and 41 atm respectively. The $V_{c}$ is:

A

\frac{400 R}{41}

B

\frac{50 R}{41}

C

\frac{41 R}{41}

D

\frac{300 R}{41}

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Solution

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The critical temperature is the temperature above which a gas cannot be liquified. It is denoted as $T_{c}$ .

$P_{c}$ is the Critical pressure, $V_{c}$ is the critical molar volume

$V_{c} = 3 b$ (from Ideal gas law)

and $b = \frac{R T_{c}}{8 p_{c}} = \frac{R \times 400}{8 \times 41} = \frac{50 R}{41}$

$V_{c} = 3 \times \frac{50 R}{41} = \frac{150 R}{41}$

Hence, option $B$ is correct.

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