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Liquefaction of Gases and Critical Constants

If excluded v...

Question

If excluded volume per mole of gases $A, B, C$ is same. If boyle's temperature of $A, B$ and $C$ are in ratio $1 : 3 : 5$ , then ratio of there critical pressure is:

5 : 3 : 1

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2 : 8 : 15

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3 : 9 : 25

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1 : 3 : 5

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Solution

The correct option is A $1 : 3 : 5$
According to the formula
$V_{C} = \frac{3 R T_{C}}{8 P_{C}}$
Where $V_{C}$ is the critical volume
$T_{C}$ is the critical temperature
$P_{C}$ is the critical pressure
$R$ is the gas constant
Temperature of $A = x$
Temperature of $B = 3 x$
Temperature of $C = 5 x$
Volume is same for A,B and C
$P_{C}$ of A $= \frac{3 R x}{8 V_{C}}$
$P_{C}$ of B $= \frac{3 R \times 3 x}{8 V_{C}}$
$P_{C}$ of C $= \frac{3 R \times 5 x}{8 V_{C}}$
Ratio of the critical pressure of A, B and C is
$P_{C} A : P_{C} B : P_{C} C$ = $\frac{3 R x}{8 V_{c}} : \frac{3 R \times 3 x}{8 V_{C}} : \frac{3 R \times 5 x}{8 V_{C}} = 1 : 3 : 5$

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