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Liquefaction of Gases and Critical Constants

The van der w...

Question

The van der waals constant for $H C l$ are $a = 371.843 {kPa.dm}^{6} {mol}^{- 2}$ and $b = 40.8 {cm}^{3} {mol}^{- 1}$ . Find the critical pressure of this substance.

8.3 Pa

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8.3 MPa

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3.2 atm

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3.2 torr

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Solution

The correct option is B $8.3 MPa$
Given: $a = 371.843 k P a d m^{6} m o l^{- 2} = 371.843 \times 10^{3} P a d m^{6} m o l^{- 2}$

$b = 40.8 c m^{3} m o l^{- 1} = 40.8 \times 10^{- 3} d m^{3} m o l^{- 1}$

The critical pressure is given as,
$P_{c} = \frac{a}{27 b^{2}} = \frac{371.843 \times 10^{3}}{27 \times (40.8)^{2} \times 10^{- 6}}$ $= \frac{371.843 \times 10^{9}}{27 \times (40.8)^{2}} = 8.273 \times 10^{6} Pa = 8.273 MPa$

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Similar questions

Q. The van der Waals constants for

H C l

are

a = 371.843 {kPadm}^{6} {mol}^{- 2}

and

b = 40.8 {cm}^{3} {mol}^{- 1}

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H C l

Q. The van der waals constant for

H C l

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a = 371.843 {kPa.dm}^{6} {mol}^{- 2}

and

b = 40.8 {cm}^{3} {mol}^{- 1}

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For a real gas (mol. mass = 60 ),if density at critical point is 0.80 $\frac{g}{c m^{3}}$ and its $T_{c} = \frac{4 \times 10^{5}}{821} K$ .Find the Van der Waals constant “a”(in $a t m L^{2} m o l^{- 2}$ ) .

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Liquefaction of Gases and Critical Constants

Standard XII Chemistry

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The van der waals constant for HCl are a=371.843 kPa.dm6 mol–2 and b=40.8 cm3 mol−1. Find the critical pressure of this substance.

The correct option is B 8.3 MPaGiven: a=371.843 kPa dm6 mol−2=371.843×103 Pa dm6 mol−2 b=40.8 cm3 mol−1=40.8×10−3 dm3 mol−1 The critical pressure is given as, Pc=a27b2=371.843×10327×(40.8)2×10−6=371.843×10927×(40.8)2=8.273×106 Pa=8.273 MPa

The van der waals constant for $H C l$ are $a = 371.843 {kPa.dm}^{6} {mol}^{- 2}$ and $b = 40.8 {cm}^{3} {mol}^{- 1}$ . Find the critical pressure of this substance.