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Liquefaction of Gases and Critical Constants

The critical ...

Question

The critical constant for water are 374 $^{o} C$ , 218 atm and 0.0566 litre mol $^{- 1}$ . Calculate $a$ and $b$ .

a

= 1.095 litre

^{2}

atm mol

^{- 1}

b

= 0.0185 litre mol

^{- 1}

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a

= 1.92 litre

^{2}

atm mol

^{- 1}

b

= 0.0185 litre mol

^{- 1}

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a

= 2.095 litre

^{2}

atm mol

^{- 1}

b

= 0.0189 litre mol

^{- 1}

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a

= 2.95 litre

^{2}

atm mol

^{- 1}

b

= 0.1185 litre mol

^{- 1}

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Solution

The correct option is C $a$ = 2.095 litre $^{2}$ atm mol $^{- 1}$ , $b$ = 0.0189 litre mol $^{- 1}$
$T_{c} = 374^{0} C$

$P_{c} = 218 a t m$

$V_{c} = 0.0566 l i t r e m o l^{- 1}$

$a = 3 P_{c} V_{c}^{2}$

$= 3 \times 218 a t m \times (0.0566 l i t r e / m o l)^{2}$

$= 2.095 {l i t r e}^{2} a t m / m o l$

$b = \frac{V_{c}}{3}$

$= \frac{0.0566 l i t r e / m o l}{3}$

$= 0.0189 l i t r e / m o l$

Hence, option C is correct.

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The critical constant for water are 374oC, 218 atm and 0.0566 litre mol−1. Calculate a and b.

The correct option is C a = 2.095 litre2 atm mol−1 , b = 0.0189 litre mol−1Tc=3740 CPc=218 atmVc=0.0566 litre mol−1a=3PcV2c=3×218atm×(0.0566litre/mol)2=2.095litre2atm/molb=Vc3=0.0566litre/mol3=0.0189litre/molHence, option C is correct.

The critical constant for water are 374 $^{o} C$ , 218 atm and 0.0566 litre mol $^{- 1}$ . Calculate $a$ and $b$ .