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Standard XIII

Chemistry

Liquefaction of Gases and Critical Constants

For Argon, th...

Question

For Argon, the critical point occurs at a pressure $P_{C} = 4.83 m P a$ and temperature $T_{c} = 151 K$

Estimate the diameter of an Ar atom.

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Solution

$P_{C} = \frac{1}{27} \frac{a}{b^{2}} . . . . . .1 T_{C} = \frac{8}{27} \frac{a}{R b} . . . . . . .2$
Divide 2 and 1 we get,
$b = \frac{T_{C} R}{8 P_{C}} . . . . . .3$
We also have $b = 4 \times molar volume$
$b = 4 \times N_{A} \times V$
$b = 4 \times N_{A} \frac{4}{3} π (\frac{d}{2})^{3} . . . . . . .4$
where d is diameter of Ar atom
from 3 and 4
$\frac{T_{C}}{P_{C}} = \frac{8}{R} \times 4 \times N_{A} \frac{4}{3} π (\frac{d}{2})^{3}$
$\frac{151}{4.83 \times 10^{6}} = \frac{8}{8.314} \times 4 \times 6.02 \times 10^{23} \times \frac{4}{3} 3.14 (\frac{d}{2})^{3}$
$d^{3} = 25.8 \times 10^{- 30} m^{3}$
$d = 2.96 o A$

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For Argon, the critical point occurs at a pressure PC=4.83 mPa and temperature Tc=151 K Estimate the diameter of an Ar atom.

PC=127ab2......1TC=827aRb.......2 Divide 2 and 1 we get, b=TCR8PC ......3 We also have b=4×molar volume b=4×NA×V b=4×NA43π(d2)3 .......4 where d is diameter of Ar atom from 3 and 4 TCPC=8R×4×NA43π(d2)3 1514.83×106=88.314×4×6.02×1023×433.14(d2)3 d3=25.8×10−30 m3 d=2.96oA

For Argon, the critical point occurs at a pressure $P_{C} = 4.83 m P a$ and temperature $T_{c} = 151 K$

Estimate the diameter of an Ar atom.