The critical constants for water are 374- C - 217 atm and 0-05 L mol -1- calculate a and b- A- a -1-92 L 2 atm mol -2 - b -0-028 L mol -1B- a -1-64 L 2 atm mol -2 - b -0-017 L mol -1C- a -2-95 L 2 atm mol -2- b -0-011 L mol -1D- a -3-92 L 2 atm mol -2 - b -0-03 L mol -1

Given, T_c = 374 ^∘ C P_c = 218 atm V_c = 0.05 L mol^ 1 We know, a = 3P_c V_c^2 nbsp;= 3 × 218 atm × (0.05 L / mol)^2 = 1.635 L^2 atm mol^ 2 ≈ 1.64 L^2 a ...

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Standard XI

Mathematics

Jump of a Discontinuity

The critical ...

Question

The critical constants for water are $374^{\circ} C, 217$ atm and 0.05 L mol $^{- 1}$ . calculate $a$ and $b$ .

a = 3.92 L^{2} a t m m o l^{- 2}; b = 0.03 L m o l^{- 1}

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a = 1.92 L^{2} a t m m o l^{- 2}; b = 0.028 L m o l^{- 1}

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a = 1.64 L^{2} a t m m o l^{- 2}; b = 0.017 L m o l^{- 1}

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a = 2.95 L^{2} a t m m o l^{- 2} : b = 0.011 L m o l^{- 1}

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Solution

The correct option is C $a = 1.64 L^{2} a t m m o l^{- 2}; b = 0.017 L m o l^{- 1}$
Given,
$T_{c} = 374^{\circ} C$
$P_{c} = 218$ atm
$V_{c} = 0.05$ L mol $^{- 1}$

We know,
$a = 3 P_{c} V_{c}^{2}$
$= 3 \times 218 a t m \times (0.05 L / m o l)^{2}$
$= 1.635 L^{2} a t m m o l^{- 2}$
$\approx 1.64 L^{2} a t m m o l^{- 2}$
Again,
$b = \frac{V_{c}}{3} = \frac{0.05 L m o l^{- 1}}{3}$
$= 0.017 L m o l^{- 1}$

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The critical constants for water are 374 ∘C, 217 atm and 0.05 L mol−1. calculate a and b.

The correct option is C a=1.64 L2 atm mol−2; b=0.017 L mol−1Given, Tc=374 ∘C Pc=218 atm Vc=0.05 L mol−1 We know, a=3PcV2c =3×218 atm×(0.05 L/mol)2 =1.635 L2atm mol−2 ≈1.64 L2atm mol−2 Again, b=Vc3=0.05 L mol−13 =0.017 L mol−1

The critical constants for water are $374^{\circ} C, 217$ atm and 0.05 L mol $^{- 1}$ . calculate $a$ and $b$ .