For a real gas mol-mass-30 if density as critical point is 0-40 g- cm 3 and T C - 2- 10 5 821 K- then calculate van der Waals constant a in atm L 2 mol -2-

T_C=8a27Rb=2× 10^5821 ∴ b= 8 × 821a27 × 0.0821 × 2× 10^5=2a135 ⇒ PV= mMRT ⇒ P= dRTM Critical pressure= a27b^2=dRTM ∴a27(4a^2135× 1 ...

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Standard XII

Chemistry

Real Gases

For a real ga...

Question

For a real gas (mol.mass=30) if density as critical point is $0.40 g / {c m}^{3}$ and $T_{C} = \frac{2 \times 10^{5}}{821} K$ , then calculate van der Waals constant $a$ (in atm $L^{2} {m o l}^{- 2}$ ).

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Solution

$T_{C} = \frac{8 a}{27 R b} = \frac{2 \times 10^{5}}{821}$
$∴ b = \frac{8 \times 821 a}{27 \times 0.0821 \times 2 \times 10^{5}} = \frac{2 a}{135}$
$\Rightarrow P V = \frac{m}{M} R T$
$\Rightarrow P = \frac{d R T}{M}$
Critical pressure= $\frac{a}{27 b^{2}} = \frac{d R T}{M}$
$∴ \frac{a}{27 (\frac{4 a^{2}}{135 \times 135})} = \frac{10^{3} \times 0.40 \times 0.0821 \times \frac{2 \times 10^{5}}{821}}{30}$
$∴ a = 0.633 a t m L^{2} m o l^{- 2}$

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For a real gas (mol.mass=30) if density as critical point is 0.40g/cm3 and TC=2×105821K, then calculate van der Waals constant a (in atm L2mol−2).

TC=8a27Rb=2×105821∴b=8×821a27×0.0821×2×105=2a135⇒PV=mMRT⇒P=dRTMCritical pressure= a27b2=dRTM∴a27(4a2135×135)=103×0.40×0.0821×2×10582130∴a=0.633atmL2mol−2

For a real gas (mol.mass=30) if density as critical point is $0.40 g / {c m}^{3}$ and $T_{C} = \frac{2 \times 10^{5}}{821} K$ , then calculate van der Waals constant $a$ (in atm $L^{2} {m o l}^{- 2}$ ).