For a real gas MW-60 if density at critical point is 0-80 g-cm 3 and its Tc-4 - 105-821K- then van der Waals constant a in atm L2 mol-2 is -

The correct option is D 3.375 V_c= nbsp;60/0.80=75 cm^3mol^ 1 ; nbsp; b= nbsp;V_c/3=25 cm^3mol^ 1=0.025 L mol^ 1 ∴ T_c= nbsp;8a/27Rb ...

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Standard XII

Physics

Surface Energy

For a real ga...

Question

For a real gas ( $M W = 60$ ) if density at critical point is $0.80 g / c m^{3}$ and its $T_{c} = \frac{4 \times 10^{5}}{821} K$ , then van der Waals constant $a$ (in $a t m L^{2} m o l^{- 2}$ ) is :

0.3375

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3.375

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1.68

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0.025

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Solution

The correct option is D 3.375
$V_{c} = \frac{60}{0.80} = 75 c m^{3} m o l^{- 1}$ ; $b = \frac{V_{c}}{3} = 25 c m^{3} m o l^{- 1} = 0.025 L m o l^{- 1}$
$∴ T_{c} = \frac{8 a}{27 R b}$
$\frac{4 \times 10^{5}}{821} = \frac{8 \times a}{27 \times 0.0821 \times 0.025}$
$\Rightarrow a = 3.375$ .

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For a real gas (MW=60) if density at critical point is 0.80g/cm3 and its Tc=4×105821K, then van der Waals constant a (in atmL2mol−2) is :

The correct option is D 3.375Vc=600.80=75cm3mol−1; b=Vc3=25cm3mol−1=0.025Lmol−1∴Tc=8a27Rb4×105821=8×a27×0.0821×0.025⇒a=3.375.

For a real gas ( $M W = 60$ ) if density at critical point is $0.80 g / c m^{3}$ and its $T_{c} = \frac{4 \times 10^{5}}{821} K$ , then van der Waals constant $a$ (in $a t m L^{2} m o l^{- 2}$ ) is :

The correct option is D 3.375
$V_{c} = \frac{60}{0.80} = 75 c m^{3} m o l^{- 1}$ ; $b = \frac{V_{c}}{3} = 25 c m^{3} m o l^{- 1} = 0.025 L m o l^{- 1}$
$∴ T_{c} = \frac{8 a}{27 R b}$
$\frac{4 \times 10^{5}}{821} = \frac{8 \times a}{27 \times 0.0821 \times 0.025}$
$\Rightarrow a = 3.375$ .