For a real gas mol- mass -60 - if density at critical point is 0-80 g - cm 3 and its T C -4 - 105-821 K-Find the Van der Waals constant -a-in atm L 2 mol -2 -A- 0-3375B- 3-375C- 0-025D- 1-68

The correct option is B 3.375 V_c = 60/0.80 = 75 cm^3 mol^ 1 b = V_c/3 = 25 cm^3 mol^ 1 = 0.025 mol^ 1 T_c = 8a/27Rb 4 × 10^5/821 = 8 × a/27 × 0.0821 × 0.02 ...

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Byju's Answer

Standard XII

Chemistry

Liquefaction of Gases and Critical Constants

For a real ga...

Question

For a real gas (mol. mass = 60 ),if density at critical point is 0.80 $\frac{g}{c m^{3}}$ and its $T_{c} = \frac{4 \times 10^{5}}{821} K$ .Find the Van der Waals constant “a”(in $a t m L^{2} m o l^{- 2}$ ) .

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 B
3.375

$V_{c} = \frac{60}{0.80} = 75 c m^{3} m o l^{- 1}$

$b = \frac{V_{c}}{3} = 25 c m^{3} m o l^{- 1} = 0.025 m o l^{- 1}$

$T_{c} = \frac{8 a}{27 R b}$

$\frac{4 \times 10^{5}}{821} = \frac{8 \times a}{27 \times 0.0821 \times 0.025}$

$\Rightarrow a = 3.375$

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For a real gas (mol. mass = 60 ),if density at critical point is 0.80 gcm3 and its Tc=4×105821K.Find the Van der Waals constant “a”(in atm L2 mol−2 ) .

The correct option is B 3.375 Vc=600.80=75 cm3 mol−1 b = Vc3=25 cm3 mol−1 = 0.025 mol−1 Tc=8a27Rb 4×105821=8×a27×0.0821×0.025 ⇒a = 3.375

For a real gas (mol. mass = 60 ),if density at critical point is 0.80 $\frac{g}{c m^{3}}$ and its $T_{c} = \frac{4 \times 10^{5}}{821} K$ .Find the Van der Waals constant “a”(in $a t m L^{2} m o l^{- 2}$ ) .

The correct option is B
3.375

$V_{c} = \frac{60}{0.80} = 75 c m^{3} m o l^{- 1}$

$b = \frac{V_{c}}{3} = 25 c m^{3} m o l^{- 1} = 0.025 m o l^{- 1}$

$T_{c} = \frac{8 a}{27 R b}$

$\frac{4 \times 10^{5}}{821} = \frac{8 \times a}{27 \times 0.0821 \times 0.025}$

$\Rightarrow a = 3.375$