For a real gas (molar mass = 60 , g/mol),if t | vedclass

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(B) The critical molar volume $V_c$ is calculated as: $V_c = \frac{	ext{Molar mass}}{	ext{Density}} = \frac{60 \, g/mol}{0.80 \, g/cm^3} = 75 \, cm^

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(B) The critical molar volume $V_c$ is calculated as: $V_c = \frac{	ext{Molar mass}}{	ext{Density}} = \frac{60 \, g/mol}{0.80 \, g/cm^3} = 75 \, cm^3/mol = 0.075 \, L/mol$.For a van der Waals gas,the critical volume is related to the constant $b$ by: $V_c = 3b$.Therefore,$b = \frac{V_c}{3} = \frac{0.075}{3} = 0.025 \, L/mol$.The critical temperature $T_c$ is given by: $T_c = \frac{8a}{27Rb}$.Given $T_c = \frac{4 	imes 10^5}{821} \, K$ and $R = 0.0821 \, L \, atm \, K^{-1} \, mol^{-1}$,we substitute the values:$\frac{4 	imes 10^5}{821} = \frac{8 	imes a}{27 	imes 0.0821 	imes 0.025}$.Since $821 = 10^4 	imes 0.0821$,we simplify:$\frac{4 	imes 10^5}{10^4 	imes 0.0821} = \frac{8 	imes a}{27 	imes 0.0821 	imes 0.025}$.$40 = \frac{8a}{27 	imes 0.025} \Rightarrow 40 = \frac{8a}{0.675}$.$a = \frac{40 	imes 0.675}{8} = 5 	imes 0.675 = 3.375 \, atm \, L^2 \, mol^{-2}$.

For a real gas (molar mass $= 60 \, g/mol$),if the density at the critical point is $0.80 \, g/cm^3$ and its critical temperature $T_c = \frac{4 \times 10^5}{821} \, K$,then the van der Waals constant $a$ (in $atm \, L^2 \, mol^{-2}$) is:

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