JEE Advanced · Chemistry · 5. States of Matter
Molar volume \(\left(V_m\right)\) of a van der Waals gas can be calculated by expressing the van der Waals equation as a cubic equation with \(V_m\) as the variable. The ratio (in \(\mathrm{mol} \mathrm{dm}^{-3}\) ) of the coefficient of \(V_m^2\) to the coefficient of \(V_m\) for a gas having van der Waals constants \(a=6.0 \mathrm{dm}^6 \mathrm{~atm} \mathrm{~mol}^{-2}\) and \(b=0.060 \mathrm{dm}^3 \mathrm{~mol}^{-1}\) at 300 K and 300 atm is _____ .
Use: Universal gas constant \((R)=0.082 \mathrm{dm}^3 \mathrm{~atm} \mathrm{~mol}{ }^{-1} \mathrm{~K}^{-1}\)
- A -7.1
- B -7.2
- C -7.3
- D -7.4
Answer & Solution
Correct Answer
(A) -7.1
Step-by-step Solution
Detailed explanation
\(\begin{aligned}
& \left(P+\frac{a}{V_m^2}\right)\left(V_m-b\right)=R T \\
& P V_m-b P+\frac{a}{V_m}-\frac{a b}{V_m^2}-R T=0 \\
& \Rightarrow \quad P V_m^2-(b P+R T) V_m^2+a V_m-a b=0
\end{aligned}\)
Coefficient of \(\mathrm{V}_{\mathrm{m}}^2=-(\mathrm{bP}+\mathrm{RT})\)
Coefficient of \(\mathrm{V}_{\mathrm{m}}=\mathrm{a}\)
\(\text { Ratio }=-\frac{(\mathrm{bP}+\mathrm{RT})}{\mathrm{a}}=-\left[\frac{0.06 \times 300+24.6}{6}\right]=-7.1 .\)
& \left(P+\frac{a}{V_m^2}\right)\left(V_m-b\right)=R T \\
& P V_m-b P+\frac{a}{V_m}-\frac{a b}{V_m^2}-R T=0 \\
& \Rightarrow \quad P V_m^2-(b P+R T) V_m^2+a V_m-a b=0
\end{aligned}\)
Coefficient of \(\mathrm{V}_{\mathrm{m}}^2=-(\mathrm{bP}+\mathrm{RT})\)
Coefficient of \(\mathrm{V}_{\mathrm{m}}=\mathrm{a}\)
\(\text { Ratio }=-\frac{(\mathrm{bP}+\mathrm{RT})}{\mathrm{a}}=-\left[\frac{0.06 \times 300+24.6}{6}\right]=-7.1 .\)
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