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JEE Advanced
Exam
JEE Advanced
Subject
Chemistry
Chapter
States of Matter

JEE Advanced · Chemistry · States of Matter

A certain gas has the equation of state, P= RT V_ m -bT - a V_ m ^3...

A certain gas has the equation of state, P=RTVm−bT−aVm3\mathrm{P}=\frac{\mathrm{RT}}{\mathrm{V}_{\mathrm{m}}-\mathrm{bT}}-\frac{\mathrm{a}}{\mathrm{V}_{\mathrm{m}}^3}. Where a and b are constants characteristic of the gas. Choose the incorrect relationship(s). [ Pc,Vc\mathrm{P}_{\mathrm{c}}, \mathrm{V}_{\mathrm{c}} and Tc\mathrm{T}_{\mathrm{c}} are critical pressure, volume and temperature respectively]

Options

  1. Aa=RTcVc23\mathbf{a}=\frac{\mathbf{R T}_c \mathbf{V}_c^2}{3}
  2. Bb=Vc2Tcb=\frac{V_c}{2 T_c}
  3. Ca=4RTeVc3\mathrm{a}=\frac{4 \mathrm{R} \mathrm{T}_{\mathrm{e}} \mathrm{V}_{\mathrm{c}}}{3}
  4. Db=Vc3 Tc\mathbf{b}=\frac{\mathbf{V}_{\mathrm{c}}}{3 \mathrm{~T}_{\mathrm{c}}}

Correct answer

D. b=Vc3 Tc\mathbf{b}=\frac{\mathbf{V}_{\mathrm{c}}}{3 \mathrm{~T}_{\mathrm{c}}}

Step-by-step solution

P=RTVm−bT−aVm3(dPdVm)T=−RT(Vm−bT)2+3aVm4\begin{aligned} & P=\frac{R T}{V_m-b T}-\frac{a}{V_m^3} \\ & \left(\frac{d P}{d V_m}\right)_T=-\frac{R T}{\left(V_m-b T\right)^2}+\frac{3 a}{V_m^4}\end{aligned} Under critical condition \left(\frac{d P}{d V_m}\right)_T=-\frac{R T}{\left(V_m-b T\right)^2}+\frac{3 a}{V_m^4}=0...(i)(d2PdVm2)T=+2RT(Vm−bT)3−12aVm5=0...(i) \left(\frac{d^2 P}{d V_m^2}\right)_T=+\frac{2 R T}{\left(V_m-b T\right)^3}-\frac{12 a}{V_m^5}=0 ...(ii) On solving (i) and (ii), we get a=4RTcVc23a=\frac{4 R T_c V_c^2}{3} and b=Vc2Tcb=\frac{V_c}{2 T_c}
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