Chemistry · States of Matter - Gaseous State

JEE Advanced 2018 — Paper 1 — Question 23

A closed tank has two compartments $\mathbf{A}$ and $\mathbf{B}$ , both filled with oxygen (assumed to be ideal gas). The partition separating the two compartments is fixed and is a perfect heat insulator (Figure 1). If the old partition is replaced by a new partition which can slide and conduct heat but does NOT allow the gas to leak across (Figure 2), the volume (in $\mathrm{m}^{3}$ ) of the compartment $\mathbf{A}$ after the system attains equilibrium is

Answer: 2.22

Numerical answer — enter this value.

Step-by-step solution

As in fig 2, the system attains equilibrium, so,

$P_{A}=P_{B}$ and $T_{A}=T_{B}$

$\frac{\mathrm{P}_{\mathrm{A}} \mathrm{V}_{\mathrm{A}}}{\operatorname{Rn}_{\mathrm{A}} \mathrm{T}_{\mathrm{A}}}=\frac{\mathrm{P}_{\mathrm{B}} \mathrm{V}_{\mathrm{B}}}{\mathrm{n}_{\mathrm{B}} \mathrm{T}_{\mathrm{B}} \mathrm{R}}$

$\frac{\mathrm{V}_{\mathrm{A}}}{\mathrm{V}_{\mathrm{B}}}=\frac{\mathrm{n}_{\mathrm{A}}}{\mathrm{n}_{\mathrm{B}}}$

$\mathrm{n}_{\mathrm{A}}=\frac{5}{400 \mathrm{R}}$

$\mathrm{n}_{\mathrm{B}}=\frac{3}{300 \mathrm{R}}$

Due to sliding of piston vol. of $A$ will be increased by $x$ and that of $B$ will be decreased by $x$ .

$\mathrm{V}_{\mathrm{A}}=1+\mathrm{x}$

$\mathrm{V}_{\mathrm{B}}=3-\mathrm{x}$

$\frac{1+x}{3-x}=\frac{\frac{5}{400 R}}{\frac{3}{300 R}}$

$4(1+x)=5(3-x)$

$4 \mathrm{x}+5 \mathrm{x}=11 \Rightarrow \mathrm{x}=\frac{11}{9}$

Hence volume of container A will be

$\mathrm{V}_{\mathrm{A}}=1+\frac{11}{9}=\frac{20}{9}=2.22$

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2018
Paper: Paper 1
Subject: Chemistry
Chapter: States of Matter - Gaseous State
Topic: Measurable Properties of Gases and Basic Gas Laws