QJEEAdv18Chbdf3: JEE Advanced 2018 Chemistry - Class 11 JEEAdv18Chbdf3 Solution: A closed tank has two compartments and both filled with oxygen…

Given p₁ = 5 bar, V₁ = 1 m³, T₁ = 400 K

So, ${n_1} = {5 \over {400R}}$ (from pV = nRT)

Similarly, p₂ = 1 bar, V₂ = 3 m³, T₂ = 300 K,

${n_2} = {3 \over {300R}}$

Let at equilibrium the new volume of A will be (1 + x)

So, the new volume of B will be (3 $-$ x)

Now, from the ideal gas equation,

${{{p_1}{V_1}} \over {{n_1}R{T_1}}} = {{{p_2}{V_2}} \over {{n_2}R{T_2}}}$

and at equilibrium (due to conduction of heat)

${{{p_1}} \over {{T_1}}} = {{{p_2}} \over {{T_2}}}$

So, ${{{V_1}} \over {{n_1}}} = {{{V_2}} \over {{n_2}}}$ or ${V_1}{n_2} = {V_2}{n_1}$

After putting the values

$(1 + x) \times {3 \over {300R}} = (3 - x) \times {5 \over {400R}}$

or $(1 + x) = {{(3 - x)5} \over 4}$ or $4(1 + x) = 15 - 5x$

or $4 + 4x = 15 - 5x$ or $x = {{11} \over 9}$

Hence, new volume of A i.e., (1 + x) will comes as $1 + {{11} \over 9} = {{20} \over 9}$ or 2.22.