JEE Advanced 2018 · Paper 1 · Chemistry Q11

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ChemistryNumerical+3Question 29 of 54

A closed tank has two compartments A and 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 m³) of the compartment A after the system attains equilibrium is ______.

Figure 1: a closed tank split by a fixed insulating partition into A (1 m³, 5 bar, 400 K) and B (3 m³, 1 bar, 300 K). Figure 2: the same tank with a sliding, heat-conducting partition between A and B.

Your answer

Answer and worked solution

Official answer: 2.2

Moles (in units of bar m³/R):
$n_A=\frac{5\times1}{400}=0.0125,\qquad n_B=\frac{1\times3}{300}=0.01.$

Final temperature. The tank is closed and rigid, so total internal energy is conserved. Both are $\ce{O2}$ with the same $C_v$ :
$n_A(400)+n_B(300)=(n_A+n_B)T_f\implies T_f=\frac{5+3}{0.0225}=355.6\text{ K}.$

Final volumes. The sliding, conducting partition equalises $P$ and $T$ , so $V\propto n$ :
$V_A=4\times\frac{0.0125}{0.0225}=2.22\text{ m}^3.$

Answer: 2.22 (key: 2.20 to 2.23)

Topics:Ideal gas equation Gas laws

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JEE Advanced 2018 · Paper 1, Chemistry question 11