JEE Challenger
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ChemMedAdv 2018 • Paper 1

Equilibrium Volume of Compartment with Movable Conducting Partition

A closed container consists of two compartments A\mathbf{A} and B\mathbf{B}, both filled with oxygen (treated as an ideal gas). Initially, the separating partition is rigidly fixed and serves as a perfect thermal insulator (Figure 1). If this partition is replaced by a movable, heat-conducting wall that does NOT permit gas to pass through (Figure 2), determine the equilibrium volume (in m3\text{m}^3) of compartment A\mathbf{A}.

Question Diagram 1
Official Numerical Answer2.22

Topics & Concepts

States of MatterIdeal Gas Law

Step-by-Step Solution

To find the equilibrium volume of compartment A\mathbf{A}, we use the ideal gas law and the conditions for thermal and mechanical equilibrium.

Step 1: Calculate the initial moles of gas in each compartment

Using the ideal gas equation PV=nRTPV = nRT, we determine the number of moles in compartments A\mathbf{A} and B\mathbf{B}:

  • For compartment A\mathbf{A}: nA=PAVARTA=5 bar×1 m3R×400 K=5400R=180Rn_A = \frac{P_A V_A}{R T_A} = \frac{5\text{ bar} \times 1\text{ m}^3}{R \times 400\text{ K}} = \frac{5}{400 R} = \frac{1}{80 R}

  • For compartment B\mathbf{B}: nB=PBVBRTB=1 bar×3 m3R×300 K=3300R=1100Rn_B = \frac{P_B V_B}{R T_B} = \frac{1\text{ bar} \times 3\text{ m}^3}{R \times 300\text{ K}} = \frac{3}{300 R} = \frac{1}{100 R}

The total volume of the closed container is: Vtotal=VA+VB=1 m3+3 m3=4 m3V_{\text{total}} = V_A + V_B = 1\text{ m}^3 + 3\text{ m}^3 = 4\text{ m}^3

Step 2: Establish the equilibrium conditions

When the partition is replaced by a movable, heat-conducting wall:

  1. Thermal equilibrium ensures that the final temperature in both compartments is the same: TA′=TB′=TfT_A' = T_B' = T_f
  2. Mechanical equilibrium ensures that the final pressure in both compartments is the same: PA′=PB′=PfP_A' = P_B' = P_f

Step 3: Relate volumes to the number of moles

For the two compartments at equilibrium: PfVA′=nARTfP_f V_A' = n_A R T_f PfVB′=nBRTfP_f V_B' = n_B R T_f

Dividing the two equations gives: VA′VB′=nAnB\frac{V_A'}{V_B'} = \frac{n_A}{n_B}

Substitute the values of nAn_A and nBn_B: VA′VB′=180R1100R=10080=54\frac{V_A'}{V_B'} = \frac{\frac{1}{80 R}}{\frac{1}{100 R}} = \frac{100}{80} = \frac{5}{4}

Thus, VB′=45VA′V_B' = \frac{4}{5} V_A'

Step 4: Solve for VA′V_A'

Since the total volume remains constant: VA′+VB′=VtotalV_A' + V_B' = V_{\text{total}} VA′+45VA′=4 m3V_A' + \frac{4}{5} V_A' = 4\text{ m}^3 95VA′=4 m3\frac{9}{5} V_A' = 4\text{ m}^3 VA′=209 m3≈2.22 m3V_A' = \frac{20}{9}\text{ m}^3 \approx 2.22\text{ m}^3

The equilibrium volume of compartment A\mathbf{A} is 2.22 m32.22\text{ m}^3 (or 209 m3\frac{20}{9}\text{ m}^3).

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