GATE BT 2025 • Q57 NAT • 2 Marks

Section 4: Bioprocess Engineering and Process Biotechnology

Mass transfer in bioprocesses (kLa)

Subtopic: Fundamentals of Mass transfer in bioprocesses (kLa)

Question Statement:

A binary mixture of benzene and toluene under vapour-liquid equilibrium at 80 C follows ideal Raoult's law. At this condition, the saturation pressures of benzene and toluene are 101 kPa and 40 kPa, respectively. If the mole fraction of benzene in the liquid phase is 0.6, the corresponding mole fraction of benzene in the vapour phase will be _______. (Round off to two decimal places)

Official Answer Key:

0.76 to 0.82

Step-by-Step Solution & Explanation

### **Step-by-Step Solution & Mathematical Derivation (GATE BT 2025 Q.57)** **1. Governing Concept & Definition**: - **Topic**: Thermodynamics & Vapor-Liquid Equilibrium (Raoult's Law) - For an ideal binary solution under vapor-liquid equilibrium: $P_i = x_i P_i^{\text{sat}}$ $P_{\text{total}} = x_1 P_1^{\text{sat}} + x_2 P_2^{\text{sat}}$ - By Dalton's Law of partial pressures, the mole fraction in the vapor phase ($y_i$) is: $y_i = \frac{P_i}{P_{\text{total}}} = \frac{x_i P_i^{\text{sat}}}{P_{\text{total}}}$ **2. Detailed Step-by-Step Calculation**: - **Given**: - Component 1 (Benzene): $P_1^{\text{sat}} = 101\text{ kPa}$, $x_1 = 0.6$ - Component 2 (Toluene): $P_2^{\text{sat}} = 40\text{ kPa}$, $x_2 = 1 - 0.6 = 0.4$ - **Step 1: Calculate partial pressures**: $P_1 = x_1 P_1^{\text{sat}} = 0.6 \times 101\text{ kPa} = 60.6\text{ kPa}$ $P_2 = x_2 P_2^{\text{sat}} = 0.4 \times 40\text{ kPa} = 16.0\text{ kPa}$ - **Step 2: Calculate total vapor pressure**: $P_{\text{total}} = P_1 + P_2 = 60.6 + 16.0 = 76.6\text{ kPa}$ - **Step 3: Calculate vapor phase mole fraction of benzene ($y_1$)**: $y_1 = \frac{P_1}{P_{\text{total}}} = \frac{60.6}{76.6} \approx 0.7911 \approx 0.79$ **3. Final Answer**: - **Calculated Value**: **0.79** - **Official GATE Accepted Range**: **0.76 to 0.82**

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