Question:
K value is defined as the ratio of mole fraction of a component in the vapor phase to that in the liquid phase. The K values of propane and iso-butane at 293 K are given in the table.
Pressure (kPa) K value: Propane K value: iso-Butane 1100 1.8 0.85 1200 1.6 0.75 1300 1.4 0.65 1400 1.25 0.5 1500 1.1 0.45
Which one of the following is closest to the bubble point pressure (in kPa) of an equimolar mixture of propane and iso-butane at 293 K?
K value is defined as the ratio of mole fraction of a component in the vapor phase to that in the liquid phase. The K values of propane and iso-butane at 293 K are given in the table.
Which one of the following is closest to the bubble point pressure (in kPa) of an equimolar mixture of propane and iso-butane at 293 K?
| Pressure (kPa) | K value: Propane | K value: iso-Butane |
|---|---|---|
| 1100 | 1.8 | 0.85 |
| 1200 | 1.6 | 0.75 |
| 1300 | 1.4 | 0.65 |
| 1400 | 1.25 | 0.5 |
| 1500 | 1.1 | 0.45 |
Which one of the following is closest to the bubble point pressure (in kPa) of an equimolar mixture of propane and iso-butane at 293 K?
Show Hint
The bubble point pressure is where the sum of (K value times liquid mole fraction) for all components equals 1; interpolate between the tabulated pressures that bracket this condition.
Updated On: Aug 10, 2026
- 1110
- 1190
- 1310
- 1390
Show Solution
Verified By Collegedunia
The Correct Option is C
Solution and Explanation
Step 1: State the bubble point condition.
At the bubble point of a liquid mixture, the liquid composition equals the feed composition, and the bubble point pressure is the pressure at which:
\[ \sum_i K_i x_i = 1 \]
Step 2: Set the liquid composition.
Equimolar mixture, so \(x_{propane} = x_{iso-butane} = 0.5\).
Step 3: Compute \(\sum K_i x_i\) at each tabulated pressure.
At \(P=1100\): \(0.9+0.425 = 1.325\)
At \(P=1200\): \(0.8+0.375 = 1.175\)
At \(P=1300\): \(0.7+0.325 = 1.025\)
At \(P=1400\): \(0.625+0.25 = 0.875\)
At \(P=1500\): \(0.55+0.225 = 0.775\)
Step 4: Locate the pressure where the sum equals 1.
The sum is above 1 at \(P=1300\) (\(1.025\)) and below 1 at \(P=1400\) (\(0.875\)), so the bubble point pressure lies between 1300 and 1400 kPa.
Step 5: Linearly interpolate.
\[ P_{bubble} = 1300 + \frac{0.025}{0.150}\times 100 = 1300+16.7 = 1316.7 \text{ kPa} \]
Step 6: Compare with the given options.
1316.7 kPa is closest to option (C) 1310 kPa.
\[ \boxed{P_{bubble} \approx 1310 \text{ kPa (Option C)}} \]
At the bubble point of a liquid mixture, the liquid composition equals the feed composition, and the bubble point pressure is the pressure at which:
\[ \sum_i K_i x_i = 1 \]
Step 2: Set the liquid composition.
Equimolar mixture, so \(x_{propane} = x_{iso-butane} = 0.5\).
Step 3: Compute \(\sum K_i x_i\) at each tabulated pressure.
At \(P=1100\): \(0.9+0.425 = 1.325\)
At \(P=1200\): \(0.8+0.375 = 1.175\)
At \(P=1300\): \(0.7+0.325 = 1.025\)
At \(P=1400\): \(0.625+0.25 = 0.875\)
At \(P=1500\): \(0.55+0.225 = 0.775\)
Step 4: Locate the pressure where the sum equals 1.
The sum is above 1 at \(P=1300\) (\(1.025\)) and below 1 at \(P=1400\) (\(0.875\)), so the bubble point pressure lies between 1300 and 1400 kPa.
Step 5: Linearly interpolate.
\[ P_{bubble} = 1300 + \frac{0.025}{0.150}\times 100 = 1300+16.7 = 1316.7 \text{ kPa} \]
Step 6: Compare with the given options.
1316.7 kPa is closest to option (C) 1310 kPa.
\[ \boxed{P_{bubble} \approx 1310 \text{ kPa (Option C)}} \]
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