Question

An equimolar mixture of carbon dioxide and water vapor at 1 atm and \(60^{\circ} \mathrm{C}\) enter a dehumidifying section where the entire water vapor is condensed and removed from the mixture, and the carbon dioxide leaves at 1 atm and \(60^{\circ} \mathrm{C}\). The entropy change of carbon dioxide in the dehumidifying section is \((a)-2.8 \mathrm{kJ} / \mathrm{kg} \cdot \mathrm{K}\) \((b)-0.13 \mathrm{kJ} / \mathrm{kg} \cdot \mathrm{K}\) \((c) 0\) \((d) 0.13 \mathrm{kJ} / \mathrm{kg} \cdot \mathrm{K}\) \((e) 2.8 \mathrm{kJ} / \mathrm{kg} \cdot \mathrm{K}\)

Short answer

a) Positive b) Negative c) Zero d) Cannot be determined Answer: c) Zero

Step-by-step solution

Calculate the initial entropy of CO2

As the initial and final temperatures are the same, the initial entropy of CO2, \(S_{initial}\), will be the same as final entropy, \(S_{final}\). Hence, we don't need to calculate the initial entropy of CO2.

Calculate the final entropy of CO2

Since the final temperature is the same as the initial temperature, the final entropy of CO2, \(S_{final}\), will be the same as the initial entropy, \(S_{initial}\). Hence, we don't need to calculate the final entropy of CO2 either.

Calculate the entropy change of CO2

Now, we can calculate the entropy change of carbon dioxide using the formula: \(\Delta S = S_{final} - S_{initial}\) But we know that \(S_{final} = S_{initial}\), so the equation becomes: \(\Delta S = S_{final} - S_{initial} = 0\) Therefore, the entropy change of carbon dioxide in the dehumidifying section is: \(\Delta S = 0\) The correct answer is \((c) 0\).