Question

For 3 mol of an ideal gas \(\left[ C_{p,m} = \frac{5}{2}R \right]\), being heated from 300 K to 600 K, the change in entropy at constant pressure is

• A. 48.22 J K\(^{-1}\)
• B. 43.22 J K\(^{-1}\)
• C. 33.22 J K\(^{-1}\)
• D. 38.22 J K\(^{-1}\)

Hint:

For an ideal gas, the change in entropy during heating at constant pressure depends on the heat capacity \( C_p \) and the temperature ratio.

Answer 1

The Correct Option is A

The entropy change (\( \Delta S \)) under constant pressure conditions is determined by the formula:\[\Delta S = n C_p \ln \left( \frac{T_2}{T_1} \right)\]Given values are:- \( n = 3 \) moles- \( C_p = \frac{5}{2}R = \frac{5}{2} \times 8.314 \, \text{J/mol·K} \)- \( T_1 = 300 \, \text{K} \)- \( T_2 = 600 \, \text{K} \)Substitution into the formula yields:\[\Delta S = 3 \times \left( \frac{5}{2} \times 8.314 \right) \times \ln \left( \frac{600}{300} \right)\]This simplifies to:\[\Delta S = 3 \times 20.785 \times \ln 2\]Calculating the result:\[\Delta S \approx 48.22 \, \text{J/K}\] Final Answer: \[\boxed{48.22 \, \text{J/K}}\]