If the molar enthalpy change for vaporization of 1 mol of water at 1 bar and 100°C is 41 kJ mol⁻¹, c

Q: What is the correct answer for CBSE11?

The correct answer is B. The problem is similar to Problem 5.5. The reaction is H₂O(l) → H₂O(g). Δn_g = 1 (for 1 mole of gas product). T = 100°C = 373 K. ΔH = 41 kJ mol⁻¹. Using ΔH = ΔU + Δn_gRT, we get ΔU = ΔH - Δn_gRT. ΔU = 41 kJ mol⁻¹ - (1 mol * 8.314 J K⁻¹ mol⁻¹ * 373 K). Convert R to kJ: 8.314 J = 0.008314 kJ. ΔU = 41 kJ mol⁻¹ - (1 * 0.008314 kJ K⁻¹ mol⁻¹ * 373 K) = 41 - 3.096 = 37.904 kJ mol⁻¹.

Q: If the molar enthalpy change for vaporization of 1 mol of water at 1 bar and 100°C is 41 kJ mol⁻¹, calculate the internal energy change (ΔU) for this process. Assume water vapor behaves as an ideal…

The correct answer is B. The problem is similar to Problem 5.5. The reaction is H₂O(l) → H₂O(g). Δn_g = 1 (for 1 mole of gas product). T = 100°C = 373 K. ΔH = 41 kJ mol⁻¹. Using ΔH = ΔU + Δn_gRT, we get ΔU = ΔH - Δn_gRT. ΔU = 41 kJ mol⁻¹ - (1 mol * 8.314 J K⁻¹ mol⁻¹ * 373 K). Convert R to kJ: 8.314 J = 0.008314 kJ. ΔU = 41 kJ mol⁻¹ - (1 * 0.008314 kJ K⁻¹ mol⁻¹ * 373 K) = 41 - 3.096 = 37.904 kJ mol⁻¹.

Q: What are the options for this CBSE11 question?

A. 41.00 kJ mol⁻¹ · B. 37.904 kJ mol⁻¹ · C. 44.096 kJ mol⁻¹ · D. 33.00 kJ mol⁻¹

Question

If the molar enthalpy change for vaporization of 1 mol of water at 1 bar and 100°C is 41 kJ mol⁻¹, calculate the internal energy change (ΔU) for this process. Assume water vapor behaves as an ideal gas. (R = 8.314 J K⁻¹ mol⁻¹)

FirstInTest · Accepted Answer

The problem is similar to Problem 5.5. The reaction is H₂O(l) → H₂O(g). Δn_g = 1 (for 1 mole of gas product). T = 100°C = 373 K. ΔH = 41 kJ mol⁻¹. Using ΔH = ΔU + Δn_gRT, we get ΔU = ΔH - Δn_gRT. ΔU = 41 kJ mol⁻¹ - (1 mol * 8.314 J K⁻¹ mol⁻¹ * 373 K). Convert R to kJ: 8.314 J = 0.008314 kJ. ΔU = 41 kJ mol⁻¹ - (1 * 0.008314 kJ K⁻¹ mol⁻¹ * 373 K) = 41 - 3.096 = 37.904 kJ mol⁻¹.

Answer

41.00 kJ mol⁻¹

Answer

37.904 kJ mol⁻¹

Answer

44.096 kJ mol⁻¹

Answer

33.00 kJ mol⁻¹

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