Question What are the values of q, w, ∆U , ∆H, ∆Ssys , ∆Ssurr , and ∆Suniv for the following constant pressure process for a system containing 0.760 moles of CH3 OH? CH3 OH(g, 119.0◦ C, 1.00 atm) −→ CH3 OH(l, 34.0◦ C, 1.00 atm) Assume that CH3 OH behaves as an ideal gas and the volume of CH3 OH(l) is much less than that of CH3 OH(g). Also, assume that the temperature of the surroundings is 34.0◦ C. Data: • Molar heat capacity for CH3 OH(g), Cp,m = 44.1 J mol−1 K−1 • Molar heat capacity for CH3 OH(l), Cp,m = 81.1 J mol−1 K−1 • Enthalpy of vaporization, ∆vap H = 35.2 kJ mol−1 at 64.7◦ C and 1.00 atm Enter answers that are accurate to three (3) or more significant figures. Don’t use rounded values when calculating answers that combine multiple variables (e.g., ∆Suniv ). Do not use rounded values for ∆Ssys and ∆Ssurr . Use accurate values and round to enter the final answer. Solution Step 1: Data and Assumptions n = 0.760 mol T1 = 119.0◦ C = 392.15 K T2 = 34.0◦ C = 307.15 K ∆Hvap = 35.2 kJ/mol = 35200 J/mol Cp,m (CH3 OH(g)) = 44.1 J/mol·K Cp,m (CH3 OH(l)) = 81.1 J/mol·K Tsurroundings = 307.15 K Step 2: Calculate q (Heat Transfer) 1. Cooling of CH3 OH(g): q1 = nCp,m (CH3 OH(g))(T2 − T1 ) q1 = 0.760 · 44.1 · (307.15 − 392.15) q1 = 0.760 · 44.1 · (−85) = −2845.32 J 2. Condensation of CH3 OH(g) to CH3 OH(l): q2 = n∆Hvap q2 = 0.760 · 35200 = 26752 J 1 3. Cooling of CH3 OH(l): q3 = nCp,m (CH3 OH(l))(T2 − T1 ) q3 = 0 4. Total Heat Transfer: q = q1 + q2 + q3 = −2845.32 + 26752 + 0 = 23906.68 J Step 3: Calculate w (Work Done) Since this is a constant-pressure process: w = 0J Step 4: Calculate ∆U (Internal Energy Change) Using the first law of thermodynamics: ∆U = q + w ∆U = 23906.68 + 0 = 23906.68 J Step 5: Calculate ∆H (Enthalpy Change) At constant pressure: ∆H = q = 23906.68 J Step 6: Calculate ∆Ssys (Entropy Change of System) 1. Entropy change for cooling of CH3 OH(g): T2 ∆S1 = nCp,m (CH3 OH(g)) ln T1 307.15 ∆S1 = 0.760 · 44.1 · ln 392.15 ∆S1 = 0.760 · 44.1 · (−0.246) = −8.246 J/K 2. Entropy change for condensation: ∆S2 = ∆S2 = −n∆Hvap Tcondensation −0.760 · 35200 = −87.128 J/K 307.15 3. Total entropy change: ∆Ssys = ∆S1 + ∆S2 = −8.246 + (−87.128) = −95.374 J/K 2 Step 7: Calculate ∆Ssurr (Entropy Change of Surroundings) ∆Ssurr = ∆Ssurr = −q Tsurroundings −23906.68 = −77.86 J/K 307.15 Step 8: Calculate ∆Suniv (Entropy Change of Universe) ∆Suniv = ∆Ssys + ∆Ssurr ∆Suniv = −95.374 + (−77.86) = −173.234 J/K Final Answers q w ∆U ∆H ∆Ssys ∆Ssurr ∆Suniv = 23906.68 J = 0J = 23906.68 J = 23906.68 J = −95.374 J/K = −77.86 J/K = −173.234 J/K 3