1) (5 pts) A mole of benzene (CH with an initial tempcrature of 0.0°C is placed
ID: 3308336 • Letter: 1
Question
1) (5 pts) A mole of benzene (CH with an initial tempcrature of 0.0°C is placed in contact with a hcat bath (the surroundings) at 10.0 C. Calculate Hsys, ASsys. AA,r, Ssurr. AH nav, andAS iv ("univ" refers to the universe, meaning the sum of the values for the sys and surroundings). Is this a revcrsible process? The thermodynamic data for benzene is: AH 10.04 kJ/mol C, (solid) 100.4 J/(mol K) C, liquid) 133.0 J(mol K) you can assume thesc valucs apply at all tempcratures. Hints: ) To calculate ATI and AS, you should consider the reversible path connecting the initial to final states. This path may involve multiple steps and, through TTess's law, you can add the values obtained at each step. (ii) lhe heat bath is sufficiently large that it can absorb or emit heat while maintaining temperature of 10 C, and it absorbs emits heat in a reversible manner.Explanation / Answer
1.
No. of moles of benzene, n = 1
Melting point of benzene at atmospheric pressure, T1 = 5.5 °C = 278.65 K
Initial temperature of solid benzene, T0 = 0 °C
Final temperature of benzene, T2 = 10 °C
Total heat transferred, Qrev = Q1 + Q2 + Q3
Heat gained by solid benzene, Q1 = n * Cp(solid) * (T1 – T0)
= 1 * 100.4 * (5.5 – 0) J
= 552.2 J = 0.552 kJ
Heat gained during phase change, Q2 = n * Hfus
= 1 * 10.04 kJ = 10.04 kJ
Heat gained by liquid benzene, Q3 = n * Cp(liquid) * (T2 – T1)
= 1 * 133 * (10 – 5.5) J
= 598.5 J = 0.5985 kJ
Hsys = Qrev = Q1 + Q2 + Q3
= 11.19 kJ
Hsurr = - Qrev = -11.19 kJ
Huniv = Hsys + Hsurr
= 0
Total entropy change of system, Srev = S1 + S2 + S3
Entropy change during heat gain by solid benzene, S1 = n * Cp(solid) * ln(T1 / T0)
= 1 * 100.4 * ln(278.65 / 273.15) J/K
= 2 J/K = 0.002 kJ/K
Entropy change during phase change, S2 = n * Hfus / T1
= 1 * 10.04 / 278.65 kJ/K
= 0.03603 kJ/k
Entropy change during heat gained by liquid benzene, S3 = n * Cp(liquid) * ln(T2 / T1)
= 1 * 133 * ln(283.15 / 278.65) J/K
= 2.13 J/K = 0.00213 kJ/K
Ssys = Srev = S1 + S2 + S3
= 0.04016 kJ/K
Ssurr = - Qrev / T2
= -11.19 / 283.15 kJ/K
= - 0.0395 kJ/K
Suniv = Ssys + Ssurr
= 0.00064 kJ/K
= 0.64 J/K
Since, Suniv > 0, the process is spontaneous and not reversible.
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