Homework 9.19 [ENDORSED]

[Dang Lam]

Calculate the standard entropy of vaporization of water at 85 C, given that its standard entropy of vaporization at 100. C is 109.0 J K 1 mol 1 and the molar heat capacities at constant pressure of liquid water and water vapor are 75.3 J K 1 mol 1 and 33.6 J K 1 mol 1, respectively, in this range.

Can someone explain to me how to approach to this problem. Why do we need molar heat capacities for this problem.

[Chem_Mod]

We need molar heat capacities because they are used to determine q which in turn comprises the numerator of our definition of macroscopic entropy change dS = q_rev/T.

If we want to know the entropy associated with the vaporization of a material at a temperature other than that at standard pressure, we can use the procedure of calculating the entropy associated with bringing the material from its non standard boiling point to its standard (normal) boiling point, calculating the entropy associated with this vaporization phase change at that standard (normal) boiling point using the enthalpy of vaporization as q_rev, then calculating the entropy change associated with cooling that material back down to the non standard (non normal) boiling temperature we are interested in.

Recall that substances can be made to boil at a large range of temperatures depending on that substance's phase diagram as long as we increase or reduce the ambient pressure. The boiling point is defined as that temperature where the vapor pressure of the material matches the ambient pressure.

[jillian1k]

This problem wants you to find the standard entropy of vaporization of water at 85 C reversibly. This takes much more work by the system than an irreversible vaporization of water at 85 C, which consists of one step and lots of energy lost as heat.

To do it reversibly:
1.) reversibly heat the water (reactants) to 100 C (aka standard boiling point of water) from 85 C [use molar heat capacity (C) of liquid water here in eqn for dS derived from q=CdT: dS=Cln(T2/T1)]
2.) calculate dS for phase change (dHvap) at 100 C (this happens to be given)
3.) reversibly cool the vapor (products) back down to 85 C [use molar heat capacity (C) of water vapor here in dS=Cln(T2/T1)]
4.) add steps 1-3 together to get the TOTAL standard entropy of vaporization of water at 85 C.

[Austin Ho 1E]

This is a 3 part entropy calculation, because we need to (1) heat the liquid water to 100°C, (2) vaporize the liquid water into gas at 100°C, and finally (3) cool the water vapor down to 85°C. We need the specific heats to find ΔS for parts 1 and 3, using the heat for liquid water in part 1 and the heat for water vapor in part 3.

[Anne 2L]

This problem wants you to find the standard entropy of vaporization of water at 85 C reversibly. This takes much more work by the system than an irreversible vaporization of water at 85 C, which consists of one step and lots of energy lost as heat.

To do it reversibly:
1.) reversibly heat the water (reactants) to 100 C (aka standard boiling point of water) from 85 C [use molar heat capacity (C) of liquid water here in eqn for dS derived from q=CdT: dS=Cln(T2/T1)]
2.) calculate dS for phase change (dHvap) at 100 C (this happens to be given)
3.) reversibly cool the vapor (products) back down to 85 C [use molar heat capacity (C) of water vapor here in dS=Cln(T2/T1)]
4.) add steps 1-3 together to get the TOTAL standard entropy of vaporization of water at 85 C.

How do we know that the question is asking to analyze the system reversibly? Is there a key word in the question that hints at this? I can't seem to find any indication that the system was mean to be reversible.

[jillian1k]

We know because the eqns for dS all come from dS=qrev/T. qrev is a greater value than qirev because q=-w for reversible systems. Since we know w is always greater in reversible systems than in irreversible systems, this difference translates to q in q=-w.