Is Work Meant to Be Positive and Heat Negative During Ethanol-Water Mixing?

[AXidenT]

Homework Statement

Not a homework question exactly, but for a lab report I've looked at the changes in temperature and volume mixing different ratios (with same total mass) of ethanol and water. From this in the report I've been analysing questions such as "whether the work due to the volume change or the heat due to the temperature change is more significant". To answer this part of my investigation I've calculated the heat from the change in temperature and the work from the change in volume.

For my calculations for heat, I get a negative result which I somewhat set myself as I had to calculate an equilibrium temperature based on the heat exchange between the ethanol and water as if there had been no mixing and found the difference between that and the maximum temperature (so when the ethanol and water had fully mixed but heat had not been lost to the environment yet). As a result I could have stated it as ΔT=T_eq/-T_max or as ΔT=T_max-T_eq. As the max temperature is greater than the equilibrium temperature I determined the heat should be negative as the solution is giving off heat, thus I chose to make it the former option (implying T_Eq is the initial weighted average temperature BEFORE the mixing process has occurred. I'm pretty happy with this conclusion.

However for the pressure volume work, ΔV will be negative as the "expected volume" V_ex is higher than the measured volume V_me. Intuitively, I would calculate:

ΔV= V_ex- V_me

Since the expected volume is basically acting as the initial volume (analogously to the expected equilibrium temperature acting as the initial temperature). However as:

W=-pΔV

This would mean the work is positive which is the opposite sign to the heat. So my questions are:
0) Is work meant to be positive and heat negative?
1) What is doing the work on the solution?
2) Is whatever factor this is working "against" the spontaneity/driving of the mixing?

From the Gibbs Free energy equation:

ΔG=ΔH-TΔS

As no chemical or electrical work etc.. are being done:

ΔH = ΔU+ pΔV = Q + W+ pΔV = Q - pΔV + pΔV = Q

3) Does this imply the work done has no affect on the spontaneity of the mixing? Ie no matter the magnitude or "direction" of the work (so if its being done on the system or against the system) as long as its only PV work, the components would still mix/not mix regardless of this?

4) If so what significance does the work have then?

Homework Equations

W=-pΔV
ΔH = ΔU+ pΔV
ΔG=ΔH-TΔS

The Attempt at a Solution

I've already included most of my working/logic above but for question 4 I was thinking that since the heat is being lost by the system this is somehow offset by the work done on the system? So somehow and for some reason mixing in this case favours the loss of heat for a certain amount of work done on it? I can't think of a physical explanation for this, but the maths makes sense. :S

Thank you for any help! :)

 

[Chestermiller]

I assume you are learning about solution thermodynamics in your course now, and this lab is designed to give you some practical understanding. The question you have been focusing on, "whether the work due to the volume change or the heat due to the temperature change is more significant", is this your own construction, or is it actually what you were asked to answer in your lab report?

Your experiment is associated with "heat of mixing" and "volume change on mixing." Have you studied partial molar properties in your course yet? If so, then you should be considering these in your data analysis. Have you made plots of specific enthalpy and specific volume of the mixture as a function of mole fraction? From such plots, have you constructed graphs of partial molar enthalpy and partial molar volume as function of mole fraction?

Regarding your questions about heat and work, you should be examining or approximating the enthalpy change and volume change at constant temperature and pressure. In that case, the enthalpy change would be equal to the heat added, and the work done would be equal to PΔV. Then, these results could be used to determine ΔU. There would be no inconsistencies here.

As far as free energy and entropy are concerned, you do realize you are dealing with an irreversible process, correct? So, trying to relate the heat transferred to the entropy change won't be fruitful for the path that you are dealing with. For example, even if there were no heat of mixing and no volume change, there still would be a substantial increase in entropy. But, to measure what that entropy change would be, you would have to work out some complicated reversible path from the initial pure components to the final mixture.

Check out your textbook for the changes in free energy and entropy for ideal gases when they are mixed. For liquids, the situation is more complicated.

Chet

 

[AXidenT]

Thanks for the reply! The question is one of three in the topic I chose to cover - the other two points of investigation were "How far from ideal is this mixing process? " and "Is the spontaneous mixing driven more by the entropy increase from simple intermingling (ideal) or by an energy decrease due to molecular interactions?"

To answer the first of those two there - I was going to compare the enthalpy of each result (at different mass fractions of water:ethanol) to the ideal case where enthalpy =0 J. For the second of those points I was thinking of comparing the Gibbs energy of my solutions to the ideal. So for my solution I was going to calculate:
ΔG = ΔH-TΔS = Q-T(ΔS_mixing+ΔS_temp)
Where I was going to have:
ΔS_temp=Cp*ln(Tf/Ti) (Not sure how valid my use of this one is but I figured it should give a somewhat decent approximation?)
and ΔS_mixing=-nR(xlnx-(1-x)ln(1-x))

For all of these I was going to graph the desired value on the y-axis and the mass fraction on the x-axis.

We haven't studied partial molar properties extensively - it could be possible we would be studying these later. The prac sheet did talk about partial specific volumes (particularly for the other investigation question which was about the volume changes specifically) - the first investigation seemed to be about constructing a relationship similar to this.

I'm assuming specific enthalpy like specific volume is just the enthalpy divided by the mass? So if I'd used a constant total mass in my calculations the graph of enthalpy vs mass fraction would show the same relationship as specific enthalpy vs mass fraction? (Since dividing the enthalpy by the total mass would be dividing each value by a constant?)

What would be the purpose in calculating the change in internal energy? I found on first inspection that work done and heat released were of opposite sign, but the work done on the liquid was less than the heat released - so the solution was losing internal energy.

To summarise what I've been plotting so far:
Heat and Work against mass fraction
by extension enthalpy against mass fraction (same as heat - maybe I could change this to specific enthalpy?)
Change in Gibbs energy for an equivalent ideal process and the Gibbs energy given earlier against mass fraction
If I have time I may also try and plot volume against the mass fraction though I'm doubting I will at this rate. :S

Thank you again!

EDIT: Here's something like I'd hope to end up with. :)