# Questions About Work of Mixing

**1. The problem statement, all variables and given/known data**

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?

**2. Relevant equations**

W=-pΔV

ΔH = ΔU+ pΔV

ΔG=ΔH-TΔS

**3. 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! đź™‚

http://ift.tt/1iVdGFK

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