[tadisc]

Hello,

I am somewhat versed in Chemistry, and have some thermodynamics background, but it has been a while since I have practiced it, so I am looking for help.  I have a theoretical situation I am mulling over and this problem is stumping me.

If I take the compound Acetaldehyde and boil it, what is the expansion ratio from when it is a liquid to when it is a gas? such as - liquid:gas. I have looked all around the internet but cannot find answers. Is there a way to figure this out? I am also trying to figure the same thing out with Chlorine.  I used the Ideal Gas law, but I dont seem the get answers that make much sense. If anybody could help me here or point me in the right direction I would appreciate it.

thanks,

Tadisc

[Hunter2]

Ideal gas law is a good start up.

If you have the volume of your liquid Acetaldehyde, you can convert via density to the mass. With the molar mass you can calculate the moles. Now you have to settle a temperature and a pressure and you can calculate V = nRT/P.

[tadisc]

Ok, did a little work on this and here is what I got, but it seems off.

Following what you said: I used an arbitrary volume (1 cm³) and found the mass based on the density at 10°C from wikipedia. 

V = 1 cm³
d = 0.784 g/cm³
m = d × V
m = 0.784 × 1 = 0.784 g

Now the molar mass also per wikipedia is 44.05 g/mol so....

mol = m/mm = 0.7904 g/44.05 g/m = 0.1794

Now, I have mols (0.1794 mol), Vol (0.000001 m³) , R (8.314 J/molK), and T (10°C or 283.15 K). So I figured I will solve for P, and then keep the pressure constant and solve for the volume at 30°C, which should be when the compound is in a gaseous form, and then I will be able to see the volume change.  This is what I found:

P = nRT/V = 401.14 Mpa

First off, that seems way too big.  I was thinking it should be around 1 atm? But, I could be wrong....

Next, Solving for V with the new characteristics....

V = nRT/P = ((0.1794 mol)(8.314 J/molK)(303.15 K))/401140190.6 Pa = 1.127 cm³

Alright.  So according to this, it expanded by 12%, which doesn't seem right. I was fairly certain that most liquids expand quite alot when going through the phase change from liquid to gas.  I am assuming I am either missing something or have flawed logic in how this should work.  Any thoughts or suggestions?

Thanks,

Tadisc

[Hunter2]

No, Don't mix the liquid volume with gaseous volume. Read what I said. Until to the moles you are right.

Now you have to choose a fixed pressure and temperature. V = nRT/P.

[tadisc]

Ok, so reevaluating....

 Based on the 1 cm³ volume of liquid we found that the moles equals...

mol = 0.1794

Now I know the mols, R, and set a T and a P.

I will choose about 1 atm (101 kPa) and 303 K (30°C). These are the results:

V = nRT/P = (0.1794 × 8.134 × 303 K)/101 kPa = 0.0043777 m³ or 4377.7 cm³

This gives me a expansion ratio of 1:4377.  Does this seem right?

Thanks again!

Tadisc

[Hunter2]

Yes I would say so. Question is under which conditions the substance is liquid or gaseous? Not every pressure and temperature gives a gas.

[tadisc]

I believe that it boils at 20.1 degrees Celsius at normal atmospheric pressure.  This being the case, the calculation at 10 degrees should be liquid and the calculation for volume at 30 degrees should be gaseous.  So, to keep this train moving, how can we find out how much energy it would take to warm this compound up from 10 to 30 degrees, converting it from liquid to gas?

[tadisc]

Alright so I did some digging and this is what I have found.  Well be using Q = m×C×ΔT and Q = m×ΔH_vaporization with these constants:

Heat of Vaporization (ΔH) = 26.11 kJ/mol
Specific Heat Capacity for Gases (C) = 1.255 J/g K
Specific Heat Capacity for Liquids (C) = 1.38 J/g K

So lets say we start at 15.1 °C and raise by 5 to 20.1 which is the boiling point. 

Q_l = m C ΔT
Q_l = (0.7904 g)(1.38 J/gK)(5 K)
Q_l = 5.45 J

Now we do heat of vaporization. First we need to convert though:

(26.11 kJ/mol) / (44.05 g/mol) = 0.5927 kJ/g

Q_v = m ΔH_v
Q_v = (0.7904 g)(0.5927 kJ/g)
Q_v = 468.5 J

Now for Gases...

Q_g = m C ΔT
Q_g = (0.7904 g)(1.255 J/g K)(5 K)
Q_g = 4.960 J

Now we add them all up....

Q_T = Q_l + Q_v + Q_g
Q_T = 5.45 J + 468.5 J + 4.960 J
Q_T = 478.9 J

So I guess that's it.  Does that process/answer look correct?

Thank you!

Tadisc

 

[cseil]

It seems right to me (you brought the gas to 25.1°C).
Let's wait for Hunter2 to confirm 

[mjc123]

Going back to the original question, there is an error in the number of moles.
0.7904/44.05 = 0.01794
so the expansion ratio is 1:437.7, which looks more sensible.
Also, 0.784g seems to have changed to 0.7904g somehow...

[tadisc]

Oh good catch! So my answer for energy are relatively the same, but redoing the first part does then give me this expansion ratio:

1:437.7

Which does seem more reasonable. So to keep my train of thought....if I took that volume of gas in a container and put it underground a few feet where its about 50 degrees F, how long would it take to condense completely? whats the best way to figure that out?

Thanks!

Tadisc

[mjc123]

It won't condense completely; it will have a substantial vapour pressure at 10K below the boiling point. (0.67 bar in fact. So less than half would condense in a constant volume set-up.)
How long it would take is a matter of kinetics, not thermodynamics. It will condense pretty rapidly once thermal equilibrium is reached. How long that takes depends on things like thermal contact with surroundings, thickness and thermal conductivity of container wall etc.