[rebeccak]

Hello, this is a problem I missed on an exam:

"In an isothermal process, the pressure on one mole of an ideal monatomic gas suddenly (i.e. irreversibly) changes from 9.40 atm to 3.20 atm (the external pressure) at 298K.

Calculate q."

I understand that because the process is isothermal, the change in energy (and enthalpy) is 0.  So,

q=-w (where work will be negative since the gas is expanding)
w=delta(PV)

I tried using the PV=nRT relation to solve for the volumes at 9.40 and 3.20 atm,

V=(nRT)/P
V=(1mole*.0821 Latm/molK*298K)/9.40 atm,

 then plugging the difference between those two pressures and volumes into the work equation, but I am not getting the correct answer of 1.63 kJ.  Any clarification would be appreciated.  Thanks a lot!

[integral0]

Hello again rebeccak.  Sorry for your unfortunate circumstances.  I didn't finish the problem since I had thought that I had done enough to show you the way but apparently the problem is more involved after seeing your calculations go astray of the real answer.  I'll ponder on it some more and get back to you A.S.A.P. but now I'm off to school so till then, see ya!

[rebeccak]

no problem. Thanks for your *delete me*

[integral0]

Hey Rebeccak,

Ok, first I would like to say that my derivation of q was wrong.  Sorry but I am only in AP Chemistry and I thought I could modify the first law of thermodynamics and the equation for Enthalpy to fit your situation.  

After struggling with your problem I finally took it to my teacher and what he said we would have to do is quite difficult.

Now, I am unsure of your background but I am sure of mine  .  I haven't taken Calculus yet but according to my teacher it involves principles used in physics that involve calc-applied math.

This is what he said to do.

First of all . . .

I was right in that . . .

isothermal process ?E=0  b/c T is proportional to internal E of system in isothermal process

I was not correct in my modification of work .  . . here is why

?E = q + w
0 = q + w
q = -w
=-(-?(PV))
=?(PV)          

in the 2 last lines, you can't make the assumption that volume is changing because you do not know the volume and that one can't use this enthalpy formula for a changing volume and changing pressure at the same time.  This formula can only be utilized when pressure is constant.

To account for a changing volume, it becomes more involved:

First of all, you use the ideal gas law equation to find the two different volumes

then once you find the two volumes, you plug it into P?V (constant pressure)

he comes the tough party (for me at least)


then you use this equation to determine the amount of energy = work performed

w/n = RT (integrate symbol w/ superscript Vf and subscript Vi) dv / V = RT ln Vf/Vi

R will be 8.314 and T will be in kelvins, n = moles

use the two volumes you found and plug it into the equation above


this should enable you to find what you're looking for . . . although I don't know how to use integrals yet so GOOD LUCK!

Please will other MEMBERS EVALUATE THE METHOD I USED!

Thanks

With regards,

Integral0