[Ax3shr3dd3r]

Ok so I'm having a problem solving for maximum work. The problem calls to find the maximum work for the expansion of an ideal gas. I'm pretty sure the formula for work (ω=PΔV) would require a negative sign for the pressure as it is expanding? Anyways, what I'm stuck on is that the problem provides:

-moles of the gas at a given temperature
-initial and final volume of the expanded system

How would I use this information given to find the maximum work?
I notice that volume change is simple to find and is needed to calculate work, but what about the pressure?

[clinz63]

Which produces greater work: reversible or irreversible expansion?

[Ax3shr3dd3r]

Which produces greater work: reversible or irreversible expansion?

Reversible?

[Ax3shr3dd3r]

Would I use the idea gas law to find the pressure? (PV=nRT)
If so, what volume would be used?

[clinz63]

Hmmm... I guess you really have no idea. The general equation for expansion work is
dw = -p_extdV.
In a reversible expansion, p_ext = p_gas (orsimple p). Substituting the ideal gas law,
dw = -nRTdV/V
Upon integration,
w = -nRTln(V₂/V₁)

[Ax3shr3dd3r]

Hmmm... I guess you really have no idea. The general equation for expansion work is
dw = -p_extdV.
In a reversible expansion, p_ext = p_gas (orsimple p). Substituting the ideal gas law,
dw = -nRTdV/V
Upon integration,
w = -nRTln(V₂/V₁)

Thank you for providing me with a bunch of equations, but i don't understand kind of..is that ln as in natural log in the integration equation? I've never seen that.

[idest]

http://en.wikipedia.org/wiki/Isothermal_process#Calculation_of_work

clinz63 said the stuff I linked. Actually, The solution reduced to find the area under a hyperbola.
So, we probably have to use calculus to solve it.



There might exist calculus-free method to find the area, because a hyperbola is a special figure in geometry. However, this is a subject of chemistry, and the integration is a general way to find work involved in a thermodynamical process.

General chemistry doesn't usually treat this (maybe it's a more advanced concept), and Physical chemistry does.

[Ax3shr3dd3r]

http://en.wikipedia.org/wiki/Isothermal_process#Calculation_of_work

clinz63 said the stuff I linked. Actually, The solution reduced to find the area under a hyperbola.
So, we probably have to use calculus to solve it.



There might exist calculus-free method to find the area, because a hyperbola is a special figure in geometry. However, this is a subject of chemistry, and the integration is a general way to find work involved in a thermodynamical process.

General chemistry doesn't usually treat this (maybe it's a more advanced concept), and Physical chemistry does.

Hm I thought this was too hard to be general chem question..Anyways, thank you guys for the help.

[Ax3shr3dd3r]

Can someone please help me as I am stuck on what the units will be when using the equations provided above.

[clinz63]

Oh. Sorry. I thought this was physical chemistry.

Yes, that is ln as in natural logarithm.

Your units should be J.

[Ax3shr3dd3r]

Oh. Sorry. I thought this was physical chemistry.

Yes, that is ln as in natural logarithm.

Your units should be J.

How does one get Joules if nothing being plugged in to: w = -nRTln(V2/V1) is in the units of J?

[TheSodesa]

Oh. Sorry. I thought this was physical chemistry.

Yes, that is ln as in natural logarithm.

Your units should be J.

How does one get Joules if nothing being plugged in to: w = -nRTln(V2/V1) is in the units of J?

You're just going to have to look up SI-conversions of your original units and use those instead. Wolfram Alpha might help with this. From memory, R is roughly 8.314 J/molK.