[gsel]

I need guidance on how to solve this problem:

"One mole of an ideal gas undergoes an isothermal expansion at 300 K from 1.00atm to a final pressure while performing 200 J of expansion work. Calculate the final pressure if the external pressure is 0.2atm."

My work so far:
T = 300K
Pin = 1.00atm
Pex. = 0.2atm
w = 200J

I've tried with the formula w = -nRln(P1/P2) and solve for P2 = e^(-w/nR) * V1, whereas V1 = nRT/P1 = 24.62L, but P2 isn't actually P(final) is it?

I understand this has to occur:
Pin. = Pex.

Then I tried to put up an equation for it:

nRT/Vin = -(w/dW) -> From ideal gas law and w = -Pex * dV, but then i need V2 in order to calculate dV.

For V2 i got:

V2 = e^(-w/nR) * V1 -> From w = nR*ln(V2/V1) and the answer i got was no were near probable.

What to do? 

[gsel]

How about this being a solution to my problem:

P1 = 0.2atm
P2 = final pressure

w = -nRTln(P1/P2)   <=>    P2 = P1 - e^[-(w/nRT)]

P2 = 0.2atm - e^[-(200J/1mol*8.3145J/mol*K*300K)] = 0.722atm

According to my book the answer is supposed to be 7.1atm and I think I'm fairly close.

[Enthalpy]

Could it possibly be that you try to apply formulas without first thinking with words to what happens? I've seen similar attempts fail often.

A few suggestions:
- Here you have numbers. That must help you. How big must the final pressure be versus the inital one? Versus the external pressure? That's the kind of checkings that avoid mistakes. You have to do many ones, always.
- What bodies contribute to the work?
- You don't need to learn formulas just to forget them in a year. If you understand what happens in the expansion, you can rewrite the formulas immediately. That's less risky.

Hint: Pin=Pext is wrong.

[gsel]

Lovely advice, appreciate it!
I'll come back stronger!

[ddazzel]

Hi,
Actually I like to discuss on Isothermal expansion, because I like this topic very much.
Also question given above is also very interesting.

[gsel]

I asked my  teacher about it, she had to think about it she said. I admit i felt satisfied she did not pull the answer out of the air like this was some easy going problem.

I have thought so far that I need to find either dV or final volume. The only thing affected in the process which can tell how far the gas has expanded are either work or volume. Since Pfinal will be somewhere in between P1 and Pex. The gas expands only to a certain level requiring 200J of work. The question is: How far will the gas expand for that amount of work? Since PV relation will remain constant, it will be possible to find final pressure for there.

What are your suggestions?
 

[Enthalpy]

Agreed with final P versus initial and external.

I'm not quite sure about "performing 200J work". Is this the work by the gas only? Or the work available despite the external pressure? Let's admit the second version.

The you must evaluate the work by the gas, by the external pressure, and deduce the final pressure.

Much of thermodynamics consists in choosing the proper variable to make computations simpler. What is the best variable to pick, to express both works: P, V, T, W, Q, U, S, H, G...? Then you can solve for this variable, and if needed convert it to a final pressure.

The best variable depends on the process. In adiabatic processes of perfect gasses, the temperature is often the best. Here it's an isothermal process.

[Benzene]

The original volume you can find using PV=nRT 
because initially you know all the variables except V

w=fd

f=20,265 Kg/m*s² w=200kg*m²/s² d= 0.009869m³

so the original volume, plus the 0.009869m³ is the new volume.

so again you can use PV=nRT and now you have the everything except the final pressure.

is this correct?

[Enthalpy]

Are you kidding?

Matching the units is already a good thing, sure. But it won't provide you the proper formula.

Take a little time to think at what a gas is: if you change the pressure at constant temperature, the volume changes - the temperature at constant volume, the pressure changes - and so on. PV=nRT already tells that P, V, T are interdependent.

How could a product like P*(V2-V1) possibly give some work while you know that P depends on V?
You intuition should shout that there's an integral somewhere, from which for instance the Log results.

[Benzene]

my first thought was that we have:
initial pressure=1atm
T constant=300K
mols=1

so I thought I could find the initial volume using PV=nRT

then I assumed that the expansion of the gas was against a fixed external pressure of 0.2 atm.

Is this where I have made my mistake? assuming the external pressure is constant?

because against a constant pressure you could be certain of the expansion volume using w=fd

thanks in advance 

[Enthalpy]

Work is done or received by several entities here:
- The expanding gas
- The external gas at 0.2 atm
- And something else!

It needs something else, like an alternator, a machine-tool... to balance the forces. Imagine that the expansion operates against a piston (after all, thermodynamics was invented for that purpose):
- pressure from 1 atm down to unknown at the inner face
- pressure 0.2 atm at the outer face
- some additional force on the piston must balance the other two! Have a connecting rod and a crankshaft a the piston, and a use at the shaft.

Now, the initial question isn't perfectly clear, but I'd rather understand that the "work" is the one available at the shaft, that is, the work provided by the gas minus the work absorbed by the atmosphere.

Both are accessible through formulas written as function of the proper variable (is it P V, T, U, H, Q, W...?). Write the difference, solve (probably not an algebraic solution) for this variable to get 200J, and convert this solution to a pressure.

In case you get two solutions, choose the best. I'd stop the expansion as soon as the desired work is obtained.