[Tedjn]

I'm quite confused with this problem from Zumdahl's:

6. Consider 5.5L of a gas at a pressure of 3.0atm in a cylinder with a movable piston. The external pressure is changed so that the volume changes to 10.5L

a) Calculate the work done, and indicate the correct sign.
b) Use the preceding data but consider the process to occur in two steps. At the end of the first step, the volume is 7.0L. The second step results in a final volume of 10.5L. Calculate the work done and indicate the correct sign.
c) Similar to b), not important

--Zumdahl's 4th Ed. Pg 279

As it's an even numbered question without an answer in the back, so I can't even check for the answer and crunch some numbers  !

I'm not too sure where to start with this problem. Section 6.1 (which we were told to read) emphasized that work = -P?V, but P is the external pressure and it was always constant when the book examples were looking for the amount of work. In this problem, I'm assuming that 3.0atm is the external pressure, and not the pressure of the gas. Am I correct in doing that?

But even doing that, the problem says that the external pressure varies, and the volume varies as well. The two seem to be varying at the same time, and work shouldn't be held constant (we're looking for it, right?) so I have no clue where to begin.

It's probably something simple I'm missing, but any hints, guys?  

[Donaldson Tan]

assume perfect gas.

assume isothermal process, so T is a constant.

work done = integrate -p.dv from initial volume to final volume

substitute perfect gas equation into p-term of the above integral.
           

[Tedjn]

Thanks for the response To tell the truth, this is only an honors chem course, not even AP chem. I haven't even had physics not to mention calculus, so pardon me for my elementary understanding of your post -_-

subsitute perfect gas equation into p-term of the above integral.

Does this mean I should assume internal pressure = external pressure (only way I can see using ideal gas law as substitute for P)? "If I do that w = -P?V = 0 which is definitely false." (EDIT: I still can see this, since PV is constant both at initial and final volume. The only explanation for this and still maintaining what I wrote below is that w = -P?V is only applicable for constant external pressure, I conjecture) (EDIT2: Ok, I just can't understand. P?V = PV_f - PV_i = 0 if external pressure = internal pressure, but the whole integral thing makes sense as well. What am I missing?) Or am I misunderstanding again? :blush:

work done = integrate -p.dv from initial volume to final volume

How exactly does this work? I've only self-studied differentiation and don't completely understand integration, so I may have gotten this wrong, but is the ?V is a constant? Thus, after we substitute k/V for P where k = nRT, we only worry about the V^-1?

I can kind of picture it, if I graph a graph with V as x-axis and P as y-axis, then P?V is the area of the curve underneath the graph, but because P and V are inversely proportional, the graph should be linear (right?), and thus we can merely take the average of the beginning and ending pressures and multiply that to ?V?

So, for a), we would have 3(5.5) = x(10.5) and P_avg = 2.3 atm and our work is (2.3)(10.5 - 5.5) = 12 atm.L = 1.2 kJ (if we don't round the whole way through).

Is that how it should be done?

Sorry about the rambling response, I was thinking while typing.

[Donaldson Tan]

You cannot use -P?V directly because this is not an isobaric process.

Pressure is not constant. It is stated in question that gas in the piston expands when the external pressure is changed. The gas expansion was not due to heat supplied (which is none) but due to reduction in external pressure.

Therefore you have to integrate.

PV = nRT => P = nRT/V

work = - integrate from V_ini to V_final nRT/V dV = - nRT integrate from V_ini to V_final 1/V dV = - nRT.ln(V_final/V_ini) = nRT.ln(V_ini/V_final)

[Tedjn]

Thanks once again for your patience. Since n and T aren't given for the problem, I think it is safe to assume that the question was most likely written wrong -- perhaps it was meant to refer to something else.

Anyhow, if you still have any patience with me at all, can you explain the relationship between external pressure and the work? I understand that I most likely do not have the knowledge at this level to understand any explanation (I'm still a sophomore in high school), but if you're willing to give a more detailed explanation than you usually need to, to accommodate me, I would be extremely grateful. I take it that w = integrate(...) is a completely different formula than -P?V? What does integrating do exactly.

One of the conceptual barriers that I need a bit of clarification on is your use of PV = nRT. Does it indicate that the internal pressure of the gas is equivalent to the external pressure (since the piston is movable)?

And from the integration formula, because of my meager at best understanding of calculus, I can only picture integration as the area under a curve, namely P = nRT/V, which is linear (is it?). Thus, the area underneath that curve would be

( (P_final - P_initial)/2 ) * ?V

which would then be the work (am I correct in this? my understanding of integration is very shaky).

Of course, we do not know the final external pressure, but if I was correct in assuming that internal pressure = external pressure (now, this I'm really unsure of, see above), we can find the final pressure with P₁V₁ = P₂V₂.

Which in this case, for (a), would be 11.43 atm.L = 1.16 kJ as the work (ignoring sig figs and signs).

Is this thought process correct? If not, where is it wrong? And if it is somehow correct, is it possible to still explain to me about why we need to integrate (how it ties into the real world data, for example).

Thank you so, so much for sticking with me. I'm very grateful, and if you find that even the most simplest explanation would be above my level, just let me know, and I'll leave it for when I'm ready

[Donaldson Tan]

Assuming constant temperature allows us to use Boyle's Law to find the final pressure.

In fact, work is given as:
W = - integrate p dV = - integrate nRT/V dV = -nRT ln(V_final/V_ini) = nRT ln(V_ini/V_final)

Although n and T are unknown, we know the value of nRT (by virtue of the perfect gas equation), ie. P₁V₁ = nRT

W = nRT.ln(V_ini/V_final) = P₁V₁.ln(V_ini/V_final)