4A.3: Why is "expansion work" seperate from Work?

[Nancy Romo 1E]

When I first attempted the problem (see below), I had tried using the general equation of W = F * ΔDistance. Since we were given the change in distance, and could solve for the force F from Pressure = Force / Area -> Force = Pressure (which we were given) * (Area) (which I thought we could solve for since we were given a diameter).

But when I used the previous technique, I didn't get the same answer as when I used the expansion work equation W_exp = -P_ext * ΔV.

If someone could clear up the difference, or troubleshoot my errors in the first method, it would be much appreciated! Is it because the Pressure equation doesn't account for a "changing" area?


**Problem: 4A.3 Air in a bicycle pump is compressed by pushing in the handle. The inner diameter of the pump is 3.0 cm and the pump is depressed 20. cm with a pressure of 2.00 atm. (a) How much work is done in the compression? (b) Is the work positive or negative with respect to the air in the pump? (c) What is the change in internal energy of the system?**

[Allison Peng 1D]

I actually think it should be the same thing?

If we calculate work=force*distance=(pressure)*(area)*(distance) = (2.00atm)*(1.5^2*pi cm^2) * (20cm) --> this is work done on the system, around 90*pi (units in atm*cm^3)

Then if we do work = -Pext*delta V, we get work = -(2.00atm) * (-20cm*1.5^2*pi cm^3) which is also 90*pi atm*cm^3.

[Allison Peng 1D]

^ Obviously for the final answer, you should convert the atm*cm^3 to Joules, but this is kind of a rough outline

[Mia Yamada 1J]

This would make sense since expansion work refers to the specific work done against an external pressure when the volume of a gas expands. And work, more generally speaking, is done when a force displaces an object by a distance. The latter is speaking more in mechanical terms, but essentially they refer to the same concept.