An ideal gas is enclosed in a cylinder with a movable piston on top of it. The piston has a mass of 800 g and an area of `5.00 cm^(2)` and is free slide up and down, keeping the pressure of the gas constant. How much work is done on the gas as the temperature of 0.200 mol of the gas is raised from `20.0^(@)C` to `300^(@)C` ?

Explanation

The gas is expanding and doing positive quantity of work in lifting the piston. The work on the gas in this number of joules but negative. The original internal energy of the gas in some multiple of `nRT_(i) = (0.2 mol) (8.314 J//mol K) (293 K) = 10^(3) J`. The wor done by the gas does not come just from the internal energy, but from some of the energy that must be put into the sample by heating to raise its temperature. But still we can estimate the work as on the order of - kJ.
The integal of PdV is easy because the pressure is constant. Then the equation of state will let us evaluate what we need about the pressure and the change in volume.
For constant pressure,
`W = - int_(i)^(f) PdV = - P Delta V = - P (V_(f) - V_(i))`
Rather than evalating the pressure numerically form atmospheric pressure plus the pressure due to the weight of the piston, we can just use the ideal gas law to express in the volumes obtaining
`W = - P ((nRT_(h))/(P) - (nRT_(c ))/(P)) - nR (T_(h) - T_(c ))`
Therefore, `W = - nR Delta T`
`= - (0.200 mol) (8.314 J//mol//K) (280 )`
`= - 466 J`
We did not need to use the mass of the piston or its face area. The answer would be the same if the gas had to lift a heavier load over a smaller volume change a lighter load through a larger volume change. The calculation turned out to be simple, but only because the pressure stayed constant during the expansion. Do not try to use `- P Delta V` for work in any case other than constant pressure, and never try to use `-V Delta P`