Question

The initial state of certain gas is `(P_(i), V_(i), T_(i))`. It undergoes expansion till its volume become `V_(f)`. Consider the following two cases :
(a) the expansion takes place at constant temperature.
(b) the expension takes place at constant pressure.
Plot the `P-V` diagram for each case. In which of the two cases, is the work done by the gas more?

Answer 1

In f(figure) A represents initial state of the gas.
(a) When expansion takes place at a constant temp. `T_(1)`, the `PV` curve is an isothermal curve AB. Work done by gas, `W_(1)`= area under the curve AB
(b) When expansion takes place at constant pressure, the `P-V` curve is straight line AC parallel to OX.
Work done by the gas, `W_(2)`= area under the curve AC.
Clearly, `W_(2) gt W_(1)`
i.e., work done is more when the gas expands at constant pressure.