Question

The initial state of a certain gas is `(p_(i),V_(i),T_(i))`. It undergoes explansion till its volume becomes `V_(f)`. Consider the following two cases
(a) the expansion takes place at constant temperature.
(b) the expansion 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

Consider the diagram `p-V`, where variation is shown for each process.

Process 1 is isobaric and process 2 is isothermal.
Since, work done=area under the `p-V` curve. Here, area under the `p-V` curve 1 is more. So, work done is more when the gas expands in isobaric process.