A gaseous sample is generally allowed to do only expansion`//`compression type work against its surroundings The work done in case of an irreversible expansion ( in the intermediate stages of expansion`//`compression the states of gases are not defined). The work done can be calculated using
`dw= -P_(ext)dV`
while in case of reversible process the work done can be calculated using
`dw= -PdV` where `P` is pressure of gas at some intermediate stages. Like for an isothermal reversible process. Since `P=(nRT)/(V)`, so
`w=intdW= - underset(v_(i))overset(v_(f))int(nRT)/(V).dV= -nRT ln(V_(f)/(V_(i)))`
Since `dw= PdV` so magnitude of work done can also be calculated by calculating the area under the `PV` curve of the reversible process in `PV` diagram.
If four identical samples of an ideal gas initially at similar state `(P_(0),V_(0),T_(0))` are allowed to expand to double their volumes by four different process.
I: by isothermal irreversible process
II: by reversible process having equation `P^(2)V=` constant
III. by reversible adiabatic process
IV. by irreversible adiabatic expansion against constant external pressure.
Then, in the graph shown in the final state is represented by four different points then, the correct match can be

Options

A. work done in first process will be greater than work in second process (magnitude wise)
B. The order of values of work done can not be compared unless we know the value of `K_(1)` and `K_(2)`.
C. Value of work done (magnitude) in second process is greater in above expansion irrespective of the values of `K_(1)` and `K_(2)`
D. `I^(st)` process is not possible

Answer

C

Explanation

work done in isothermal process will be more than `PV^(2)=` const, whatever be the value of `K_(1)` and `K_(2)` as is shown in digaram.