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. 1-I, 2-II,3-III,4-IV
B. I-II,2-I,3-IV,4-III
C. 2-III,3-II,4-I,1-IV
D. 3-II,1-I,3-IV,4-II

Answer

B

Explanation

For isothermal irreversible process `(T="constant")`
`P^(@)V^(@)=P'2V^(@)rArrP=(P^(@))/(2)`
For `P^(2)V="constant"`
`P_(o)^(2)V^(@)=P^(2)2V^(@)rArrP=(P^(@))/(sqrt(2))`
Final pressure of isothermal process `gt` Final pressure of adiabatic process. and Final pressure of irreversible adiabatic process `gt` Final pressure of reversible adiabatic process. So, final pressure order `II gt I gt IV gt III`