A gaseous sample is generally allowed to do only expansion/compression type of 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=-int_(V_(i))^(V_(f))(nRT)/(V).dV=-nRT ""In((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. An ideal gaseous sample at time state i(P_(0)V_(0)T_(0)) is allowed to expand to volume 2V_(0) using two different processes, in the first process the equation of process is PV^(2)=K_(1) and in second process the equation of the process is PV=K_(2). Then :

A gaseous sample is generally allowed to do only expansion/compression

1.	A gaseous sample is generally allowed to do only expansion/compression type of 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=-int_(V_(i))^(V_(f))(nRT)/(V).dV=-nRT ""In((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. An ideal gaseous sample at time state i(P_(0)V_(0)T_(0)) is allowed to expand to volume 2V_(0) using two different processes, in the first process the equation of process is PV^(2)=K_(1) and in second process the equation of the process is PV=K_(2). Then :
Answer» WORK done in the first PROCESS will be greater than work in second process ( magnitude WISE) the order of VALUES of work done cannot be compared unless we know the value of `K_(1)" and "K_(2)` value of work done ( magnitude) in second process is greater in above expansion irrespective of the value of `K_(1)" and "K_(2)` 1st process is not possible Answer :c