An ideal gaseous sample at intial state `i(P_(0),V_(0),T_(0))` is allowed to expand to volume `2 V_(0)` using two different process, in the first process, thje equation of process is `2PV^(2)=K_(1)` and in the second process the equation of the process is `PV=K_(2)`. Then:

An ideal gaseous sample at initial state I (P_0, V_0, T_0) is allowed to expand to volume 2V_0 using two different process, 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 radiactive sample decays by two different processes .Half - life for the first process is t_(1) and for the second process is t_(2) . The effective half-life is

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

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

A radioactive sample can decay by two different processes. The half-life for the first process is T_1 and that for the second process is T_2 Find the effective half - life T of the radioactive sample.