The volume of an ideal gas increases 8 times and temperature becomes (1 / 4)^ th of initial temperature during a reversible change. If there is no exchange of heat in this process ( Q =0) then identify the gas from the following options (Assuming the gases given in the options are ideal gases):

Step-by-step solution

For an adiabatic process ( Q = 0 ) with an ideal gas: TV^ -1 = constant Given: V_f = 8V_i and T_f = 1 4 T_i Applying the adiabatic relation: T_i V_i^ -1 = T_f V_f^ -1 T_i V_i^ -1 = T_i 4 (8V_i)^ -1 1 = 1 4 8^ -1 4 = 8^ -1 2^2 = 2^ 3( -1) 2 = 3( -1) = 5 3 = 1.67 This value corresponds to a monatomic gas. Among the options, only He is monatomic ( = 5/3 for monatomic ideal gases). O₂ and N₂ are diatomic ( = 7/5 ), CO₂ is polyatomic ( 1.30 ), and NH₃ is polyatomic.