One mole of an ideal monoatomic gas undergoes the process A B C D A as shown in the graph. The work done during the process is

Step-by-step solution

Process undergone by 1 mole gas is as shown For an ideal monoatomic gas, ratio of specific heat = 5 3 For adiabatic process A Bp_A V_A^ =p_B V_B^ Substituting values of p_A, p_B and V_B from graph, 32 10^5 V_A^ =1 10^5 (1)^ V_A^ = ( 1 32 )^ Here, = 5 3 (V_A )^ 5 3 = 1 32 or V_A= ( 1 32 )^ 3 5 = 1 (2^5 )^ 3 5 = 1 8 ~m ^3 Similarly for adiabatic compression C D , p_C V_C^ =p_D V_D^ We substitute values from graph, array ll & (1 10^5 ) (8)^ 5 3 = (32 10^5 ) (V_D )^ 5 3 & V_D^ 5 / 3 =1 & V_D=1^ 3 / 5 =1 ~m ^3 array Now