Two moles of helium gas undergo cyclic processes shown as assuming the gas to be ideal.calculate the following quantities in this process. (a) The net change in the heat energy (b) The network done (C) The net change in internal energy. (T_A = 300K, T_B = 400K)

Two moles of helium gas undergo cyclic processes shown as assuming the

1.	Two moles of helium gas undergo cyclic processes shown as assuming the gas to be ideal.calculate the following quantities in this process. (a) The net change in the heat energy (b) The network done (C) The net change in internal energy. (T_A = 300K, T_B = 400K)
Answer» Solution :We shall first calculate here work done in the cyclic process. As AB is isobaric process, HENCE work done during this process from A to B is `W_(AB)= P_A(V_B -V_A)=P_BV_B-P_AV_A(:.P_A =P_B )` According to perfect gas EQUATION,` P_A V_A= nRT _Band P_BV_B= nRT_B ` ` thereforeW_(AB )= mu RT(T_B -T_A) 2 xx 8.32 xx ( 400-300= 2 xx 832 = 1664` joule workdoneduringisothermalprocessfromB to C ` W_(BC )= nRTlog_e((V_f)/(V_i))= mu RT_Clog_e((V_C)/(V_B))=muRT _C log((P_B)/(P_C))[ :.P_BV_B=P_C V_C ]` `= 2 xx 8.32 xx 400 xx log _e21[ :.P_B= 2AND P_C= 1 atm]= 4068` joule Work done during isobaric process from C to D `W_(CD)= nR(T_D -T_C ) 2 xx 8.32 xx (300 -400 ) =- 1664` joule Work done during isothrmal process from D to A ` W_(DA ) = nRTlog_e((V_f)/(V_i))= nRT _Dlog_e((V_A)/(V_D)) = nRT_D log_e (P_d //P_A) [:.P_AV_A=P_DV_D ]` `=- 2 xx 8.32 xx 300 xxlog_e 2 =- 3456` joule ` therefore` Networkdone ` =W_(AB) +W_(BC) +W_(CD) +W_(DA) = 1664+ 4608- 1664- 3456= 1152` joule (a)INA cyclicprocess `DeltaU=0therefore `Fromfirstlawof thermodynamics Netchangein heatenergy ` Delta Q = Deltaw= 1152` joule (b )Networkdone ` DeltaW= 11.52` joule `(c )` Netchange ininternalenergy ` Delta U=0`.