An ideal gas is taken through the cycle A B C A , as shown in figure. If the net heat supplied to the gas in the cycle is 5 ~J the work done by the gas in the process A B is

Step-by-step solution

Process B C occurs at constant volume Hence, W_ B C =0 Process C A occurs at constant process. Hence the work done in the process is W_ C A =P V=(1) (V_A-V_C )=(1)(4-2)=2 ~J For a cyclic process, U=0 According to the first law of thermodynamics Q= W+ U Q= W=W_ A B +W_ B C +W_ C A 5=W_ A B +0+2 or W_ A B =3 ~J