3 moles of an ideal monoatomic gas performs A B C D A cyclic process as shown in figure below. The gas temperatures are T_A=400 ~K , T_B=800 ~K , T_C=2400 ~K and T_D=1200 ~K . The work done by the gas is (approximately) (R=8.314 ~J / mol K )

Step-by-step solution

Processes A to B and C to D are parts of straight line graphs of form y=m x . Also, p= R V T( =3)p T So, volume remains constant for the graphs A B and C D . So, no work is done during processes for A to B and C to D . W_ A B =W_ C D =0 and W_ B C =p₂ (V_C-V_B )= R (T_C-T_B )=3 R(2400-800)=3 R 1600=4800 RW_ D A =p₁ (V_A-V_D )= R (T_A-T_D )=3 R(400-1200)=-2400 R Work done in complete cycle W=W_ A B +W_ B C +W_ C D +W_ D A =0+4800 R+0+(-2400) R=2400 R=19.944 ~J =20 ~kJ