Three moles of an ideal monoatomic gas undergoes a cyclic process as shown in the figure. The temperature of the gas in different states marked as 1, 2, 3 and 4 are 400 ~K , 700 ~K , 2500 ~K and 1100 ~K respectively. The work done by the gas during the process 1-2-3-4-1 is (universal gas constant is R )

Step-by-step solution

We knows d Q=d u+d w and we also known d u=0 for cyclio process so that d Q=d w Here, in given condition the work done during is a basic process aligned & w_ 2-3 =P₂ (v₃-v₂ ) & w_ 4-1 =p₁ (v₁-v₄ ) aligned Total work done =p₂ (v₃-v₂ )+p₁ (v₁-v₄ ) From gas equation p V=n R T= 3 T 2 Hence, total work done aligned & = 3 R 2 (400+2500-700-1100) & = 3 2 R(2900-1800) & = 3 2 R(1100)= 3300 R 2 & =1650 R aligned