An ideal gas is subjected to cyclic process involving four thermodynamic states, the amounts of heat (Q) and work (W) involved in each of these states are aligned & Q₁=6000 ~J , Q₂=-5500 ~J ; Q₃=-3000 ~J ; & Q₄=3500 ~J & W₁=2500 ~J ; W₂=-1000 ~J ; W₃=-1200 ~J ; & W₄=x ~J . aligned The ratio of the net work done by the gas to the total heat absorbed by the gas is . The values of x and respectively are

Step-by-step solution

From first law of thermodynamics aligned & Q= U+W & or U=Q-W & U₁=Q₁-W₁=6000-2500=3500 ~J & U₂=Q₂-W₂=-5500+1000=-4500 ~J aligned aligned & U₃=Q₃-W₃=-3000+1200=-1800 ~J & U₄=Q₄-W₄=3500-x aligned For cyclic process U=0 aligned & 3500-4500-1800+3500-x=0 & or & x=700 ~J & Efficiency, = output input 100 & = W₁+W₂+W₃+W₄ Q₁+Q₄ 100 & = (2500-1000-1200+700) 6000+3500 100 & = 1000 9500 100 & =10.5 % & aligned