An ideal gas is subjected to cyclic process involving four thermodynamics states, the amounts of heat `(Q)` and work `(W)` involved in each of these states.
`Q_1=6000J,Q_2=-5500J,Q_3=-3000J,Q_4=3500J`
`W_1=2500J,W_2=-1000J,W_3=-1200J,W_4=xJ` The ratio of the net work done by the gas to the total heat absorbed by the gas is `eta`. The values of x and `eta` respectively are

Options

A. 500, 7.5%
B. 700, 10.5%
C. 1000: 21%
D. 1500, 15%

Answer

B

Explanation

Since, `Q=Delta U+W implies DeltaU=Q-W`
`DeltaU_1=Q_1-W_1=6000-2500=3500 J`
`DeltaU_2=Q_2-W_2= -5000+1000= -4500 J`
`DeltaU_3=Q_3-W_3= -3000+1200= -1800 J`
`DeltaU_4=Q_4-W_4=3500-x`
For cyclic process, `DeltaU=0`
`3500-4500-1800+3500-x=0`
`implies x = 700 J`
Now, `eta=(W_("net"))(Q)=(W_1+W_2+W_3+W_4)/(Q_1+Q_4)=(1000)/(9500)`
`implies eta=0.105=10.5%`