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

`Q = Q_(1) + Q_(2) + Q_(3) + Q_(4)`
`=6000 - 5500 - 3000 + 3500 = +1000 J`
`W = W_(1) + W_(2) + W_(3) + W_(4)`
`=+300 + x`
In cyclic process,
`DeltaU =0`
Now, `Q= DeltaU +W` or 1000=0 + (300 +x)
`therefore x= 700 J, eta =W/(Q_(1) + Q_(4)) =1000/(6000 + 3500) = 10.5%`