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
`Q_(1)= 6000 J` ,`Q_(2)= -5500 J,` `Q_(3) = -3000 J,``Q_(4) =3500 J`
`W_(1) = 2500 J, ``W_(2) = -1000 J,``W_(3) =-1200 J:``W_(4) = x J `
The ratio of the net work done by the gas to the total heat absorbed 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

From first law of thermodynamics,
Net heat `(SigmaQ)` = Net work `(SigmaW)`
`ointSigmaQ=ointSigmaW`
`6000-5500-3000 + 3500 = 2500 - 1000 - 1200 +x`
`x = 700`
`eta = ("Network")/("Heat absorbed") xx 100 =(W_(1) + W_(2) + W_(3) + W_(4))/(Q_(1)+Q_(4)) xx100`
=10.5%