An ideal gas after going through a series of four thermodynamic states in order, reaches the initial state again (cyclic process). The amounts of heat (Q) and work (W) involved in the states are,
`Q_(1) = 6000J, Q_(2)= -5500 J,`
`Q_(3)=-3000 J, Q_(4)=3500 J`
`W_(1)=2500J, W_(2)=-1000 J,`
`W_(3)=-1200 J, W_(4)=xxJ`
The ratio of net work done by the gas to the total heat absorbed by the gas in η. The value of x and `eta` are nearly :-

Options

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

Answer

B

Explanation

In a Cyclic proces , `dU_("cyclic")=0`
`Delta Q_("cyclic")=Delta W_("cyclic")`
`rArr Q_(1)+Q_(2)+Q_(3)+Q_(4)=W_(1)+W_(2)+W_(3)+W_(4)`
` rArr 6000-5500-3000+3500`
`=2500-1000-1200+X`
`rArr X=700J`
`W=W_(1)+W_(2)+W_(3)+W_(4)`
`=2500-1000-1200+700=1000J`
`Q= Q_(1)+Q_(2)`(Positive)
`=6000 + 3500`
`=9500J`
Efficiency `eta=(W)/(Q_(1)) xx 100%`
`eta=(1000)/(9500) xx 100%`
`eta=10.5%`