An ideal gas is taken through a cyclic thermodynamic process through four steps. The amounts of heat involved in these steps are `Q_1=5960J, Q_2=-5585J, Q_3=-2980J and Q_4=3645J`, respectively. The corresponding quantities of work involved are `W_1=2200J, W_2=-825J, W_3=-1100J and W_4` respectively.
(1) Find the value of `W_4`.
(2) What is the efficiency of the cycle

Options

A. `1315 J`
B. `275 J`
C. `765 J`
D. `675 J`

Answer

C

Explanation

c. `Delta Q = Q_(1) + Q_(2) + Q_(3) + Q_(4)`
`= 5960 - 5585 - 2980 + 3645 = 1040 J`
`Delta W = W_(1) + W_(2) + W_(3) + W_(4)`
`= 220 + 825 - 1100 - W_(4) = 275 + W_(4)`
For a cyclic process, `U_(f) = U_(f)`
`Delta U = Uf - U_(i) = 0`
Form the first law of thermodynamics,
`Delta Q = Delta U + Delta W`
`1040 = 0 = 275 + W_(4)` or `W_(4) = 765 J`