A monoatomic ideal gas through a cyclic process as shown in the figure . The efficiency of this process is

Options

A. `19 . 04 %`
B. `40 . 04 %`
C. `50.00 %`
D. `10.00 %`

Explanation

(No otptions match)
we know that
`W = P Delta V`
Now,
`W_(AB) = 0 `
`W_(BC) = 3P (2V - V) = 3PV`
`W_(CD) = 0 `
`W_(DA) = - PV`
Total work done ,
`W = W_(AB) + W_(BC) + W_(CD) + W_(DA)`
`= 3 PV - PV - 2PV`
Heat given to the system from A to B
`= nC_(n) Delta T`
`= n(3)/(2) R DeltaT = (3)/(2) xx V xx Delta P`
`= (3)/(2) xx V xx (3P - P)`
`= (3)/(2) xx V xx 2P`
= 3 PV
Similarly ,
Heat, given to the system from B to C
`= nC_(P) Delta T`
`= n ((5)/(2) R) Delta T `
`= (5)/(2) (3P) (Delta V)`
`= (5)/(2) xx (3P) . (2V - V)`
`= (5)/(2) xx 3P xx V`
`= (15)/(2) PV`
The heat is released from C to D and D to A .
Efficiency `(eta) = ("Work done by gas")/("Heat given to gas ") xx 100`
`= (2PV)/(3PV +(15)/(2)PV) xx 100`
` = (2PV)/(PV(3 +(15)/(2)))xx100`
`= (2)/(((6 + 15)/(2)))xx100`
`= (2 xx2)/(21) xx100`
` = (4)/(21) xx 100 = 19 . 04 %`