N moles of a monotomic gas is carried round the reversible rectangular cycle ABCDA as shown in the diagram. The temperature at A is `T_sigma`. The thermodynamic efficiency of the cycle is

Options

A. 0.15
B. 0.5
C. 0.2
D. 0.25

Explanation

First of all note that point C is at highest temperature and A is at lowest temperature So during process A to B and B to C heat is being added to the cycle.
Work output can be calculated by calculating the area under the cycle of P.V diagram.
Then efficiency can be calcutated. as work output upon heat input.
Note that the temperuture at A is `T_o`, that at B is `2T_0`and that at C is `4T_0`
Also note `C_v` for moatomic gas is `(3R)/2`
For `A- B`
`DeltaQ = NC_bDeltaT` (`:.` constant volume process)
`= N (3R)/2(2T_0-T_0)=N/2 3RT_0`
For B - C
`Delta=NC_pDeltaT` (`:.` constant pressuce process)
`=N (5R)/2 (4T_0- 2T_0)`
`=5NRT_0`
Now, Total heat input `Q = (3/2+5) NRT_0 = 13/2 NRT_0`
Work output = Area Under the cycle
`=(2P_0-P_0)(2V_0-V_0)`
`= P_0V_0`
`NRT_0`
`:.` Efficiency `=W/Q =(NRT_0)/(13/2NRT_0)`
`= 2/13 or 15.38%`
Note : Just as an exercise do the same problem for diatomic gas and find the answer 10.5% for efficiency.