An engine operates by taking n moles of an ideal gas through the cycle ABCDA shown in figure. The thermal efficiency of the engine is :(Take Cv= 1.5 R, where R is gas constant) 0.24, 0.15, 0.08, 0.32

Solution:As we know the formula of work done which is equal to the W=P0V0
Heat given will be equal to, =QAB+QBC
Now by using the heat given formula mentioned above, we can write it as⇒ncvdtAB+ncpdtBC
As we already know the value of Cv=1.5R
Therefore substituting this value in the heat given equation, we get, =32(nRTB−nRTA)+52(nRTC−nRTB)
For the monatomic gas, the value will be 3
Now putting the values we had calculated above, we get⇒32(2P0V0−P0V0)+52(4P0V0𕒶P0V0)
On simplifying the solution, we get⇒132P0V0
As we know the formula for efficiency can be given by η=Wheat gain
It can also be written in the following way and also substituting the values, we get⇒P0V0132P0V0
On solving,⇒213⇒η=0.15
Hence C is the correct option