Thermodynamics-3

Efficiency is work/heat supplied...

work=area of the square and heat can be found out by using C_p and C_v respectively BC,AD and AB,CD respectively...
answer is (b)2/21

I am getting (b)

Oh...got it...
Work done=Area of square
= 2
Heat =MCΔT
For AB and CD---use Cv
and for BC and DA use Cp
It will come as 21/2. [Cv=fR/2 and Cp=(f+2)/2*R. Where f=degree of freedom which in this case=3]
Since efficiency=W/Q
=4/21

efficiency = work done / total heat supplied to the system
work done from graph is 2 p₀v_o clearly
heat is supplied in process AB and BC
during AB heat supplied ={(3p_ov₀-p₀v_o)/R}*3R/2
during BC heat supplied ={(6p_ov_o-3p_ov_o/R)}*5R/2
so total heat supplied is 21p_ov₀/2
therefore answer is (a) 4/21