**A hypothetical thermodynamic cycle is shown in the figure. Calculate the work done in 25 cycles.**

**Given** : a = \(\frac{ΔV_{max}}{2}=\frac{6-2}{2}×10^{-3}m^3\) = 2 x 10^{−3} m^{3},

b = \(\frac{ΔP_{max}}{2}=\frac{11-1}{2}×10^{5}Pa\) = 5 x 10^{5 }Pa

Cycles = 25

The work done in one cycle, \(\oint PdV\)

= πab = (3.142) (2 x 10^{−}^{3}) (5 x 10^{5})

= 3.142 x 10^{3}J

Hence, the work done in 25 cycles

= (25) (3.142 x 10^{3}) = **7.855 x 10 ^{4}J**

This is the work done in 25 cycles