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

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Given : a = \(\frac{ΔV_{max}}{2}=\frac{6-2}{2}×10^{-3}m^3\)  = 2 x 10−3 m3,

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

Cycles = 25

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

= πab = (3.142) (2 x 103) (5 x 105)

= 3.142 x 103J

Hence, the work done in 25 cycles

= (25) (3.142 x 103) = 7.855 x 104J

This is the work done in 25 cycles

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