1.0 mol of a perfect monatomic gas is put through the cycle shown in the figure. The total work (in J) done during the cycle is

(use 1 L-bar = 100 J, R = 8.3 J K⁻¹ mol⁻¹ = 0.083 L-bar K⁻¹ mol^−1, ln 2 = 0.7)

Concept:

It follows cyclic process as the initial and final stages are equal. So the equation goes as follows,

Here  in a cyclic process,

So,  (Work done by the system)

How to calculate work done in cyclic process.

Explanation:

(i)  Path AB , Isobaric expansion process as P= const.

= -2 (23.24-46.48) =-46.48

(ii) Path BC , Isothermal

at B, P₁=2, V₁=46.48=0

at C, P₁= 1, V₂= 92.96

so here  , T is constant. So it's an isothermal process.

   =

        = 

       =  

      = -65.072

(iii) path CD, Isobaric process

         = 69.62

(iv) path DA, V=constant, Isochoric process

W_DA=0

Total work done W_Total= -41.9100J=-4190J

Conclusion:

The total work (in J) done during the cycle is -4190J