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,

\(\Delta E=Q+W\)

Here \( \Delta E= 0\) in a cyclic process,

So, \(Q=-W\) (Work done by the system)

How to calculate work done in cyclic process.

\(Total\: work \: done \: in\:cyclic\:process = \sum work\: done\: single\: steps \)

Explanation:

\(W_{Total}= W_{AB}+ W_{BC} + W_{CD} + W_{DA}\)

(i)  Path A\(\rightarrow \)B , Isobaric expansion process as P= const.

\(W_{AB}=-P\Delta V\)= -2 (23.24-46.48) =-46.48

(ii) Path B\(\rightarrow \)C , Isothermal

at B, P₁=2, V₁=46.48=0

at C, P₁= 1, V₂= 92.96

so here \(P_{1}V_{1}=P_{2}V_{2}\) , T is constant. So it's an isothermal process.

\(PV=nRT\)

\(T=\frac{PV}{nR}\)

   =\(\frac{2\times 46.48}{1\times 0.083}=1120\)

\(W_{BC}= -nRT \: ln\frac{V_2}{V_1}\)

        = \(-1\times 0.083\times 1120\times ln\frac{92.96}{23.24}\)

       =  \(-0.083\times 1120 \times ln2\)

      = -65.072

(iii) path C\(\rightarrow \)D, Isobaric process

\(W_{CD} = -P\Delta V\)

         = 69.62

(iv) path D\(\rightarrow \)A, V=constant, Isochoric process

W_DA=0

Total work done W_Total= -41.9\(\times\)100J=-4190J

Conclusion:

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