Three moles of an ideal gas are taken through a cyclic process ABCA as shown on T-V diagram in Fig. The gas loses 2510 J of heat in the complete cycle. If ${\mathrm{T}}_{\mathrm{A}}=100\mathrm{K}\text{ and }{\mathrm{T}}_{\mathrm{B}}=200\mathrm{K}$. The work done by the gas during the process BC is….. (Take R = 8.3 JmolK^-1)

In the process AB, the volume V increases linearly with temperature T. hence AB is isobaric (constant pressure). Therefore, work done in this process is 

${\mathrm{W}}_{\mathrm{AB}}=\mathrm{P}∆\mathrm{V}=\mathrm{nR}∆\mathrm{T} (\because \mathrm{PV}=\mathrm{nRT})$

       $=\mathrm{nR}({\mathrm{T}}_{\mathrm{B}}-{\mathrm{T}}_{\mathrm{A}})$

      $=3\times 8.3\times (200-100)=2490\hspace{0.17em}\mathrm{J}$

Process CA is isochoric (constant volume). Hence work done in this process W_CA = 0. Since the whole process ABCA is cyclic, the change in internal energy in the complete cycle is zero, i.e., $∆\mathrm{U}=0$. Now, from the first law of the thermodynamics, (Given Q = -2510 J)

$\mathrm{Q}=∆\mathrm{U}+\mathrm{W}=∆\mathrm{U}+{\mathrm{W}}_{\mathrm{AB}}+{\mathrm{W}}_{\mathrm{BC}}+{\mathrm{W}}_{\mathrm{CA}}$

     $-2510=0+2490+{\mathrm{W}}_{\mathrm{BC}}+0$

     ${\mathrm{W}}_{\mathrm{BC}}=-2510-2490=-5000 \mathrm{J}$

The negative sign shows that the work is done by the gas.