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 TA=100K and TB=200K. 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 WAB=P∆V=nR∆T     (∵PV=nRT)       =nR(TB-TA)      =3×8.3×(200-100)=2490 JProcess CA is isochoric (constant volume). Hence work done in this process WCA = 0. Since the whole process ABCA is cyclic, the change in internal energy in the complete cycle is zero, i.e., ∆U=0. Now, from the first law of the thermodynamics, (Given Q = -2510 J)Q=∆U+W=∆U+WAB+WBC+WCA     -2510=0+2490+WBC+0     WBC=-2510-2490=-5000 JThe negative sign shows that the work is done by the gas.