First slide
Thermodynamic processes

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=PV=nRT     (PV=nRT)



Process 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)



     WBC=-2510-2490=-5000 J

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

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