Questions
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 . The work done by the gas during the process BC is….. (Take R = 8.3 JmolK-1)
detailed solution
Correct option is B
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.Talk to our academic expert!
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