# NEET Physics Thermodynamics Questions Solved

A cyclic process for 1 mole of an ideal gas is shown in figure in the V-T, diagram. The work done in AB, BC and CA respectively

(1) $0,\text{\hspace{0.17em}}R{T}_{2}\mathrm{ln}\left(\frac{{V}_{1}}{{V}_{2}}\right)\text{\hspace{0.17em}},\text{\hspace{0.17em}}R\text{\hspace{0.17em}}\left({T}_{1}-{T}_{2}\right)$

(2) $R\left({T}_{1}-{T}_{2}\right),\text{\hspace{0.17em}}0,\text{\hspace{0.17em}}R{T}_{1}\mathrm{ln}\frac{{V}_{1}}{{V}_{2}}$

(3) $0,\text{\hspace{0.17em}}R{T}_{2}\mathrm{ln}\left(\frac{{V}_{2}}{{V}_{1}}\right)\text{\hspace{0.17em}},\text{\hspace{0.17em}}R\text{\hspace{0.17em}}\left({T}_{1}-{T}_{2}\right)$

(4) $0,\text{\hspace{0.17em}}R{T}_{2}\mathrm{ln}\left(\frac{{V}_{2}}{{V}_{1}}\right)\text{\hspace{0.17em}},\text{\hspace{0.17em}}R\text{\hspace{0.17em}}\left({T}_{2}-{T}_{1}\right)$

(3) Process AB is isochoric, ∴ ${W}_{AB}=P\text{\hspace{0.17em}}\Delta V=0$

Process BC is isothermal

${W}_{BC}=R{T}_{2}.\mathrm{ln}\left(\frac{{V}_{2}}{{V}_{1}}\right)$

Process CA is isobaric

${W}_{CA}=-\text{\hspace{0.17em}}P\Delta V=-\text{\hspace{0.17em}}R\Delta T=-\text{\hspace{0.17em}}R\left({T}_{1}-{T}_{2}\right)=R\left({T}_{2}-{T}_{1}\right)$

(Negative sign is taken because of compression)

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