# NEET Physics Thermodynamics Questions Solved

A thermodynamic process is shown in the figure. The pressures and volumes corresponding to some points in the figure are : ${P}_{A}=3×{10}^{4}Pa,\text{\hspace{0.17em}}{P}_{B}=8×{10}^{4}Pa$ and ${V}_{A}=2×{10}^{-3}{m}^{3},\text{\hspace{0.17em}}{V}_{D}=5×{10}^{-3}{m}^{3}$. In process AB, 600 J of heat is added to the system and in process BC, 200 J of heat is added to the system. The change in internal energy of the system in process AC would be

(1) 560 J

(2) 800 J

(3) 600 J

(4) 640 J

(1) By adjoining graph ${W}_{AB}=0$ and

${W}_{BC}=8×{10}^{4}\left[5-2\right]×{10}^{-3}=240\text{\hspace{0.17em}}J$

${W}_{AC}={W}_{AB}+{W}_{BC}=0+240=240\text{\hspace{0.17em}}J$

Now, $\Delta {Q}_{AC}=\Delta {Q}_{AB}+\Delta {Q}_{BC}=600+200=800\text{\hspace{0.17em}}J$

From FLOT $\Delta {Q}_{AC}=\Delta {U}_{AC}+\Delta {W}_{AC}$

$800=\Delta {U}_{AC}+240$$\Delta {U}_{AC}=560\text{\hspace{0.17em}}J.$

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