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NEET PHYSICSThermodynamicsMedium

Question

Figure below shows two paths that may be taken by a gas to go from a state AA to a state CC. In process ABAB, 400 J400\text{ J} of heat is added to the system and in process BCBC, 100 J100\text{ J} of heat is added to the system. The heat absorbed by the system in the process ACAC will be:

A

380 J380\text{ J}

B

500 J500\text{ J}

C

460 J460\text{ J}

D

300 J300\text{ J}

Step-by-Step Solution

According to the first law of thermodynamics, ΔU=QW\Delta U = Q - W. For the path ABCABC, the total heat added is QABC=QAB+QBC=400 J+100 J=500 JQ_{ABC} = Q_{AB} + Q_{BC} = 400\text{ J} + 100\text{ J} = 500\text{ J}. From the standard PP-VV diagram given in this PYQ, the coordinates are: A(2×103 m3,2×104 Pa)A \equiv (2 \times 10^{-3}\text{ m}^3, 2 \times 10^4\text{ Pa}) B(2×103 m3,6×104 Pa)B \equiv (2 \times 10^{-3}\text{ m}^3, 6 \times 10^4\text{ Pa}) C(4×103 m3,6×104 Pa)C \equiv (4 \times 10^{-3}\text{ m}^3, 6 \times 10^4\text{ Pa}) Work done in path ABAB (isochoric process, ΔV=0\Delta V = 0) is WAB=0W_{AB} = 0. Work done in path BCBC (isobaric process) is WBC=PB(VCVB)=6×104×(42)×103=120 JW_{BC} = P_B(V_C - V_B) = 6 \times 10^4 \times (4 - 2) \times 10^{-3} = 120\text{ J}. Total work done along path ABCABC is WABC=WAB+WBC=120 JW_{ABC} = W_{AB} + W_{BC} = 120\text{ J}. Change in internal energy is ΔU=QABCWABC=500 J120 J=380 J\Delta U = Q_{ABC} - W_{ABC} = 500\text{ J} - 120\text{ J} = 380\text{ J}. Since internal energy is a state function, ΔU\Delta U for path ACAC is also 380 J380\text{ J}. Work done along path ACAC is the area of the trapezium under the curve ACAC: WAC=12(PA+PC)(VCVA)=12(2×104+6×104)(42)×103=12(8×104)(2×103)=80 JW_{AC} = \frac{1}{2} (P_A + P_C)(V_C - V_A) = \frac{1}{2}(2 \times 10^4 + 6 \times 10^4)(4 - 2) \times 10^{-3} = \frac{1}{2}(8 \times 10^4)(2 \times 10^{-3}) = 80\text{ J}. Therefore, heat absorbed in path ACAC is QAC=ΔU+WAC=380 J+80 J=460 JQ_{AC} = \Delta U + W_{AC} = 380\text{ J} + 80\text{ J} = 460\text{ J}.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from Thermodynamics. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSThermodynamicsfigureprocesssystemprocesssystem

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