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NEET PHYSICSThermodynamicsMedium

Question

During an isothermal expansion, a confined ideal gas does 150 J-150\text{ J} of work against its surrounding. This implies that:

A

300 J300\text{ J} of heat has been added to the gas.

B

no heat is transferred because the process is isothermal.

C

150 J150\text{ J} of heat has been added to the gas.

D

150 J150\text{ J} of heat has been removed from the gas.

Step-by-Step Solution

According to the First Law of Thermodynamics, the relationship between heat (QQ), internal energy (DeltaU\\Delta U), and work (WW) is given by Q=DeltaU+WQ = \\Delta U + W (using the Physics sign convention where WW is work done by the gas).

  1. Isothermal Process: Since the process is isothermal (constant temperature) and involves an ideal gas, the internal energy change is zero (DeltaU=0\\Delta U = 0).
  2. First Law Application: Substituting DeltaU=0\\Delta U = 0 into the First Law equation gives Q=WQ = W.
  3. Calculation: The problem states the gas does 150 J-150\text{ J} of work against its surroundings. Thus, W=150 JW = -150\text{ J}. Q=150 JQ = -150\text{ J} The negative sign for heat (QQ) indicates that heat is released or removed from the system.

Therefore, 150 J150\text{ J} of heat has been removed from the gas.

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.

PHYSICSThermodynamicsduringisothermalexpansionconfinedagainst

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