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JEE Main 202431 JanuaryMorning Shift
Exam
JEE Main
Year
2024
Date
31 January
Shift
Morning Shift
Subject
Physics
Chapter
Thermodynamics

JEE Main 2024 · 31 January · Morning Shift · Physics · Thermodynamics

The given figure represents two isobaric processes for the same...

The given figure represents two isobaric processes for the same mass of an ideal gas, then

The given figure represents two isobaric processes for the same mass of an ideal gas, then

Options

  1. AP2≥P1{P}_{2} \ge {P}_{1}
  2. BP2>P1{P}_{2} > {P}_{1}
  3. CP1=P2{P}_{1} = {P}_{2}
  4. DP1>P2{P}_{1} > {P}_{2}

Correct answer

D. P1>P2{P}_{1} > {P}_{2}

Step-by-step solution

The given figure represents two isobaric processes for the same mass of an ideal gas, then

As we know, PV=nRTPV = nRT

⇒V=(nRP)T⇒ V = \left(\dfrac{nR}{P}\right)T

Therefore, slope =nRP= \dfrac{nR}{P}

Or slope ∝1P∝ \dfrac{1}{P}

As (Slope)2>(Slope)1{\left(Slope\right)}_{2} > {\left(Slope\right)}_{1}

⇒P2<P1⇒ {P}_{2} < {P}_{1}

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