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JEE Main 202523 JanuaryMorning Shift
Exam
JEE Main
Year
2025
Date
23 January
Shift
Morning Shift
Subject
Physics
Chapter
Thermodynamics

JEE Main 2025 · 23 January · Morning Shift · Physics · Thermodynamics

Match the List-I with List-II \[ |l|p 4cm | List-I List-II A....

Match the List-I with List-II\text{Match the List-I with List-II}<table class="qx-match-list qx-match-table" border="1" cellpadding="6" cellspacing="0"><tr><td>|l|p{4cm}|} List-I</td><td></td><td>List-II</td><td></td></tr><tr><td>A.</td><td>Pressure varies inversely with volume of an ideal gas.</td><td>I.</td><td>Adiabatic process</td></tr><tr><td>B.</td><td>Heat absorbed goes partly to increase internal energy and partly to do work.</td><td>II.</td><td>Isochoric process</td></tr><tr><td>C.</td><td>Heat is neither absorbed nor released by a system.</td><td>III.</td><td>Isothermal process</td></tr><tr><td>D.</td><td>No work is done on or by a gas.</td><td>IV.</td><td>Isobaric process</td></tr></table> Choose the correct answer from the options given below:\text{Choose the correct answer from the options given below:}

Options

  1. AA-III, B-IV, C-I, D-II
  2. BA-I, B-IV, C-II, D-III
  3. CA-III, B-I, C-IV, D-II
  4. DA-I, B-III, C-II, D-IV

Correct answer

A. A-III, B-IV, C-I, D-II

Step-by-step solution

A→P∝1V⇒PV= constant ⇒nRT= const. ⇒T= const. \begin{aligned} & \mathrm{A} \rightarrow \mathrm{P} \propto \frac{1}{\mathrm{ V}} \\ & \Rightarrow \mathrm{PV}=\text{ constant } \\ & \Rightarrow \mathrm{nRT}=\text{ const. } \Rightarrow \mathrm{T}=\text{ const. } \end{aligned} Hence Isothermal III B \rightarrowIVW≠0,ΔU≠0,ΔQ≠0IV \mathrm{W} \neq 0, \Delta \mathrm{U} \neq 0, \Delta \mathrm{Q} \neq 0 [only isobaric] C\mathrm{C} \rightarrow I\mathrm{I} \Delta Q=\mathrm{Q}=0AdiabaticD→Adiabatic \mathrm{D} \rightarrow II w =0=0 Isochoric III, IV, I, II
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