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JEE Mains · Physics · STD 11 - 11. thermodynamics

किसी गैस को दो प्रक्रमों \(A\) तथा \(B\) से गुजारा जाता है जेसा कि चित्र में दिखाया गया है। यदि \(\Delta Q _{ A }\) तथा \(\Delta Q _{ B }\) तंत्र द्वारा शोषित ऊष्मा है तथा \(\Delta U _{ A }\) और \(\Delta Q _{ B }\) आंतरिक ऊर्जा में परिवर्तन है, तो सही विकल्प चुनें

  1. A \(\Delta {Q_A}\, = \,\Delta {Q_B}\,;\,\Delta {U_A} = \Delta {U_B}\)
  2. B \(\Delta {Q_A}\, > \,\Delta {Q_B}\,;\,\Delta {U_A} = \Delta {U_B}\)
  3. C \(\Delta {Q_A}\, < \,\Delta {Q_B}\,;\,\Delta {U_A} < \Delta {U_B}\)
  4. D \(\Delta {Q_A}\, > \,\Delta {Q_B}\,;\,\Delta {U_A} > \Delta {U_B}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\Delta {Q_A}\, > \,\Delta {Q_B}\,;\,\Delta {U_A} = \Delta {U_B}\)

Step-by-step Solution

Detailed explanation

Initial and final states for both the processes are same, \(\therefore \Delta \mathrm{U}_{\mathrm{A}}=\Delta \mathrm{U}_{\mathrm{B}}\) Work done during process \(A\) is greater than in process \(B\). Because area is more By First law of thermodynamics \(\Delta Q=\Delta U+W\)…
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