For isobaric process and diatomic gas we can write, ΔQ=nCPΔT=735J
⇒n27RΔT=735J⇒nRΔT=210
Now internal energy will be,
ΔU=nCVΔT=n25RΔT=25×210
=525J
JEE Main 2023 — Physics Thermodynamics
Heat energy of 735J is given to a diatomic gas allowing the gas to expand at constant pressure. Each gas molecule rotates around an internal axis but do not oscillate. The increase in the internal energy of the gas will be:
Held on 31 Jan 2023 · Verified 6 Jul 2026.
525J
441J
572J
735J
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