From ideal gas equation, PV=nRT
⇒PV=NANRT, where N= total number of molecules
⇒P=VN(NAR)T⇒P=VNkT
⇒1.38×(1.01×105)=(2×1025)×(1.38×10−23)×T
⇒1.01×105=2×102×T
T=21.01×103≈500K
JEE Main 2024 — Physics Thermodynamics
The temperature of a gas having 2.0×1025 molecules per cubic meter at 1.38atm (Given, k=1.38×10−23JK−1) is :
Held on 29 Jan 2024 · Verified 6 Jul 2026.
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