Since its given that volume is constant, P∝T.
Ti=50°C=323 K
Tf=2×50=100°C=373 K
PiPf=TiTf
3.23 kPaPf=323373
Pf=3.23×323373=3.73 kPa =3730 Pa
JEE Main 2026 — Physics Thermodynamics
A gas of certain mass filled in a closed cylinder at a pressure of 3.23 kPa has temperature 50∘C. The gas is now heated to double its temperature. The modified pressure is ____ Pa. Note: Volume is constant.
Held on 24 Jan 2026 · Verified 6 Jul 2026.
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