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JEE Main 2026 (21 Jan, Shift 1) · Thermodynamic Processes

Thermodynamic Processes question from JEE Main 2026 (21 Jan, Shift 1)

Q26JEE Main 2026 · 21 JAN · SHIFT 1Medium

Thermodynamic Processes

10 mole of oxygen is heated at constant volume from 30∘C30^\circ\text{C} to 40∘C40^\circ\text{C}. The change in the internal energy of the gas is ______ cal. (The molecular specific heat of oxygen at constant pressure, CP=7 cal/mol. ∘CC_P = 7\text{ cal/mol.}\,{}^\circ\text{C} and R=2 cal./mol. ∘C.)R = 2\text{ cal./mol.}\,{}^\circ\text{C.})

Numerical answer type. Work it out, then open the answer.

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Answer: 500

  1. Cv=Cp−R=7−2=5 cal/mol∘CC_v = C_p - R = 7 - 2 = 5\text{ cal/mol}^\circ\text{C}
  2. ΔU=nCvΔT=10×5×10=500 cal\Delta U = nC_v\Delta T = 10 \times 5 \times 10 = 500\text{ cal}
  3. So the answer is 500 cal.
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