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JEE Main 2026 (2 Apr, Shift 2) · Thermodynamic Processes

Thermodynamic Processes question from JEE Main 2026 (2 Apr, Shift 2)

Q25JEE Main 2026 · 2 APR · SHIFT 2Medium

Thermodynamic Processes

5 moles of unknown gas is heated at constant volume from 10°C to 20°C. The molar specific heat of this gas at constant pressure cp=8c_p = 8 cal/mol.°C and R=8.36R = 8.36 J/mol.°C. The change in the internal energy of the gas is __________ calorie.

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

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Answer: 300

  1. Convert RR to calories: R=8.364.18=2 cal/mol∘CR = \frac{8.36}{4.18} = 2\text{ cal/mol}^\circ\text{C}.
  2. Cv=Cp−R=8−2=6 cal/mol∘CC_v = C_p - R = 8 - 2 = 6\text{ cal/mol}^\circ\text{C}
  3. ΔU=nCvΔT=5×6×10=300 cal\Delta U = nC_v\Delta T = 5 \times 6 \times 10 = 300\text{ cal}
  4. So the answer is 300 cal.
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