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JEE Main 2026 (24 Jan, Shift 2) · Gas Laws & Ideal Gas Equation

Gas Laws & Ideal Gas Equation question from JEE Main 2026 (24 Jan, Shift 2)

Q9JEE Main 2026 · 24 JAN · SHIFT 2Medium

Gas Laws & Ideal Gas Equation

When 300 J of heat given to an ideal gas with Cp=72RC_p = \frac{7}{2}R its temperature raises from 20∘C20^\circ\text{C} to 50∘C50^\circ\text{C} keeping its volume constant. The mass of the gas is (approximately) ______ g. (R=8.314 J/mol.K)(R = 8.314\text{ J/mol.K})

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

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Answer: 481

  1. Volume constant, so Q=nCVΔTQ = nC_V\Delta T with CV=Cp−R=52RC_V = C_p - R = \frac{5}{2}R.
  2. ΔT=50−20=30 K\Delta T = 50 - 20 = 30\text{ K}.
  3. n=30052×8.314×30=300623.6≈0.481n = \frac{300}{\frac{5}{2} \times 8.314 \times 30} = \frac{300}{623.6} \approx 0.481 mol.
  4. Expressed as asked (in units of 10−310^{-3}), the answer is 481.
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