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JEE MainPhysicsThermal Properties of Matter

A rigid cylinder of volume 67.2 L initially contains a mixture of helium and nitrogen gases at standard temperature and pressure. The total mass of the gas mixture is 36 g . The amount of heat required to raise the temperature of the gas mixture in the cylinder by 10.0^ C is (Given gas constant R=8.3 J K ⁻¹ mol ⁻¹ )

Options

  1. A456.5 J
  2. B373.5 J
  3. C622.5 J
  4. D705.5 J

Correct answer

A. 456.5 J

Step-by-step solution

At STP, the volume occupied by one mole of an ideal gas is 22.4 L . The total number of moles in the mixture is: n_ total = 67.2 22.4 = 3 mol Let n_ He and n_ N ₂ be the number of moles of helium and nitrogen, respectively. n_ He + n_ N ₂ = 3 The total mass of the mixture is 36 g . The molar mass of He is 4 g/mol and of N₂ is 28 g/mol . 4n_ He + 28n_ N ₂ = 36 Substituting n_ N ₂ = 3 - n_ He into the mass equation: 4n_ He + 28(3 - n_ He ) = 36 4n_ He + 84 - 28n_ He = 36 24n_ He = 48 n_ He = 2 mol Thus, n_ N ₂ = 1 mo

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