Moderate

1 liter of a gas weighs 2 g at 300 K and 1 atm pressure. If the pressure is made 0.75 atm, at which of the following temperatures will 1 L of the same gas weigh 1 g?

Correct answer: A. 450 K

  • A. 450 K
  • B. 800 K
  • C. 600 K
  • D. 900 K

Explanation

So we know that, P₁V₁ / P₂V₂ = W₁T₁ / W₂T₂ Substituting in the above formula we get, => 1 × 1 / 0.75 × 1 = 2 × 300 / 1 × T₂ => 1 / 0.75 = 600 / T₂ Cross Multiplying we get, => T₂ × 1 = 600 × 0.75 => T₂ = 450 K

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About Ideal Gas Equation

The ideal gas equation, PV = nRT, relates pressure, volume, amount and absolute temperature when gas particles are assumed to have negligible volume and no intermolecular forces. Applications include finding molar mass, density or an unknown gas variable, with careful use of Kelvin temperature and consistent units, and comparison with real-gas behaviour.

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