Two vessels are labeled A and B. Vessel A contains NH3 gas at 70°C, and vessel B contains He gas at the same temperature. If the average kinetic energy of NH3 is 7.1x10^(-21) J/molecule, calculate the mean-square speed of He atoms in m^2/s^2.

Respuesta :

Answer:

  • v² = 2.14 × 10⁶ m²/s²

Explanation:

1) Since the average kinetic energy of a gas is only dependent on its temperature, you conclude that, at the same temperature of 70°C, the average kinetic energy of He atoms is the same that the average kinetic energy of NH₃ molecules: 7.1 × 10⁻²¹ J / molecule.

2) Now, you can use the equation that relates the average kinetic energy with the mean-square speed:

  • KE avg = (1/2)mv²

Where:

  • Ke avg is the average kinetic energy,
  • m is the mass, and
  • v² is the mean-square speed.

You need to use the atomic mass of helium to find the mass (average) of one atom of helum:

  • Atomic mass of helium: 4.003 g / mol
  • average mass of 1 atom:

        m = 4.003 g/ mol / (6.022 × 10²³ atoms/mol)

        m = 6.647 × 10⁻²⁴ g = 6.647 10 ⁻²⁷ kg / atom

3) Solve for v², substitute, and compute:

Note: since He is a monoatomic molecule, kg/atom = kg/molecule.

  • KE avg = (1/2)mv² ⇒ v² = 2KE avg / m

  • v² = 2 × 7.1 × 10⁻²¹ J/ molecule /  (6.647 × 10 ⁻²⁷ kg / molecule)

            = 2.14 × 10⁶ m²/s² ← answer

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