Question

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.

Answer 1

• v² = 2.14 × 10⁶ m²/s²

Step-by-step 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