Q2 · Kinetic Theory

Problem

2 mol He (4 amu) + 1 mol Ar (40 amu) at 300 K. Find $v_{\mathrm{rms}}$,He / $v_{\mathrm{rms}}$,Ar.

Options

Option	Value
(A)	$0.32$
(B)	$0.45$
(C)	$2.24$
(D)	$3.16$

Key insight: RMS speed depends only on temperature and molar mass — not on moles or pressure for an ideal gas.

Setup

flowchart LR
  T["Same T = 300 K"] --> He["He: M = 4 g/mol"]
  T --> Ar["Ar: M = 40 g/mol"]
  He --> R["v_rms ∝ 1/√M"]
  Ar --> R
  R --> Ratio["v_He / v_Ar = √(M_Ar/M_He) = √10"]

1. RMS speed formula

For ideal gas at temperature T, molar mass M.

$$v_{rms} = \sqrt{\frac{3RT}{M}}$$

2. Ratio at same T

Temperature cancels.

$$\frac{v_{He}}{v_{Ar}} = \sqrt{\frac{M_{Ar}}{M_{He}}}$$

3. Calculate

√(40/4) = √10 ≈ 3.16.

$$\sqrt{10} \approx 3.16 \;\Rightarrow\; (D)$$

Deep dive

The Maxwell-Boltzmann distribution gives v_rms = √(3RT/M). Both gases share the same T = 300 K, so the speed ratio is √(M_Ar/M_He) = √(40/4) = √10 ≈ 3.16. Helium is faster because it’s lighter.

Common pitfalls

• Inverting the ratio (getting 1/3.16)
• Using number of moles in the ratio
• Confusing rms with most probable speed

Hints

• v_rms = √(3RT/M) — same T, ratio depends on molar mass.

Answer

(D)