Problem 1

In 1894 Lord Rayleigh found that nitrogen prepared chemically (mean mass in a fixed vessel: 2.2990 g) was lighter than \“atmospheric nitrogen\” freed of

\ce{O2}

(2.3102 g), at 15.0 \°C and

1.013 \times 10^{5}

Pa; the difference was later attributed to argon. (a) Calculate the vessel volume and the mole fraction of Ar in atmospheric nitrogen. (b) Ramsay isolated helium from cleveite; it emits the D

_3

line at 587.7 nm. Calculate the photon energy and identify the corresponding transition in the He orbital diagram (answer: the highest-energy allowed transition, 3d \→ 2p) and the nuclear process that produces He in uranium minerals (

\alpha

-decay) and Ar in rocks (electron capture/positron decay of

\ce{^{40}K}

). (c) An unknown gas at 15.0 \°C and atmospheric pressure has density

0.850 \pm 0.005\ \text{kg m}^{-3}

; sound in it has

\lambda = 0.116

m at

f = 3520

Hz, with

v_s = f\lambda = \sqrt{\gamma RT/M}

. Find

M

and

\gamma

and identify the gas among HCl, HF, Ne, Ar.

Step 2 of 4: Argon mole fraction

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\dfrac{28.02(1-x) + 39.95\,x}{28.02} = \dfrac{2.3102}{2.2990} \Rightarrow x = \dfrac{28.02}{39.95-28.02}\,\dfrac{2.3102-2.2990}{2.2990} = 1.14 \times 10^{-2}

Analysis

The same vessel contains the same number of moles, so the mass ratio equals the ratio of mean molar masses. Solving for the Ar fraction in a $\ce{N2}$ \–Ar mixture gives $x = 1.14 \times 10^{-2}$ (1.14 %).

Common pitfall. Equal

V

p

T

means equal mole numbers — compare mean molar masses, not densities directly.