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 1 of 4: Vessel volume from pure nitrogen

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n(\ce{N2}) = \dfrac{2.2990}{28.02} = 8.205 \times 10^{-2}\ \text{mol};\quad V = \dfrac{nRT}{p} = \dfrac{8.205 \times 10^{-2} \times 8.314 \times 288.15}{1.013 \times 10^{5}} = 1.940 \times 10^{-3}\ \text{m}^3

Analysis

Chemical nitrogen is pure $\ce{N2}$ ( $M = 28.02$ g mol $^{-1}$ ); the ideal gas law at $T = 288.15$ K gives $V = 1.940 \times 10^{-3}$ m $^3$ (about 2 dm $^3$ ).