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 4 of 4: Identify the unknown gas

← Previous Next →

Intuition

Molar mass alone cannot distinguish HF from Ne; the heat-capacity ratio $\gamma = C_p/C_V$ carries the molecular structure information.

M = \dfrac{\rho RT}{p} = \dfrac{0.850 \times 8.314 \times 288.15}{1.013 \times 10^{5}} = 2.01 \times 10^{-2}\ \text{kg mol}^{-1};\quad \gamma = \dfrac{M(f\lambda)^2}{RT} = \dfrac{2.01 \times 10^{-2} \times (3520 \times 0.116)^2}{8.314 \times 288.15} = 1.40

Analysis

Combining $\rho = pM/RT$ gives $M = 20.1$ g mol $^{-1}$ , compatible with HF (20.01) or Ne (20.18). The measured $\gamma = 1.40 \neq 5/3$ rules out a monatomic gas, so the gas is HF.