Question

Helium (He) is the lightest noble gas component of air, and xenon (Xe) is the heaviest. Perform the following calculations, using R = 8.314 J/(mol·K) and ℳ in kg/mol. (a) Find the rms speed of He in winter (0.°C) and in summer (30.°C). Enter your answers in scientific notation.

Answer 1

Answer:

$\large \boxed{\text{(a) } 1.30 \times 10^{3}\text{ m/s}; \text{(b) }1.37 \times 10^{3}\text{ m/s}}$

Explanation:

$v_{\text{rms}} = \sqrt{\dfrac{3RT}{M}$

(a) Speed in winter

T = (0. + 273.15) K = 273.15 K

$v_{\text{rms}} = \sqrt{\dfrac{3\times 8.314\text{ J}\cdot\text{K}^{-1} \text{mol}^{-1} \times 273.15 \text{ K}}{4.013 \times 10^{-3} \text{ kg}\cdot \text{mol}^{-1}}}\\\\=\sqrt{1.846\times 10^{6} \text{ (m/s)}^{2}} = 1.30 \times 10^{3}\text{ m/s}\\\text{The rms speed in winter is \large \boxed{\mathbf{1.30 \times 10^{3}}\textbf{ m/s}} }$

(b) Speed in summer

T = (30. + 273.15) K = 303.15 K

$v_{\text{rms}} = \sqrt{\dfrac{3\times 8.314\text{ J}\cdot\text{K}^{-1} \text{mol}^{-1} \times 303.15 \text{ K}}{4.013 \times 10^{-3} \text{ kg}\cdot \text{mol}^{-1}}}\\\\=\sqrt{1.884\times 10^{6} \text{ (m/s)}^{2}} = 1.37 \times 10^{3}\text{ m/s}\\\text{The rms speed in summer is \large \boxed{\mathbf{1.37 \times 10^{3}}\textbf{ m/s}} }$