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

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

Step-by-step explanation:

`v_{\text{rms}} = \sqrt{(3RT)/(M)`

(a) Speed in winter

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

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

(b) Speed in summer

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

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