Answer:
[tex]\large \boxed{\text{(a) } 1.30 \times 10^{3}\text{ m/s}; \text{(b) }1.37 \times 10^{3}\text{ m/s}}[/tex]
Explanation:
[tex]v_{\text{rms}} = \sqrt{\dfrac{3RT}{M}[/tex]
(a) Speed in winter
T = (0. + 273.15) K = 273.15 K
[tex]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}}$}[/tex]
(b) Speed in summer
T = (30. + 273.15) K = 303.15 K
[tex]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}}$}[/tex]
