Here are the solutions to the problems. 1. Given an expression relating the Celsius scale and Kelvin scale. Write an expression for the root mean square speed of a gas. Relationship between Celsius and Kelvin scales: The temperature in Kelvin ($T_K$) is related to the temperature in Celsius ($T_C$) by the following expression: $$T_K = T_C + 273.15$$ Expression for the root mean square (RMS) speed of a gas: The root mean square speed ($v_{rms}$) of gas molecules is given by the formula: $$v_{rms} = \sqrt{\frac{3RT}{M}}$$ where: $R$ is the ideal gas constant ($8.314 \text{ J mol}^{-1} \text{ K}^{-1}$) $T$ is the absolute temperature in Kelvin $M$ is the molar mass of the gas in $\text{kg mol}^{-1}$ 2. Calculate the root mean square speed of hydrogen molecule at 300K (molar mass of hydrogen is 2.016 g/mol, R = 8.314 J mol⁻¹ K⁻¹). Step 1: Identify the given values and convert units if necessary. Temperature, $T = 300 \text{ K}$ Molar mass of hydrogen (H₂), $M = 2.016 \text{ g/mol}$ Ideal gas constant, $R = 8.314 \text{ J mol}^{-1} \text{ K}^{-1}$ Convert the molar mass from g/mol to kg/mol: $$M = 2.016 \text{ g/mol} \times \frac{1 \text{ kg}}{1000 \text{ g}} = 0.002016 \text{ kg/mol}$$ Step 2: Use the formula for root mean square speed. $$v_{rms} = \sqrt{\frac{3RT}{M}}$$ Step 3: Substitute the values into the formula and calculate. $$v_{rms} = \sqrt{\frac{3 \times 8.314 \text{ J mol}^{-1} \text{ K}^{-1} \times 300 \text{ K}}{0.002016 \text{ kg mol}^{-1}}}$$ $$v_{rms} = \sqrt{\frac{7482.6 \text{ J mol}^{-1}}{0.002016 \text{ kg mol}^{-1}}}$$ Since $1 \text{ J} = 1 \text{ kg m}^2 \text{ s}^{-2}$: $$v_{rms} = \sqrt{\frac{7482.6 \text{ kg m}^2 \text{ s}^{-2}}{0.002016 \text{ kg}}}$$ $$v_{rms} = \sqrt{3711607.142857 \text{ m}^2 \text{ s}^{-2}}$$ $$v_{rms} \approx 1926.558 \text{ m/s}$$ Step 4: Round the answer to an appropriate number of significant figures (e.g., 4 significant figures). $$v_{rms} \approx 1927 \text{ m/s}$$ The root mean square speed of hydrogen molecules at 300 K is: $$\boxed{1927 \text{ m/s}}$$