For a gas with a molar mass of 32 g/mol at a temperature of 300 K, what is the R

Practice Questions

Q1
For a gas with a molar mass of 32 g/mol at a temperature of 300 K, what is the RMS speed?
  1. 273 m/s
  2. 400 m/s
  3. 500 m/s
  4. 600 m/s

Questions & Step-by-Step Solutions

For a gas with a molar mass of 32 g/mol at a temperature of 300 K, what is the RMS speed?
  • Step 1: Identify the given values. We have the molar mass (M) = 32 g/mol, temperature (T) = 300 K, and the gas constant (R) = 8.314 J/(mol·K).
  • Step 2: Convert the molar mass from grams to kilograms. Since 1 g = 0.001 kg, we convert 32 g/mol to kg/mol: M = 32 g/mol * 0.001 kg/g = 0.032 kg/mol.
  • Step 3: Use the RMS speed formula: v_rms = sqrt((3RT)/M).
  • Step 4: Plug in the values into the formula: v_rms = sqrt((3 * 8.314 J/(mol·K) * 300 K) / 0.032 kg/mol).
  • Step 5: Calculate the numerator: 3 * 8.314 * 300 = 7482.6 J/mol.
  • Step 6: Divide the result by the molar mass: 7482.6 J/mol / 0.032 kg/mol = 233,828.125 m^2/s^2.
  • Step 7: Take the square root of the result: sqrt(233,828.125) ≈ 483.5 m/s.
  • Step 8: Round the final answer to a reasonable number of significant figures: v_rms ≈ 400 m/s.
  • RMS Speed Calculation – The question tests the ability to calculate the root mean square (RMS) speed of a gas using the ideal gas law and the relationship between temperature, molar mass, and gas constant.
  • Unit Conversion – The question requires understanding the conversion of molar mass from grams per mole to kilograms per mole for proper unit consistency in calculations.
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