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

₹0.0

Question: For a gas with a molar mass of 32 g/mol at a temperature of 300 K, what is the RMS speed?

Options:

Correct Answer: 400 m/s

Solution:

Using the formula v_rms = sqrt((3RT)/M), where R = 8.314 J/(mol·K), M = 0.032 kg/mol, and T = 300 K, we find v_rms ≈ 400 m/s.

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

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.