If the RMS speed of a gas is 300 m/s and its molar mass is 28 g/mol, what is the

₹0.0

Question: If the RMS speed of a gas is 300 m/s and its molar mass is 28 g/mol, what is the temperature of the gas?

Options:

Correct Answer: 600 K

Solution:

Using the formula v_rms = sqrt((3RT)/M), we can rearrange to find T = (v_rms^2 * M)/(3R). Plugging in the values gives T = 600 K.

If the RMS speed of a gas is 300 m/s and its molar mass is 28 g/mol, what is the temperature of the gas?

If the RMS speed of a gas is 300 m/s and its molar mass is 28 g/mol, what is the temperature of the gas?

Step 1: Write down the formula for RMS speed: v_rms = sqrt((3RT)/M).
Step 2: Rearrange the formula to solve for temperature (T): T = (v_rms^2 * M) / (3R).
Step 3: Identify the values needed: v_rms = 300 m/s, M = 28 g/mol, and R = 8.314 J/(mol·K).
Step 4: Convert the molar mass from grams to kilograms: M = 28 g/mol = 0.028 kg/mol.
Step 5: Calculate v_rms squared: v_rms^2 = (300 m/s)² = 90000 m²/s².
Step 6: Plug the values into the rearranged formula: T = (90000 m²/s² * 0.028 kg/mol) / (3 * 8.314 J/(mol·K)).
Step 7: Calculate the numerator: 90000 * 0.028 = 2520.
Step 8: Calculate the denominator: 3 * 8.314 = 24.942.
Step 9: Divide the numerator by the denominator: T = 2520 / 24.942 ≈ 101.1 K.
Step 10: Check the calculations to ensure accuracy.

RMS Speed and Temperature Relationship – The relationship between the root mean square speed of gas molecules, their molar mass, and temperature is described by the equation v_rms = sqrt((3RT)/M).
Gas Laws – Understanding the ideal gas law and how temperature, pressure, volume, and number of moles relate to each other.