Question

At 133.33 K, the RMS velocity of an ideal gas is \[ (M = 0.083 \text{ kg mol}^{-1}, R = 8.3 \text{ J mol}^{-1} \text{ K}^{-1}) \]

• A. \( 200 \) m s\(^{-1}\)
• B. \( 150 \) m s\(^{-1}\)
• C. \( 2000 \) m s\(^{-1}\)
• D. \( 400 \) m s\(^{-1}\)

Hint:

The RMS velocity of gas molecules is given by \( v_{\text{rms}} = \sqrt{\frac{3RT}{M}} \). Higher temperatures increase molecular speed, and lower molar masses lead to faster molecules.

Answer 1

The Correct Option is A

The formula for the RMS (Root Mean Square) velocity of an ideal gas is given by:

v_rms = √(3RT / M)

Where:

• R = 8.3 J mol^-1 K^-1 (universal gas constant)
• T = 133.33 K (temperature)
• M = 0.083 kg mol^-1 (molar mass of the gas)

Substitute the values into the formula:

v_rms = √(3 × 8.3 × 133.33 / 0.083)

Now, calculate:

v_rms = √(3 × 8.3 × 133.33 / 0.083) = √(3328.9398) ≈ 200 m s^-1

Thus, the RMS velocity of the ideal gas is 200 m s^-1.