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

To find the root mean square (RMS) velocity of an ideal gas, we use the formula:

\[ v_{rms} = \sqrt{\frac{3RT}{M}} \]

where:

• \( v_{rms} \) is the RMS velocity.
• \( R \) is the universal gas constant \( 8.3 \text{ J mol}^{-1} \text{ K}^{-1} \).
• \( T \) is the temperature in Kelvin \( 133.33 \text{ K} \).
• \( M \) is the molar mass in kg/mol \( 0.083 \text{ kg mol}^{-1} \).

Substituting the given values into the formula:

\[ v_{rms} = \sqrt{\frac{3 \times 8.3 \times 133.33}{0.083}} \]

Calculate the numerator:

\[ 3 \times 8.3 \times 133.33 = 3320.493 \]

Then divide by the molar mass:

\[ \frac{3320.493}{0.083} = 40005.9 \]

Finally, take the square root to find the RMS velocity:

\[ v_{rms} = \sqrt{40005.9} \approx 200.01 \text{ m/s} \]

Therefore, the RMS velocity of the gas at 133.33 K is approximately \( 200 \text{ m s}^{-1} \).

Answer 2

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

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)

Substituting the values:

v_rms = √(3 × 8.3 × 133.33 / 0.083)

Calculating:

v_rms = √(3328.9398) ≈ 200 m s⁻¹

Conclusion:
The RMS velocity of the ideal gas is 200 m s⁻¹.