# Question #eacef

Mar 4, 2017

$\approx 445.3 \setminus m {s}^{- 1}$

#### Explanation:

The most probable velocity follows from the Maxwell-Boltzman distribution and is:

$v = \setminus \sqrt{\frac{2 R T}{M}} \approx 445.3 \setminus m {s}^{- 1}$

With:

• gas constant $R = 8.314$

• molar mass $M = 109 g$

Rest of question is missing information.