Calculate the root mean square velocity of gaseous argon atoms at 26 ∘c.

1 answer:

5 0

By substitution in this formula :
V= √3RTM
when R = 8.31451 J.mol^-1K^-1
T is the tempreture in K = 26 °C + 273 = 299 k
M is the molar mass of argon Kg= 39.9 g = 39.9/1000= 0.0399 Kg
∴V = √3*8.31451 * 299 / 0.0399
∴ V = 432 m/s

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