The ideal gas temperature increased from t1 = 100 to t2 = 300. In this case, the mean square velocity

The ideal gas temperature increased from t1 = 100 to t2 = 300. In this case, the mean square velocity of the gas molecules is?

Problem data: Tn (initial temperature of the taken ideal gas) = 100 ºС (absolute temperature Tn = 373 K); Тк (final temperature) = 300 ºС (absolute temperature Тк = 573 К).

We calculate the desired change in the mean square velocity of the molecules of the taken ideal gas from the ratio: k = Vav2 / Vav1 = √ (3 * R * Tc / M) / √ (3 * R * Tn / M) = √ (Tc / Tn).

Let’s calculate: k = √ (573/373) = 1.24 p.

Answer: The mean square velocity of the molecules of the taken ideal gas increased by 1.24 times.