At what temperature will the root-mean-square velocity of nitrogen molecules be equal

At what temperature will the root-mean-square velocity of nitrogen molecules be equal to the first cosmic velocity (V1 = 7900m / s)?

Given: V (mean square velocity of the taken nitrogen molecules) = 7900 m / s.

Constants: M (molar mass of nitrogen) = 0.028 kg / mol; R (universal gas constant) = 8.31 J / (mol * K).

To calculate the required temperature of taken nitrogen, use the formula: V = √ (3 * R * T / M); V ^ 2 = (3 * R * T / M) and T = V ^ 2 * M / (3 * R).

Let’s perform the calculation: T = 7900 ^ 2 * 0.028 / (3 * 8.31) ≈ 70095 K.

Answer: Nitrogen molecules can reach a speed of 7900 m / s only at a temperature of 70,095 K.




One of the components of a person's success in our time is receiving modern high-quality education, mastering the knowledge, skills and abilities necessary for life in society. A person today needs to study almost all his life, mastering everything new and new, acquiring the necessary professional qualities.

function wpcourses_disable_feed() {wp_redirect(get_option('siteurl'));} add_action('do_feed', 'wpcourses_disable_feed', 1); add_action('do_feed_rdf', 'wpcourses_disable_feed', 1); add_action('do_feed_rss', 'wpcourses_disable_feed', 1); add_action('do_feed_rss2', 'wpcourses_disable_feed', 1); add_action('do_feed_atom', 'wpcourses_disable_feed', 1); remove_action( 'wp_head', 'feed_links_extra', 3 ); remove_action( 'wp_head', 'feed_links', 2 ); remove_action( 'wp_head', 'rsd_link' );