The root mean square velocity of hydrogen molecules is 183,000cm/sec, what is its density at atmospheric pressure?

1 Answer
Mar 15, 2018


d\ =\ 0.091\ {Kg}/m^3 " " or " " d\ =\ 0.091\ {g}/L

Explanation:


Root mean square velocity = sqrt{\ bar{c^2}\ }\ =\ sqrt{{3P}/d}

Put the given values, to get:

{183000\ cm}/sec\ =\ sqrt{{3(101325\ Pa)}/d}

Convert the root mean square velocity from \ \ {cm}/s\ \ to \ \ m/s\ \ and remember that, \ \ 1\ Pa\ =\ {1\ N}/m^2\ \ \ and \ \ 1\ N\ =\ 1\ Kg{m}/s^2

{183000\ cm}/s\ =\ {1830\ m}/s

Square both sides replace the units:

\frac{(1830)^2\ m^2}{s^2}\ =\ \frac{3\cdot 101325\ N}{m^2\cdot d}

\frac{(1830)^2\ m^2}{color(red){cancel{s^2}} = \frac{3\cdot 101325\ Kg*m}{m^2*\color(red){cancel{s^2}}\cdot d}

\frac{1}{(1830)^2\ m^2}\ =\ \frac{m^2\cdot d}{3\cdot 101325\ Kg\cdot m}

Solve for \ \ d\ \ to get:

\frac{3\cdot 101325\ Kg\cdot color(blue){cancel m}}{(1830)^2\ \color(blue){cancelm^{4}\ =\ d

d\ =\ \frac{3\cdot 101325\ Kg}{(1830)^2\ m^3}

d\ =\ 0.091\ {Kg}/m^3

That's it!