Question #633fc

1 Answer
Mar 6, 2016

#"1060 km h"^(-1)#

Explanation:

The root-mean-square speed of a molecule in a gas is calculated using the absolute temperature of the gas, the molar mass of the gas, and the universal gas constant.

#color(black)(|bar(ul(color(blue)(v_"rms" = sqrt((3RT)/M_M))))|) " "#, where

#v_"rms"# - the root-mean-square speed of the molecule
#R# - the universal gas constant, used as #"8.314 J mol"^(-1)"K"^(-1)#
#T# - the absolute temperature of the gas
#M_M# - the molar mass of the gas

You're dealing with krypton, a noble gas that exists as atoms in the gaseous state. The molar mass of krypton is equal to #"83.798 g mol"^(-1)#.

Now, something important to keep track of before doing the calculations. The root-mean-square speed is expressed is meters per second, #"m s"^(-1)#.

This means that you're going to have to manipulate some units under the square root to get #"m s"^(-1)# as the units for #v_"rms"#.

More specifically, you're going to have to use the fact that

#"1 J" = 1 "kg m"^2"s"^(-2) " " " "color(orange)("(*)")#

So, plug in your values into the equation - do not forget to convert the temperature from degrees Celsius to Kelvin!

#v_"rms" = sqrt((3 * "8.314 J" color(red)(cancel(color(black)("mol"^(-1)))) * color(red)(cancel(color(black)("K"^(-1)))) * (273.15 + 20.0)color(red)(cancel(color(black)("K"))))/("83.798 g" color(red)(cancel(color(black)("mol"^(-1))))))#

#v_"rms" = sqrt((3 * 8.314 * 293.15)/83.798) * sqrt("J g"^(-1))#

Now focus on the units first. Use the conversion factor

#"1 kg" = 10^3"g"#

to write

#"J"/color(red)(cancel(color(black)("g"))) * (10^3color(red)(cancel(color(black)("g"))))/"1 kg" = 10^3 "J kg"^(-1)#

Use conversion factor #color(orange)("(*)")# to write

#10^3"J kg"^(-1) = 10^3 color(red)(cancel(color(black)("kg"))) "m"^2"s"^(-2) * color(red)(cancel(color(black)("kg"^(-1)))) = 10^3"m"^2"s"^(-2)#

Plug this back to find #v_"rms"#

#v_"rms" = sqrt((3 * 8.314 * 293.15)/83.798 * 10^3"m"^2"s"^(-2)#

#v_"rms" = "295.4 m s"^(-1)#

To convert this to kilometers per hour, #"km h"^(-1)#, use the conversion factors

#"1 km" = 10^3"m" " "# and #" " "1 h " = " 3600 s"#

You will thus have

#295.4 color(red)(cancel(color(black)("m")))/color(red)(cancel(color(black)("s"))) * "1 km"/(10^3color(red)(cancel(color(black)("m")))) * (3600color(red)(cancel(color(black)("s"))))/"1 h" = "1063.44 km h"^(-1)#

Rounded to three sig figs, the number of sig figs you have for the temperature of the gas, the answer will be

#v_"rms" = color(green)(|bar(ul(color(white)(a/a)"1060 km h"^(-1)color(white)(a/a)))|)#