Question 633fc

Mar 6, 2016

${\text{1060 km h}}^{- 1}$

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

${v}_{\text{rms}}$ - the root-mean-square speed of the molecule
$R$ - the universal gas constant, used as ${\text{8.314 J mol"^(-1)"K}}^{- 1}$
$T$ - 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 ${\text{83.798 g mol}}^{- 1}$.

Now, something important to keep track of before doing the calculations. The root-mean-square speed is expressed is meters per second, ${\text{m s}}^{- 1}$.

This means that you're going to have to manipulate some units under the square root to get ${\text{m s}}^{- 1}$ as the units for ${v}_{\text{rms}}$.

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

$\text{1 kg" = 10^3"g}$

to write

${\text{J"/color(red)(cancel(color(black)("g"))) * (10^3color(red)(cancel(color(black)("g"))))/"1 kg" = 10^3 "J kg}}^{- 1}$

Use conversion factor $\textcolor{\mathmr{and} a n \ge}{\text{(*)}}$ to write

${10}^{3} {\text{J kg"^(-1) = 10^3 color(red)(cancel(color(black)("kg"))) "m"^2"s"^(-2) * color(red)(cancel(color(black)("kg"^(-1)))) = 10^3"m"^2"s}}^{- 2}$

Plug this back to find ${v}_{\text{rms}}$

${v}_{\text{rms" = sqrt((3 * 8.314 * 293.15)/83.798 * 10^3"m"^2"s}}^{- 2}$

${v}_{\text{rms" = "295.4 m s}}^{- 1}$

To convert this to kilometers per hour, ${\text{km h}}^{- 1}$, use the conversion factors

$\text{1 km" = 10^3"m" " }$ and $\text{ " "1 h " = " 3600 s}$

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)))|)#