# The velocity of nitrogen is 55.6 cm/s. Determine the the rate at which hydrogen sulfide would travel under these same experimental conditions?

Jun 17, 2016

${\text{51 cm s}}^{- 1}$

#### Explanation:

I assume that by velocity you mean root-mean-square speed, ${v}_{\text{rms}}$, which for an ideal gas kept under a temperature $T$ is equal to

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} {v}_{\text{rms}} = \sqrt{\frac{3 R T}{M} _ M} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Here

$R$ - the universal gas constant
${M}_{M}$ - the molar mass of the gas

Now, the root-mean-square speed is usually expressed is meters per second, ${\text{m s}}^{- 1}$, but you don't have to convert it because you're going to express the root-mean-square speed of hydrogen sulfide, $\text{H"_2"S}$, relative to that of nitrogen gas, ${\text{N}}_{2}$.

So, you know that both gases are kept under the same conditions, which can only mean that they have the same absolute temperature $T$.

You can thus say that

v_("rms N"_ 2) = sqrt( (3RT)/M_("M N"_2)) -> the root-mean-squares speed for nitrogen gas

v_("rms H"_ 2"S") = sqrt((3RT)/(M_("M H"_2"S"))) -> the root-mean-square speed for hydrogen sulfide

Divide these two equations to get

v_("rms H"_ 2"S")/v_("rms N"_ 2) = sqrt( color(red)(cancel(color(black)(3RT)))/M_("M H"_2"S") * M_("M N"_2)/color(red)(cancel(color(black)(3RT))))

v_("rms H"_ 2"S")/v_("rms N"_ 2) = sqrt(M_("M N"_ 2)/M_("M H"_2"S"))

This is equivalent to

v_("rms H"_ 2"S") = v_("rms N"_2) * sqrt(M_("M N"_ 2)/M_("M H"_2"S"))

The molar masses of the two gases are

M_("M N"_2) = "28.0134 g mol"^(-1)

M_("M H"_2"S") = "34.089 g mol"^(-1)

You will thus have

v_("rms H"_2"S") = "56 cm s"^(-1) * sqrt((28.0134 color(red)(cancel(color(black)("g mol"^(-1)))))/(34.089color(red)(cancel(color(black)("g mol"^(-1))))))

v_("rms H"_2"S") = color(green)(|bar(ul(color(white)(a/a)color(black)("51 cm s"^(-1))color(white)(a/a)|)))

The answer is rounded to two sig figs.

Now, does the result make sense?

The root-mean-square speed is inversely proportional to the square root of the molar mass of the gas, which means that the heavier the molar mass, the lower the root-mean-square speed for gases kept under the same absolute temperature $T$.