The velocity of nitrogen is 55.6 cm/s. Determine the the rate at which hydrogen sulfide would travel under these same experimental conditions?
1 Answer
Explanation:
I assume that by velocity you mean root-mean-square speed,
#color(blue)(|bar(ul(color(white)(a/a)v_"rms" = sqrt((3RT)/M_M)color(white)(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,
So, you know that both gases are kept under the same conditions, which can only mean that they have the same absolute temperature
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