Q94E - Chemistry (OpenStax)

Question

Question: What is the ratio of the average kinetic energy of a \({\rm{S}}{{\rm{O}}_{\rm{2}}}\) molecule to that of an \({{\rm{O}}_{\rm{2}}}\) molecule in a mixture of two gases? What is the ratio of the root mean square speeds, \({{\rm{u}}_{{\rm{rms}}}}\), of the two gases?

Step-by-Step Solution

Answer

Ratio of the average kinetic energy of \({\rm{S}}{{\rm{O}}_{\rm{2}}}\) molecule to that of an \({{\rm{O}}_{\rm{2}}}\) molecule is obtained as: \({\rm{1:1}}\).

Ratio of the root mean square speeds of \({\rm{S}}{{\rm{O}}_{\rm{2}}}\) molecule to that of an \({{\rm{O}}_{\rm{2}}}\) molecule is obtained as: \({\rm{0}}{\rm{.707:1}}\) 

Step 1: Define Gas

One of the four fundamental states of matter, a gas is made up of particles with no definite volume or structure.

Step 2: Evaluating the ratio

According to kinetic molecular theory (KMT), all gases have the same \({\rm{K}}{{\rm{E}}_{{\rm{avg}}}}\) for their molecules at a given temperature. So,

\({\rm{K}}{{\rm{E}}_{{\rm{avg}}}}{\rm{(S}}{{\rm{O}}_{\rm{2}}}{\rm{)  =   K}}{{\rm{E}}_{{\rm{avg}}}}{\rm{(}}{{\rm{O}}_{\rm{2}}}{\rm{)}}\)

The average kinetic energy of a \({\rm{S}}{{\rm{O}}_{\rm{2}}}\) molecule divided by the average kinetic energy of an \({{\rm{O}}_{\rm{2}}}\) molecule is:

\(\frac{{{\rm{K}}{{\rm{E}}_{{\rm{avg}}}}{\rm{(S}}{{\rm{O}}_{\rm{2}}}{\rm{)}}}}{{{\rm{K}}{{\rm{E}}_{{\rm{avg}}}}{\rm{(}}{{\rm{O}}_{\rm{2}}}{\rm{)}}}}{\rm{  =   1:1}}\)

Then, the ratio of the root mean square speed is obtained as:

\(\begin{array}{c}{{\rm{u}}_{{\rm{rms}}}}{\rm{ = }}\sqrt {\frac{{{\rm{3RT}}}}{{\rm{m}}}} \\\frac{{{{\rm{u}}_{{\rm{rms}}}}\left( {{\rm{S}}{{\rm{O}}_{\rm{2}}}} \right)}}{{{{\rm{u}}_{{\rm{rms}}}}\left( {{{\rm{O}}_{\rm{2}}}} \right)}}{\rm{ = }}\frac{{\sqrt {\frac{{{\rm{3RT}}}}{{{{\rm{m}}_{{\rm{S}}{{\rm{O}}_{\rm{2}}}}}}}} }}{{\sqrt {\frac{{{\rm{3RT}}}}{{{{\rm{m}}_{{{\rm{O}}_{\rm{2}}}}}}}} }}\\{\rm{ = }}\frac{{\sqrt {{{\rm{m}}_{{{\rm{O}}_{\rm{2}}}}}} }}{{\sqrt {{{\rm{m}}_{{{\rm{O}}_{\rm{2}}}}}} }}\quad {\rm{m  -   molar mass of the molecule}}\\{\rm{ = }}\frac{{\sqrt {{\rm{32\;g/mol}}} }}{{\sqrt {{\rm{64}}{\rm{.066\;g/mol}}} }}\\{\rm{ = }}\frac{{{\rm{5}}{\rm{.6568}}}}{{{\rm{8}}{\rm{.0041}}}}\\{\rm{ = 0}}{\rm{.707}}\end{array}\)

Therefore, we have obtained as:

The ratio of the average kinetic energy of a \({\rm{S}}{{\rm{O}}_{\rm{2}}}\) molecule to that of an \({{\rm{O}}_{\rm{2}}}\) molecule is: \({\rm{1:1}}\).

The ratio of the root mean square speeds of a \({\rm{S}}{{\rm{O}}_{\rm{2}}}\) molecule to that of an \({{\rm{O}}_{\rm{2}}}\) molecule is: \({\rm{0}}{\rm{.707:1}}\) .