# calculate the root mean square velocity and kinetic energy of CO,CO2, and SO3 at 284K.

## Question:

calculate the root mean square velocity and kinetic energy of CO,CO2, and SO3 at 284K.

## Kinetic energy of gases

A gas is made up of many atoms or molecules which move with very high speed. The kinetic energy of gases is very high. It is calculated by the formula shown below.

{eq}{\rm{K}}.{\rm{E}}. = \dfrac{3}{2}RT {/eq}

## Answer and Explanation:

**Given Data:**

- The value of temperature is: 284 K

The root mean square velocity is calculated by the formula shown below.

{eq}{v_{{\rm{rms}}}} = \sqrt {\dfrac{{3RT}}{M}} {/eq}

Here,

{eq}R {/eq} is the gas constant.

{eq}T {/eq} is the temperature.

{eq}M {/eq} is the molar mass of gas (in kg/mol).

Substitute the value of gas constant, temperature and molar mass of CO (0.028 kg/mol) in the above formula.

{eq}\begin{align*} {v_{{\rm{rms}}}} &= \sqrt {\dfrac{{3 \times 8.314\;{\rm{J/mol}} \cdot {\rm{K}} \times 284\;{\rm{K}}}}{{0.028\;{\rm{kg/mol}}}}} \\ &= \sqrt {252983.143} \\ &= 502.97\;{\rm{m/s}} \end{align*} {/eq}

Substitute the value of gas constant, temperature and molar mass of {eq}{\rm{C}}{{\rm{O}}_2}\left( {0.044\;{\rm{kg/mol}}} \right) {/eq} in the above formula.

{eq}\begin{align*} {v_{{\rm{rms}}}} &= \sqrt {\dfrac{{3 \times 8.314\;{\rm{J/mol}} \cdot {\rm{K}} \times 284\;{\rm{K}}}}{{0.044\;{\rm{kg/mol}}}}} \\ &= \sqrt {160989.30} \\ &= 401.23\;{\rm{m/s}} \end{align*} {/eq}

Substitute the value of gas constant, temperature and molar mass of {eq}{\rm{S}}{{\rm{O}}_3}\left( {0.0801\;{\rm{kg/mol}}} \right) {/eq} in the above formula.

{eq}\begin{align*} {v_{{\rm{rms}}}} &= \sqrt {\dfrac{{3 \times 8.314\;{\rm{J/mol}} \cdot {\rm{K}} \times 284\;{\rm{K}}}}{{0.0801\;{\rm{kg/mol}}}}} \\ &= \sqrt {88433.56} \\ &= 297.38\;{\rm{m/s}} \end{align*} {/eq}

The kinetic energy of a gas is calculated by the formula shown below.

{eq}{\rm{K}}.{\rm{E}}. = \dfrac{3}{2}RT {/eq}

Substitute the value of gas constant and temperature in the above formula.

{eq}\begin{align*} {\rm{K}}.{\rm{E}}. &= \dfrac{3}{2} \times 8.314\;{\rm{J/mol}} \cdot {\rm{K}} \times 284\;{\rm{K}}\\ &= 3.542 \times {10^3}\;{\rm{J/mol}}\\ &= 3.542\;{\rm{kJ/mol}} \end{align*} {/eq}

Therefore, the kinetic energy of CO, {eq}{\rm{C}}{{\rm{O}}_{\rm{2}}} {/eq} and {eq}{\rm{S}}{{\rm{O}}_{\rm{3}}} {/eq} is 3.542 kJ/mol.

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from Chemistry 101: General Chemistry

Chapter 7 / Lesson 5