The root mean square velocity one mole monoatomic gas having molar mass M is ${U_{rms}}$. The relation between average kinetic energy (E) of the gas and ${U_{rms}}$is:
A. ${U_{rms}} = \sqrt {\dfrac{{3E}}{{2M}}} $
B. ${U_{rms}} = \sqrt {\dfrac{{2E}}{{3M}}} $
C. ${U_{rms}} = \sqrt {\dfrac{{2E}}{M}} $
D. ${U_{rms}} = \sqrt {\dfrac{E}{{3M}}} $
Answer
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Hint: The root mean square velocity of one mole of monoatomic gas is ${U_{rms}} = \sqrt {\dfrac{{3RT}}{M}} $. Here, we have to substitute the value of $3RT$ from the average kinetic energy expression. Then, the required relation between ${U_{rms}}$and average kinetic energy can be obtained.
Complete step-by-step answer:
First, we write the formula of root mean square speed of one mole of monoatomic gas.
${U_{rms}} = \sqrt {\dfrac{{3RT}}{M}} $ …… (1)
Here, R is universal gas constant $\left( {8.314\;\dfrac{{\rm{J}}}{{{\rm{mol}} \cdot {\rm{K}}}}} \right)$, M is the molar mass of the gas and T is temperature in Kelvin.
Following the kinetic theory of gas, average kinetic energy of one mole of gaseous particles is directly proportional to temperature. Now, we write the formula of average kinetic energy.
$E = \dfrac{3}{2}RT$ …… (2)
Where, E is the average kinetic energy, R is gas constant and T is temperature.
Now we rearrange equation (2).
$2E = 3RT$
Now we substitute the value of 3RT in equation (1).
$\begin{array}{c}{U_{rms}} = \sqrt {\dfrac{{3RT}}{M}} \\ = \sqrt {\dfrac{{2E}}{M}} \end{array}$
So, the relation between ${U_{rms}}$and average kinetic energy is,
${U_{rms}} = \sqrt {\dfrac{{2E}}{M}} $
So, the correct option is C.
Additional Information:
Some important points of kinetic theory of gas are:-
1) In collision of molecules no loss or gain of energy takes place.
2) The space occupied by gaseous molecules is negligible if compared to the size of container they occupy.
3) The molecules are always in constant motion.
Note: For monoatomic gas average kinetic energy is, $E = \dfrac{3}{2}RT$ and for diatomic molecule, $E = \dfrac{5}{2}RT$. So, using the 3RT value from the kinetic energy equation of monoatomic gas we have to establish the relation.
Complete step-by-step answer:
First, we write the formula of root mean square speed of one mole of monoatomic gas.
${U_{rms}} = \sqrt {\dfrac{{3RT}}{M}} $ …… (1)
Here, R is universal gas constant $\left( {8.314\;\dfrac{{\rm{J}}}{{{\rm{mol}} \cdot {\rm{K}}}}} \right)$, M is the molar mass of the gas and T is temperature in Kelvin.
Following the kinetic theory of gas, average kinetic energy of one mole of gaseous particles is directly proportional to temperature. Now, we write the formula of average kinetic energy.
$E = \dfrac{3}{2}RT$ …… (2)
Where, E is the average kinetic energy, R is gas constant and T is temperature.
Now we rearrange equation (2).
$2E = 3RT$
Now we substitute the value of 3RT in equation (1).
$\begin{array}{c}{U_{rms}} = \sqrt {\dfrac{{3RT}}{M}} \\ = \sqrt {\dfrac{{2E}}{M}} \end{array}$
So, the relation between ${U_{rms}}$and average kinetic energy is,
${U_{rms}} = \sqrt {\dfrac{{2E}}{M}} $
So, the correct option is C.
Additional Information:
Some important points of kinetic theory of gas are:-
1) In collision of molecules no loss or gain of energy takes place.
2) The space occupied by gaseous molecules is negligible if compared to the size of container they occupy.
3) The molecules are always in constant motion.
Note: For monoatomic gas average kinetic energy is, $E = \dfrac{3}{2}RT$ and for diatomic molecule, $E = \dfrac{5}{2}RT$. So, using the 3RT value from the kinetic energy equation of monoatomic gas we have to establish the relation.
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