Question

A gas mixture of molecules of type 1, 2 and 3 with molar masses ${M_1} > {M_2} > {M_3}$ . What will be the relation between the Root Mean Square velocity and Kinetic energy of the gases?
A. \[{\left( {{v_{rms}}} \right)_1} < {\left( {{v_{rms}}} \right)_2} < {\left( {{v_{rms}}} \right)_3}\] and ${\left( {\bar K} \right)_1} = {\left( {\bar K} \right)_2} = {\left( {\bar K} \right)_3}$
B. \[{\left( {{v_{rms}}} \right)_1} = {\left( {{v_{rms}}} \right)_2} = {\left( {{v_{rms}}} \right)_3}\] and ${\left( {\bar K} \right)_1} = {\left( {\bar K} \right)_2} > {\left( {\bar K} \right)_3}$
C. \[{\left( {{v_{rms}}} \right)_1} > {\left( {{v_{rms}}} \right)_2} > {\left( {{v_{rms}}} \right)_3}\] and ${\left( {\bar K} \right)_1} < {\left( {\bar K} \right)_2} > {\left( {\bar K} \right)_3}$
D. \[{\left( {{v_{rms}}} \right)_1} > {\left( {{v_{rms}}} \right)_2} > {\left( {{v_{rms}}} \right)_3}\] and ${\left( {\bar K} \right)_1} < {\left( {\bar K} \right)_2} < {\left( {\bar K} \right)_3}$

Answer 1

Hint: Root mean square velocity of individual gas molecules is given by ${v_{rms}} = \sqrt {\dfrac{{3RT}}{M}} $ , where T is the temperature of gas molecules in Kelvin, M is the molar mass of the gas and $R$ is the universal gas constant having a value of $8.314{\text{ J}}{{\text{K}}^{ - 1}}{\text{mo}}{{\text{l}}^{ - 1}}$ . The kinetic energy of $n$ moles of an ideal gas at a temperature of $T$ Kelvin is given by $K.E. = \dfrac{3}{2}nRT$ .



Formula used:
RMS velocity is calculated as ${v_{rms}} = \sqrt {\dfrac{{3RT}}{M}} $
Kinetic energy of gas molecules is given by $K.E. = \dfrac{3}{2}nRT$ .


Complete answer:
RMS velocity of gas molecules is given by:
${v_{rms}} = \sqrt {\dfrac{{3RT}}{M}} $ … (1)
Here, T is the temperature of gas molecules in Kelvin,
M is the molar mass of the gas and
$R$ is the universal gas constant having a value of $8.314{\text{ J}}{{\text{K}}^{ - 1}}{\text{mo}}{{\text{l}}^{ - 1}}$.
Now, we know that Molar mass of a gas molecule is given by the ratio of molecular mass, $m$ of the gas and number of moles, $n$ , that is,
It can be easily deduced that ${v_{rms}}$ is inversely proportional to the square root of the molar mass of the gas, $M$ .
Hence,
\[{\left( {{v_{rms}}} \right)_1} < {\left( {{v_{rms}}} \right)_2} < {\left( {{v_{rms}}} \right)_3}\] … (2)
The kinetic energy of $n$ moles of an ideal gas at a temperature of $T$ Kelvin is given by:
$K.E. = \dfrac{3}{2}nRT$ … (3)
Here $R$ is the ideal gas constant.
As the kinetic energy of gas molecules depends upon the number of moles and not the molar mass, therefore,
${\left( {\bar K} \right)_1} = {\left( {\bar K} \right)_2} = {\left( {\bar K} \right)_3}$ … (4)
Hence, from (2) and (4), the correct option is A.


Thus, the correct option is A.



Note: For individual gas molecules, RMS velocity is calculated as ${v_{rms}} = \sqrt {\dfrac{{3RT}}{M}} $ . Root Mean Square velocity is directly proportional to the absolute temperature of the gas and inversely proportional to the molar mass. Kinetic energy of gas molecules is given by $K.E. = \dfrac{3}{2}nRT$ .