Question

The value of ${v_{rms}}$ for a gas X at 546$^ \circ C$ was found to be equal to the value of ${v_{mp}}$ for another gas Y at 273$^ \circ C$. Assuming ideal behavior, find the molecular mass of gas Y (in amu) if the molecular mass of the gas X is 9 amu.

Answer 1

Hint: The formula to find the root mean velocity and most probable velocity is given as below.
     \[{v_{rms}} = \sqrt {\dfrac{{3RT}}{M}} \]
     \[{v_{mp}} = \sqrt {\dfrac{{2RT}}{M}} \]

Complete answer:
We will first get some information about ${v_{rms}}$ and ${v_{mp}}$ .
- ${v_{rms}}$ is also known as root mean square velocity of the gas. We know that the particles of gas have different speeds. The distribution of velocities does not change. The root mean square velocity of a gas can be expressed by following equation
     \[{v_{rms}} = \sqrt {\dfrac{{3RT}}{M}} \]
Here, R is molar gas constant, T is temperature in K and M is the molar mass of gas in kg per mole unit.
- ${v_{mp}}$ is known as the most probable velocity of the gas. It can be expressed by the following formula.
     \[{v_{mp}} = \sqrt {\dfrac{{2RT}}{M}} \]
- Now, we are given that ${v_{rms}}$ of a gas X at 546$^ \circ C$ was found to be equal to ${v_{mp}}$ of the gas Y at ${273^ \circ }C$. We are given that the molar mass of gas X is 9 amu.
We will first convert temperature into Kelvin units.
So, we know that $^ \circ C + 273 = K$
Thus, 546$^ \circ C$ = 546 + 273 = 819 K and ${273^ \circ }C$ = 273 + 273 = 546 K
Now, we can say that for given conditions,
     \[{v_{rms}} = {v_{mp}}\]
So,
     \[\sqrt {\dfrac{{3R{T_X}}}{{{M_X}}}} = \sqrt {\dfrac{{2R{T_Y}}}{{{M_Y}}}} \]
Putting the available values into the above equation, we get
     \[\sqrt {\dfrac{{(3)(R)(819)}}{9}} = \sqrt {\dfrac{{(2)(R)(546)}}{{{M_Y}}}} \]
Now, we can write the above equation as
     \[\dfrac{{(3)(819)}}{9} = \dfrac{{(2)(546)}}{{{M_Y}}}\]
Thus, we obtain
     \[{M_Y} = \dfrac{{(2)(546)(9)}}{{(3)(819)}} = 4{\text{ amu}}\]

So, we found that the molar mass of gas Y is 4 amu.

Note:
Note that most probable velocity is the speed at which the Maxwell-Boltzmann distribution graph reaches its maximum. The average velocity is the mean of magnitudes of velocity of the molecules.