Question

By what factor does the most probable velocity of a gaseous molecule increase when the temperature in Kelvin is quadrupled?
A. 3
B. 2
C. 4
D. 16

Answer 1

The most probable speed of gas molecules described by the Maxwell-Boltzmann distribution is the speed at which distribution graph reaches its maximum. The average speed of molecules can be calculated as an integral of the Maxwell-Boltzmann distribution function multiplied by the magnitude of velocity of a molecule.

Complete answer:
The mathematical representation of the most probable velocity can be represented as:
${v_{mp}} = \sqrt {\dfrac{{2RT}}{{{M_w}}}}$
Where, ${v_{mp}} =$ most probable velocity
$R =$ universal gas constant
$T =$ temperature of the gas
${M_w} =$ molecular weight of the gas
Now, from the above mathematical representation, we can clearly see that the most probable velocity is directly proportional to the square root of the temperature of the gas. This can be represented mathematically as:
${v_{mp}} \propto \sqrt T$
Now, taking the above relation as the base relation, let us determine the most probable velocity when the temperature is increased four times or is quadrupled.
$\dfrac{{{{({v_{mp}})}_1}}}{{{{({v_{mp}})}_2}}} = \dfrac{{\sqrt {{T_1}} }}{{\sqrt {{T_2}} }}$
Now, let the value of ${T_1} = T$ .
Then, temperature ${T_2} = 4T$ , when the most probable velocity becomes equal to ${({v_{mp}})_2}$ .
Substituting these values in the above equation, we have:
$\Rightarrow \dfrac{{{{({v_{mp}})}_1}}}{{{{({v_{mp}})}_2}}} = \dfrac{{\sqrt T }}{{\sqrt {4T} }}$
Thus, ${({v_{mp}})_2} = 2{({v_{mp}})_1}$ .

Thus option B is the correct answer.

Note:
The particles move freely inside a stationary container without interacting with one another, except for very brief collisions in which they exchange energy and momentum with each other or with their thermal environment. The term "particle" in this context refers to gaseous particles only (atoms or molecules), and the system of particles is assumed to have reached thermodynamic equilibrium.