Question

The molecules of which of the following gas have higher speed?
A. Hydrogen at $- {50^o}C$
B. Methane at $298K$
C. Nitrogen at ${1000^o}C$
D. Oxygen at ${0^o}C$

Answer 1

The root mean square velocity of the molecules of a gas is defined as the square root of the average/ mean squared velocity of the molecules of the gas at a particular temperature. It is directly proportional to the temperature of the gas and inversely proportional to the molecular weight.

Complete step by step answer:
The root mean square velocity of a particular kind of gas is directly proportional to the square root of the molecular weight of the gas and inversely proportional to the square root of the temperature of the gas. Mathematically, this can be written as:
${v_{rms}} = \sqrt {\dfrac{{3RT}}{{{M_w}}}}$ …. (i)
Where, R is the universal gas constant and remains the same for all the gases.
$T =$ Temperature at which the gas is being heated
${M_w} =$ Molecular weight of the gas present
Thus, we can relate the velocity, temperature and molecular weight of any gas as:
${v_{rms}} \propto \sqrt {\dfrac{T}{{{M_w}}}}$
Now, let us consider each and every option one by one.
A. Hydrogen at $- {50^o}C$:
The molecular weight of hydrogen gas = $2g$
Temperature given = $- {50^o}C = 273 + ( - 50) = 223K$
Substituting the values in equation (i), and on solving, we have:
Velocity = $v = \sqrt {\dfrac{{3 \times 8.314 \times 223}}{2}} = \sqrt {2781.033} = 52.73m/s$
B. Methane at $298K$:
The molecular weight of methane gas = $16g$
Temperature given =$298K$
Substituting the values in equation (i), and on solving, we have:
Velocity = $v = \sqrt {\dfrac{{3 \times 8.314 \times 298}}{{16}}} = \sqrt {464.544} = 21.55m/s$
C. Nitrogen at ${1000^o}C$:
The molecular weight of nitrogen gas = $28g$
Temperature given =${1000^o}C = (1000 + 273) = 1273K$
Substituting the values in equation (i), and on solving, we have:
Velocity = $v = \sqrt {\dfrac{{3 \times 8.314 \times 1273}}{{28}}} = \sqrt {1133.97} = 33.67m/s$
D. Oxygen at ${0^o}C$:
The molecular weight of oxygen gas = $32g$
Temperature given =${0^o}C = (0 + 273) = 273K$
Substituting the values in equation (i), and on solving, we have:
Velocity =$v = \sqrt {\dfrac{{3 \times 8.314 \times 273}}{{32}}} = \sqrt {567.43} = 23.82m/s$
Thus, as it is clearly visible, the highest speed will be for hydrogen gas.
The correct option is A. Hydrogen at $- {50^o}C$.

Note:
There are three types of velocities for molecules in a gas. They are most probable speed, room mean square velocity and average velocity. The relation between these mentioned velocities is: ${v_{rms}}:{v_{mp}}:{v_{avg}} = \sqrt 3 :\sqrt 2 :\sqrt {\dfrac{8}{\pi }} = 1.224:1:1.128$