Question
Question: The rms speed (in m/s) of oxygen molecules of the gas at temperature 300 K, is A. 483 B. 504 C...
The rms speed (in m/s) of oxygen molecules of the gas at temperature 300 K, is
A. 483
B. 504
C. 377
D. 364
Solution
Hint: The formula for root mean square velocity should be known to find the correct solution. Rms velocity for a gas molecule is vrms=M3RT
Complete step by step answer:
A gas consists of molecules that are constantly moving with different speeds in different directions. So, we calculate the root mean square velocity of molecules as an average velocity of the molecules since it is difficult to calculate the particular velocity of each individual gas molecule. The root mean square speed of gas molecules is defined as the square root of mean squares of random velocities of individual gas molecules.
Formula for rms velocity is
vrms=M3RT
Here, R is the ideal gas constant whose value is
R=8.314JK−1mol−1
T is absolute temperature
M is mass of 1 mol of a gas
Here we have to take the molar mass of diatomic oxygen molecule,
Molar mass of oxygen atom = 16 g/mol
Molar mass of diatomic oxygen molecule = 32 g/mol
On substituting these values in the above formula, we get
vrms=32gmol−13×(8.314JK−1mol−1)×(300K)
vrms=0.032kgmol−13×(8.314JK−1mol−1)×(300K)
vrms=0.0323×8.314×300Kms−1
After further calculations we get,
vrms≈483ms−1
The correct answer is option A. 483
Additional Information: Students should also memorise the formulas of most probable speed i.e. the speed possessed by maximum number of gas molecules and mean speed i.e. the average speed of the gas molecules:
vm.p.=M2KT
vavg=πM8KT
If we take their ratios,
vm.p:vavg:vrms=2:π8:3
We can see that vrms is maximum and vm.p is minimum.
Note: Students must always convert temperature into Kelvin and the molar mass must be taken in kg to solve the numerical using this formula. All the SI units must be according to the MKS system of units.