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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

Answer
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Hint: The formula for root mean square velocity should be known to find the correct solution. Rms velocity for a gas molecule is vrms=3RTM

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=3RTM
Here, R is the ideal gas constant whose value is
R=8.314JK1mol1
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=3×(8.314JK1mol1)×(300K)32gmol1
vrms=3×(8.314JK1mol1)×(300K)0.032kgmol1
vrms=3×8.314×300K0.032ms1
After further calculations we get,
vrms483ms1
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.=2KTM
vavg=8KTπM
If we take their ratios,
vm.p:vavg:vrms=2:8π: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.