Explanation:
Expression for the
speed is as follows.
![v_{rms} = \sqrt{\frac{3kT}{M}}](https://tex.z-dn.net/?f=v_%7Brms%7D%20%3D%20%5Csqrt%7B%5Cfrac%7B3kT%7D%7BM%7D%7D)
where,
= root mean square speed
k = Boltzmann constant
T = temperature
M = molecular mass
As the molecular weight of oxygen is 0.031 kg/mol and R = 8.314 J/mol K. Hence, we will calculate the value of
as follows.
![v_{rms} = \sqrt{\frac{3kT}{M}}](https://tex.z-dn.net/?f=v_%7Brms%7D%20%3D%20%5Csqrt%7B%5Cfrac%7B3kT%7D%7BM%7D%7D)
= ![\sqrt{\frac{3 \times 8.314 J/mol K \times 309.02 K}{0.031 kg/mol}}](https://tex.z-dn.net/?f=%5Csqrt%7B%5Cfrac%7B3%20%5Ctimes%208.314%20J%2Fmol%20K%20%5Ctimes%20309.02%20K%7D%7B0.031%20kg%2Fmol%7D%7D)
= 498.5 m/s
Hence,
for oxygen atom is 498.5 m/s.
For nitrogen atom, the molecular weight is 0.028 kg/mol. Hence, we will calculate its
speed as follows.
![v_{rms} = \sqrt{\frac{3kT}{M}}](https://tex.z-dn.net/?f=v_%7Brms%7D%20%3D%20%5Csqrt%7B%5Cfrac%7B3kT%7D%7BM%7D%7D)
= ![\sqrt{\frac{3 \times 8.314 J/mol K \times 309.92 K}{0.028 kg/mol}}](https://tex.z-dn.net/?f=%5Csqrt%7B%5Cfrac%7B3%20%5Ctimes%208.314%20J%2Fmol%20K%20%5Ctimes%20309.92%20K%7D%7B0.028%20kg%2Fmol%7D%7D)
= 524.5 m/s
Therefore,
speed for nitrogen is 524.5 m/s.
Answer:
algun moderador puede acabar a este otro este me borró respuesta que no debió borrar
Sorry I don’t get the question Buh I’m sure you’ll get it eventually
Boyle's law
I hope this helps