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6.7 Gases in Chemical Reactions: -- Part 2
Substitute the quantities into the equation to calculate root 32.00 g O2 1 kg 32.00 * 10 - 3 kg O2 mean square velocity.
It's important to note that 1 J is 1 kilo.
The units of the answer are correct.
The magnitude of the answer seems reasonable because oxygen is heavier than nitrogen and should have a slightly lower root mean square velocity at the same temperature.
The root mean square velocity of nitrogen is 515 m>s.
The mean free path is the average distance between crashes.
The root mean square velocity of gas at room temperature is in the range of hundreds of meters per second.
It is possible that your roommate put on too much perfume in the bathroom.
The answer is that even though gaseous particles travel at tremendous speeds, they also travel in chaotic paths.
The path from the perfume bottle in the bathroom to your nose is similar to a bargain hunter's path through a shopping mall during a clearance sale.
A molecule travels only a short distance before it collides with another molecule, and then collides again, and so on.
At room temperature and atmospheric pressure, a molecule in the air experiences billions of hits per second.
The path of gas at room temperature and atmospheric pressure is a mess.
The nitrogen molecule would travel about 40 feet if it were the size of a golf ball.
The mean free path of a nitrogen molecule is hundreds of kilometers.
The root mean square velocity is related to the rate of effusion.
The amount of gas that escapes from a container into a small hole is called effusion.
RateA and rateB are the effusion rates of gases A and B, and MA and MB are their molar mass.
A gas escape is called efffusion.
It escapes from the balloon quickly because it has a low mass.
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