# How Many Oxygen Molecules Collide with a Wall Per Second at Standard Pressure?

• rhochipi
In summary, the number of collisions per second between oxygen molecules and a 19.0 cm^2 wall can be calculated using the equation N/T = PA/(2mv), where N is the number of molecules, T is the time duration, P is the pressure, A is the area, m is the mass, and v is the velocity. This approach treats the molecules as billiard balls and does not take into account their spinning or vibrating. A simple calculation using the molar mass of oxygen yields an answer of approximately 6.14 collisions per second.
rhochipi
Oxygen Molecules collide with a 19.0 cm^2 wall. Assume that all molecules travel with a speed of 490 m/s and strike head on. How many collisions are there per second if the oxygen pressure is 1.00atm?

Pressure = Force/Area
Force = 2*Mass*Velocity
Collision Frequency = PI*Molecular_Diameter^2*Average_Velocity*Number_of_Molecules_per_Unit_Volume
Collision Frequency also = Velocity_Average/Mean_Free_Path
Pressure*Volume = Number_of_Molecules*Boltzmann's_Constant*Temperature
Average Velocity = sqrt(8*Boltzmann's Constant*Temperature/(PI*Mass))

A previous problem calculated the number of molecules per unit volume from the ideal gas law: N/V = P/(KbT). However, I need temperature, which I thought of getting from the average velocity equation: v_Ave = sqrt(8KbT/(PI*m)). However, I don't know the mass, so I've become stumped.

A friend worked with momentum. By Kinetic Theory, p = 2mv = Force*Time_Duration, for all molecules, N: p = 2mvN = Force*Time. By equating that force with that from P = FA, N/T = PA/(2mv), where T = 1 second, P = 1.013e5 Pa, v = 490m/s, A = .0019 m^2. But, he assumed m to be the molar mass of Oxygen, 32g. When I tried it, I received an answer of 6.1 collisions per second, which didn't make sense and was obviously wrong.

Any Suggestions?

Maybe its asking for a real simple minded approach that treats the molecules as they were billiard balls, and not spinning, vibrating, etc.

I would just use mass here as 32gm/N, convert pressure in atm to pascals, and that in turn to total force over the 19 sq cm plate, and use your elastic collsion formula to see what the number is/second I'd not solve for mass explicitly but express it in terms of N/s (I got about 6.14 N/s where
N=6.02E23)

Thank you for your question. It seems like you and your friend are on the right track in using the ideal gas law and kinetic theory to calculate the collision frequency of oxygen molecules with a wall. However, there are a few things that need to be clarified in your calculations.

Firstly, the mass that should be used in the equations is the mass of a single oxygen molecule, not the molar mass of oxygen. The molar mass is the mass of one mole of oxygen molecules, which contains Avogadro's number (6.022 x 10^23) of molecules. In this problem, we are interested in the collision frequency of a single molecule, so we need to use its individual mass.

Secondly, the average velocity equation you used (v_Ave = sqrt(8KbT/(PI*m))) is actually the root mean square velocity, which is the average velocity of all the molecules in a gas. In this problem, we are only interested in the velocity of the molecules that are colliding with the wall, which is equal to the average velocity (v_Ave) divided by 2. This is because only half of the molecules in a gas are moving towards the wall at any given time.

With these corrections, the calculation should look like this:

1. Calculate the number of oxygen molecules per unit volume:
N/V = P/(KbT)
= (1.013e5 Pa)/(1.38e-23 J/K * 273 K)
= 2.3 x 10^25 molecules/m^3

2. Calculate the average velocity of the oxygen molecules:
v_Ave = sqrt(8KbT/(PI*m))
= sqrt((8 * 1.38e-23 J/K * 273 K)/(PI * 5.31e-26 kg))
= 493 m/s

3. Calculate the velocity of the molecules colliding with the wall:
v_wall = v_Ave/2 = 493 m/s / 2 = 246.5 m/s

4. Calculate the collision frequency:
Collision Frequency = Velocity_Average/Mean_Free_Path
= v_wall/2r
= (246.5 m/s)/(2 * (0.5 * 19.0 cm))
= 6.5 x 10^6 collisions/second

So, the collision frequency of oxygen molecules with a 19.0 cm^2 wall at a pressure of

## 1. What is the collision frequency of a gas?

The collision frequency of a gas refers to the number of collisions that occur between gas molecules per unit time. It is a measure of how often gas molecules collide with each other.

## 2. How is collision frequency related to gas pressure?

Collision frequency is directly proportional to gas pressure. This means that as gas pressure increases, the collision frequency also increases. This is because higher pressure means there are more gas molecules in a given space, leading to more collisions.

## 3. What factors affect the collision frequency of a gas?

The collision frequency of a gas is affected by factors such as temperature, pressure, and the size of the gas molecules. Higher temperatures and pressures lead to increased collision frequency, while larger gas molecules have lower collision frequencies.

## 4. How is collision frequency measured?

Collision frequency can be measured using various techniques, such as the kinetic theory of gases, which calculates collision frequency based on the size and speed of gas molecules. It can also be measured experimentally using devices such as the gas collision apparatus.

## 5. Why is collision frequency important in understanding gas behavior?

Collision frequency is important in understanding gas behavior because it affects the physical properties of gases, such as pressure, temperature, and volume. It also plays a crucial role in chemical reactions involving gases, as the frequency of collisions between gas molecules determines the rate of the reaction.

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