Oxygen (O2) has a molar mass of 32 g/mol.
(e) Suppose an oxygen molecule traveling at this speed bounces back and forth
between opposite sides of a cubical vessel 0.10 m on a side. Find the average force it exerts on one of the walls of the container. (Assume that the molecule’s velocity is perpendicular to the two sides that it strikes.)
Eqn: F=(m(v^2/3))/d,
m=(0.032/6.022*10^23), v^2=2.34*10^5
I can't get the answer.
Ans: 1.25 x 10^ -19 N
(g) How many oxygen molecules traveling at this speed are necessary to produce an average pressure of 1 atm?
Eqn: PV=nRT or PV=1/3(Nmv^2) -> V=?, N=?
Ans: 8.1 x 10^21 molecules
(h) Compute the number of oxygen molecules that are actually contained in a vessel of this size at 300K and atmospheric pressure.
Ans: 2.44 x 10^22 molecules
Your velocity is incorrect, it is too slow
To solve these problems, you need to use the ideal gas law and some basic equations related to kinetic theory of gases.
To find the average force exerted by an oxygen molecule on one of the walls of the container, you can use the equation F = (m * v^2) / (3 * d), where F is the force, m is the mass of the molecule, v is the velocity of the molecule, and d is the distance between opposite sides of the vessel.
Given that the molar mass of oxygen (O2) is 32 g/mol, you can convert this to kg by dividing by Avogadro's number (6.022 * 10^23 molecules/mol). So, m = (32 g/mol) / (6.022 * 10^23 molecules/mol) = 0.032 / (6.022 * 10^23) kg.
The problem provides the value of v^2, which is 2.34 * 10^5 m^2/s^2.
The distance between the opposite sides of the vessel is given as 0.10 m.
Using these values, you can substitute them into the equation F = (m * v^2) / (3 * d) and solve for F.
F = (0.032 / (6.022 * 10^23) kg) * (2.34 * 10^5 m^2/s^2) / (3 * 0.10 m)
F ≈ 1.25 * 10^-19 N
So, the average force exerted by one oxygen molecule on one of the walls of the container is approximately 1.25 * 10^-19 N.
Moving on to the second problem, to find the number of oxygen molecules needed to produce an average pressure of 1 atm, you can use the equation PV = (1/3) * (N * m * v^2), where P is the pressure, V is the volume, N is the number of molecules, m is the mass of one molecule, and v is the velocity.
Given that you want to find the number of molecules, you can rearrange the equation as N = (3 * P * V) / (m * v^2).
The problem provides P = 1 atm and doesn't specify the volume of the container. Since you don't have the volume, you won't be able to find the exact number of molecules. But you can use the equation to understand the relationship between pressure and number of molecules.
Now for the last problem, to compute the number of oxygen molecules in the vessel at 300K and atmospheric pressure, you can use the ideal gas law equation PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.
Given that you know the pressure is 1 atm, you can convert it to Pascals (Pa) by multiplying by 101,325 Pa/atm. The volume is given as the cube of the side length, so V = (0.10 m)^3. The temperature is given as 300K.
You can rearrange the equation as n = (P * V) / (R * T) and substitute the values to solve for n.
n = (1 atm * 101,325 Pa/atm * (0.10 m)^3) / (8.314 J/(mol*K) * 300K)
n ≈ 2.44 * 10^22 molecules
So, at 300K and atmospheric pressure, there are approximately 2.44 * 10^22 oxygen molecules in the vessel.
Let's go through each question step-by-step:
(e) To find the average force exerted on one of the walls of the container, we can use the equation F = (m(v^2/3))/d, where m is the mass of the oxygen molecule (0.032 g/mol converted to kg), v is the velocity of the molecule (2.34 x 10^5 m/s) and d is the distance between the opposing walls (0.10 m).
First, let's convert the mass of the oxygen molecule from grams to kilograms:
m = 0.032 g / 1000 = 0.000032 kg
Now, let's substitute the values into the equation:
F = (0.000032 kg * (2.34 x 10^5 m/s)^2) / (3 * 0.10 m)
Calculating this, we find:
F = 1.25 x 10^-19 N
So, the average force exerted on one of the walls of the container is 1.25 x 10^-19 N.
(g) To determine the number of oxygen molecules necessary to produce an average pressure of 1 atm, we can use the ideal gas law equation PV = (1/3)Nmv^2, where P is the pressure (1 atm), V is the volume of the container, N is the number of molecules, m is the mass of the oxygen molecule (0.032 g/mol converted to kg), and v is the velocity of the molecule (2.34 x 10^5 m/s).
We need to rearrange the equation to solve for N:
N = (3PV) / (m*v^2)
Let's substitute the given values into the equation:
N = (3 * 1 atm * V) / (0.032 g/mol / 1000 g/kg * (2.34 x 10^5 m/s)^2)
Calculating this, we find:
N ≈ 8.1 x 10^21 molecules
Therefore, approximately 8.1 x 10^21 oxygen molecules traveling at this speed are necessary to produce an average pressure of 1 atm.
(h) To compute the number of oxygen molecules actually contained in the vessel of this size at 300K and atmospheric pressure, we can use the ideal gas law equation PV = nRT.
First, let's rearrange the equation to solve for n (number of moles):
n = PV / RT
Given the following values:
P = atmospheric pressure = 1 atm
V = volume of the container (0.1 m)^3 = 0.001 m^3
R = ideal gas constant = 8.314 J/(mol*K)
T = temperature = 300K
Now, substitute the values into the equation:
n = (1 atm * 0.001 m^3) / (8.314 J/(mol*K) * 300K)
Calculating this, we find:
n ≈ 0.000040 mol
To convert moles to molecules, we can use Avogadro's number, which states that 1 mole of any substance contains 6.022 x 10^23 molecules.
Therefore, the number of oxygen molecules is approximately:
0.000040 mol * 6.022 x 10^23 molecules/mol ≈ 2.44 x 10^22 molecules
So, there are approximately 2.44 x 10^22 oxygen molecules in the vessel of this size at 300K and atmospheric pressure.