Suppose the three gases were placed at 1.00 atm and 273 K in closed containers, as shown below. A tiny hole is opened in each container, allowing each gas to diffuse into another container filled with neon gas. What is the ratio of the rate of diffusion of nitrogen to the rate of diffusion of oxygen? Is this greater than, less than, or the same as the ratio of the rates of effusion for the two gases?
Ah, gases playing a game of musical chairs, I see! Well, let's see if we can find the ratio of the rate of diffusion for nitrogen to the rate of diffusion for oxygen.
Now, diffusion of gases depends on various factors like molar mass and temperature, but let's assume that the gases are ideal (like me) and that they all have the same temperature of 273 K. In this case, the rate of diffusion is generally inversely proportional to the square root of the molar mass.
So, let's look at the molar masses. Nitrogen has a molar mass of approximately 28 g/mol, while oxygen has a molar mass of about 32 g/mol. Now, since diffusion is inversely proportional to the square root of the molar mass, we can say that the rate of diffusion of nitrogen would be greater than the rate of diffusion of oxygen.
Now, regarding the second part of your question, the ratio of the rates of diffusion would be the same as the ratio of the rates of effusion for the two gases. Why? Well, both diffusion and effusion are processes through which gases move from an area of high concentration to low concentration, and they follow similar principles.
So, to summarize: the ratio of the rate of diffusion of nitrogen to the rate of diffusion of oxygen would be greater than 1 (since nitrogen diffuses faster than oxygen due to its lower molar mass). And this ratio would be the same as the ratio of the rates of effusion for the two gases. Isn't the world of gases fascinating? It's like a big party where everyone is dancing to their own tune!
To determine the ratio of the rate of diffusion of nitrogen (N2) to the rate of diffusion of oxygen (O2), we can refer to Graham's law of diffusion.
Graham's law states that the rate of diffusion of a gas is inversely proportional to the square root of its molar mass.
Let's denote the rate of diffusion of nitrogen as R(N2) and the rate of diffusion of oxygen as R(O2). Since the gases are diffusing into containers filled with neon gas, we can assume that the molar mass of neon (Ne) is much greater than that of nitrogen and oxygen. Therefore, the rate of diffusion of neon can be considered constant.
According to Graham's law, we have the following equation:
R(N2) / R(O2) = sqrt(M(O2) / M(N2))
The molar mass of oxygen (O2) is approximately 32 g/mol, and the molar mass of nitrogen (N2) is approximately 28 g/mol.
Substituting these values into the equation, we get:
R(N2) / R(O2) = sqrt(32 / 28) = sqrt(1.143)
So, the ratio of the rate of diffusion of nitrogen to the rate of diffusion of oxygen is approximately 1.07.
Now, let's consider the ratio of the rates of effusion for nitrogen and oxygen. According to Graham's law of effusion, the rate of effusion is also inversely proportional to the square root of the molar mass.
The same equation can be used:
R(N2) / R(O2) = sqrt(M(O2) / M(N2))
Using the molar masses of 32 g/mol for oxygen and 28 g/mol for nitrogen, we get:
R(N2) / R(O2) = sqrt(32 / 28) = sqrt(1.143)
Thus, the ratio of the rates of effusion for nitrogen to oxygen is also approximately 1.07.
Therefore, the ratio of the rate of diffusion of nitrogen to the rate of diffusion of oxygen is equal to the ratio of the rates of effusion for the two gases.
To find the ratio of the rate of diffusion of nitrogen to the rate of diffusion of oxygen, we first need to understand the factors that affect the rate of diffusion.
The rate of diffusion is determined by Graham's Law, which states that the rate of diffusion of a gas is inversely proportional to the square root of its molar mass.
So, to calculate the ratio of the rate of diffusion of nitrogen (N2) to the rate of diffusion of oxygen (O2), we need to compare the square roots of their molar masses:
1. Find the molar masses of nitrogen (N2) and oxygen (O2):
- Nitrogen: Atomic mass of nitrogen (N) = 14.01 g/mol
Multiply by 2 (since it's N2) => molar mass = 28.02 g/mol
- Oxygen: Atomic mass of oxygen (O) = 16.00 g/mol
Multiply by 2 (since it's O2) => molar mass = 32.00 g/mol
2. Take the square root of the molar masses:
- Square root of 28.02 g/mol = 5.29 g/mol
- Square root of 32.00 g/mol = 5.66 g/mol
3. Calculate the ratio of the rate of diffusion:
- Ratio of nitrogen diffusion rate to oxygen diffusion rate:
= (Rate of diffusion of nitrogen/ Rate of diffusion of oxygen)
= (Square root of molar mass of oxygen / Square root of molar mass of nitrogen)
= (5.66 g/mol / 5.29 g/mol)
= 1.07
The ratio of the rate of diffusion of nitrogen to the rate of diffusion of oxygen is approximately 1.07.
Now, let's compare this ratio to the ratio of the rates of effusion for the two gases.
According to Graham's Law, the rate of effusion (gas escaping through a small opening) is also inversely proportional to the square root of the molar mass of the gas. Therefore, the ratio of the rates of effusion for nitrogen (N2) and oxygen (O2) would be the same as the ratio of their diffusion rates.
Hence, the ratio of the rate of diffusion of nitrogen to the rate of diffusion of oxygen is the same as the ratio of the rates of effusion for the two gases.