1. A spaceship is found to have a tiny leak that allows the effusion of air into space. Air in the spaceship is a mixture of nitrogen (N2), oxygen (O2), argon (Ar), and trace amounts of other gases.
a. Predict which of the three main gases will have the lowest rate of effusion and which will have the highest rate. Explain your answer. (1 point)
b. Predict how the percentages of the gases in the air will change over time. (1 point)
c. Argon is found to effuse through the hole at a rate of 1.6 × 10–3 mol in 215 s. How much O2 would effuse through the hole in the same amount of time? (2 points)
d. 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? (2 points)
N2=
O2= all equal N2
Ar=
e. Suppose the three gases in the previous problem were initially in a container that was at very low temperature. Which of the gases would you expect to show the most deviation from ideal gas behavior? Explain your answer. (1 point)
for c it is 0.001872 moles
Argon is found to effuse through the hole at a rate of 1.6 × 10–3 mol in 215 s. How much O2 would effuse through the hole in the same amount of time?
a. The lowest rate of effusion would most likely be for argon (Ar) since it has the largest atomic mass among the three main gases. The highest rate of effusion would be for nitrogen (N2) since it has the smallest atomic mass.
b. Over time, the percentages of the gases in the air would change as the gases effuse through the hole. The gas with the highest rate of effusion (nitrogen) would decrease in percentage, while the gas with the lowest rate of effusion (argon) would decrease at a slower rate. The other gases would also effuse, but their percentages would be much smaller, so their changes would be negligible.
c. To determine the amount of O2 that would effuse through the hole in the same amount of time as argon, we can use the ratio of their effusion rates:
Molar ratio of effusion rates = Molar ratio of gas rate = (rate of gas 1 / rate of gas 2)
Rate of O2 effusion = (rate of Ar effusion) x (rate of O2 / rate of Ar)
Rate of O2 effusion = (1.6 × 10–3 mol in 215 s) x (rate of O2 / rate of Ar)
d. The ratio of the rate of diffusion of nitrogen to the rate of diffusion of oxygen depends on the size of their respective molecules. Since nitrogen and oxygen have the same number of atoms (two each), they would have similar rates of diffusion. Therefore, the ratio of the rate of diffusion of nitrogen to the rate of diffusion of oxygen would be around 1:1. This ratio is the same as the ratio of the rates of effusion for the two gases since they have the same sizes.
e. Among the three gases, argon would be expected to show the most deviation from ideal gas behavior when placed at very low temperature. This is because at low temperatures, the intermolecular forces between particles become more significant, and argon has the largest atomic mass among the three gases. The larger mass leads to stronger intermolecular forces and a higher likelihood of deviations from ideal gas behavior.
a. The rate of effusion is determined by the molar mass of the gas and its velocity. According to Graham's law of effusion, the rate of effusion of a gas is inversely proportional to the square root of its molar mass. Therefore, the gas with the lowest molar mass will have the highest rate of effusion, and the gas with the highest molar mass will have the lowest rate of effusion. In this case, oxygen (O2) has the lowest molar mass (32 g/mol), followed by nitrogen (N2) (28 g/mol), and argon (Ar) has the highest molar mass (40 g/mol). Hence, oxygen will have the highest rate of effusion, while argon will have the lowest rate of effusion.
b. Over time, the percentage of gases in the air will change due to effusion. Since the rate of effusion is inversely proportional to the square root of the molar mass, gases with lower molar masses will effuse more quickly, causing their percentage in the air to decrease. In contrast, gases with higher molar masses will effuse more slowly, leading to an increase in their percentage. Therefore, the percentage of oxygen (O2) will increase, while the percentages of nitrogen (N2) and argon (Ar) will decrease.
c. To find the amount of O2 that would effuse through the hole in the same amount of time, we can use the ratio of their effusion rates, which is equal to the square root of the ratio of their molar masses. The molar mass of argon (Ar) is 40 g/mol, and its effusion rate is known to be 1.6 × 10–3 mol in 215 s. Therefore, we can set up the following proportion:
(1.6 × 10–3 mol Ar) / (215 s) = (x mol O2) / (215 s)
Simplifying, x = (1.6 × 10–3 mol Ar) * (32 g/mol O2) / (40 g/mol Ar) ≈ 1.28 × 10–3 mol O2
Therefore, approximately 1.28 × 10–3 mol of O2 would effuse through the hole in the same time.
d. The ratio of the rate of diffusion of nitrogen (N2) to the rate of diffusion of oxygen (O2) can be determined using Graham's law of diffusion, which states that the rate of diffusion is inversely proportional to the square root of the molar mass. Since the molar masses of N2 and O2 are the same, their diffusion rates will be equal, and the ratio will be 1:1. This ratio is the same as the ratio of their rates of effusion, as given by Graham's law of effusion.
e. When gases are at very low temperatures and high pressures, they do not conform to ideal gas behavior due to the intermolecular forces between the gas particles becoming significant. The ideal gas law assumes that the gas particles do not interact with each other. As a result, the gas that is most likely to deviate from ideal gas behavior is the one with the highest molar mass, as intermolecular forces are generally stronger between heavier molecules. In this case, argon (Ar) has the highest molar mass, so it would be expected to show the most deviation from ideal gas behavior.
for c i think it would be 206.7 seconds
All of these use Graham's Law. Look that up on Google. I will get you started with 1a.
molar mass = mm
mm N2 = 28
mm O2 = 32
mm Ar = about 40
Graham's Law is (rate gas1/rate gas 2) = [sqrt (mm gas 2/mm gas 1)]
The heaviest gas (Ar) will be the slowest and the lightest gas (N2) will be the fastest.
Post your work if you get stuck.