A sample of N2O effuses from a container in 45 seconds.
How long would it take the same amount of gaseous I2 to effuse from the same container under identical conditions?
See Bob Pursley's answer to a similar question here.
http://www.jiskha.com/display.cgi?id=1224180860
To determine the time it would take for the same amount of gaseous I2 to effuse from the container, we can use Graham's law of effusion. Graham's law states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass.
In this case, we need to compare the molar masses of N2O and I2.
The molar mass of N2O (dinitrogen oxide) is:
M(N2O) = 14.01 g/mol + 14.01 g/mol + 16.00 g/mol = 44.02 g/mol
The molar mass of I2 (diiodine) is:
M(I2) = 126.90 g/mol
According to Graham's law, the ratio of the rates of effusion is given by the square root of the inverse ratio of the molar masses:
Rate of N2O effusion / Rate of I2 effusion = sqrt(M(I2) / M(N2O))
Let's calculate this ratio:
Rate of N2O effusion / Rate of I2 effusion = sqrt(126.90 g/mol / 44.02 g/mol)
= sqrt(2.88)
Now, since the ratio of effusion rates is the same as the ratio of times, we can calculate the time it would take for I2 to effuse as follows:
Time for I2 to effuse = (Time for N2O to effuse) / sqrt(2.88)
Given that the time for N2O to effuse is 45 seconds, we can substitute this value into the equation:
Time for I2 to effuse = 45 s / sqrt(2.88)
Calculating this, we get:
Time for I2 to effuse ≈ 45 s / 1.7
Hence, the time it would take for the same amount of gaseous I2 to effuse from the container under identical conditions is approximately 26.47 seconds.
To determine how long it would take for the same amount of gaseous I2 to effuse from the same container under identical conditions, we need to use Graham's Law of Effusion.
Graham's Law states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass. Mathematically, it can be represented as:
Rate1 / Rate2 = √(Molar mass2 / Molar mass1)
In this case, we are comparing the effusion rates of N2O and I2. Let's denote the effusion rate of N2O as RateN2O and the effusion rate of I2 as RateI2.
Since the sample sizes are the same, we can write:
RateN2O / RateI2 = 1
Now, let's substitute the molar masses of each gas:
RateN2O / RateI2 = √(Molar massI2 / Molar massN2O)
The molar mass of N2O is approximately 44 g/mol, and the molar mass of I2 is approximately 253.8 g/mol.
RateN2O / RateI2 = √(253.8 g/mol / 44 g/mol)
Simplifying this equation, we have:
RateN2O / RateI2 = √(5.766)
To solve for the ratio of effusion rates, we square both sides of the equation:
(RateN2O / RateI2)^2 = 5.766
Now, we can solve for RateI2 by multiplying both sides of the equation by RateI2:
RateN2O^2 = 5.766 * RateI2
Finally, to find the effusion time for I2, we can use the relationship between effusion rate and time: Effusion time = Volume / Rate.
Since the volume and other conditions are identical, the ratio of their effusion times will be the inverse of their effusion rates:
Effusion timeN2O / Effusion timeI2 = RateI2 / RateN2O
Plugging in the values we have, we get:
Effusion timeN2O / Effusion timeI2 = 1 / √5.766
Therefore, the effusion time for I2 would be the inverse of the square root of 5.766 times the effusion time for N2O.