A sample of gas (2.0 mmol) effused through a pinhole in 5.0 s. It will take __________ s for the same amount of to effuse under the same conditions.
same amount of what ?
4.2
To find the time it takes for the same amount of gas to effuse under the same conditions, we can use Graham's Law of Effusion. According to Graham's Law, the rate of effusion of a gas is inversely proportional to the square root of its molar mass.
In this case, we're given that a sample of gas (2.0 mmol) effused through a pinhole in 5.0 s. Let's denote the time it takes for the same amount of gas to effuse as t.
We know that the rate of effusion is inversely proportional to the square root of the molar mass, so we can set up the following proportion:
Rate1 / Rate2 = √(Molar mass2 / Molar mass1)
Since we're trying to find t, and we're given that the sample size (moles) is the same, the molar masses cancel out:
1 / t = √(molar mass1 / molar mass1)
Simplifying this equation gives us:
1 / t = 1 / √1
Simplifying further:
1 / t = 1
To solve for t, we can reciprocate both sides of the equation:
t = 1 / 1
Therefore, t = 1.0 s.
So, it will take 1.0 second for the same amount of gas to effuse under the same conditions.