a mixture of O2 and gas "Y" (mol wt. 80) in the mole ratio a:b has a mean molecular weight 40.. what would be mean molecular weight,if the gases are mixed in the ratio b:a under identical conditions? (gases are non reacting)
72
Option d
To find the mean molecular weight of a mixture of gases, you need to consider the mole ratio of each gas and their respective molecular weights.
Let's consider the given information. We have a mixture of oxygen gas (O2) and gas "Y" in the mole ratio a:b, with a mean molecular weight of 40. The molecular weight of gas "Y" is given as 80.
We can start by assigning variables to the molar ratios:
Let n(O2) be the mole ratio of O2.
Let n(Y) be the mole ratio of gas Y.
Since the total mole ratio of gases is a + b, we can calculate the mole fractions:
X(O2) = n(O2) / (n(O2) + n(Y))
X(Y) = n(Y) / (n(O2) + n(Y))
The mean molecular weight (MW) of the mixture can be calculated using the mole fractions and molecular weights:
MW = X(O2) * MW(O2) + X(Y) * MW(Y)
Given that MW(O2) is 32 (the molecular weight of O2), and MW(Y) is 80, we have:
40 = (n(O2) / (n(O2) + n(Y))) * 32 + (n(Y) / (n(O2) + n(Y))) * 80
Now, let's consider the second case where the gases are mixed in the ratio b:a. We'll assign new variables to these mole ratios:
Let n'(O2) be the mole ratio of O2 in the new mixture.
Let n'(Y) be the mole ratio of gas Y in the new mixture.
The mole ratios in the new mixture are: n'(O2) = b and n'(Y) = a.
Using the same approach, we can calculate the new mean molecular weight (MW'):
MW' = (n'(O2) / (n'(O2) + n'(Y))) * MW(O2) + (n'(Y) / (n'(O2) + n'(Y))) * MW(Y)
Substituting in the known values:
MW' = (b / (b + a)) * 32 + (a / (b + a)) * 80
Therefore, the new mean molecular weight for the gases mixed in the ratio b:a is (b / (b + a)) * 32 + (a / (b + a)) * 80.