Question

Suppose that 0.25 mole of gas C was added to the mixture without changing the total pressure of the mixture. How does the addition of C to the mixture change the mole fraction of gas A?

Answer 1

the answer is decreases it

Answer 2

Here we have to get the effect of addition of 0.25 moles of gas C on the mole fraction of gas A in a mixture of gas having constant pressure.

On addition of 0.25 moles of C gas, the mole fraction of gas A will be
`(moles of gas A)/(moles of gas A + 0.25)`.

The partial pressure of gas A can be written as
`P_(A)` =
`x_(A)`×P (where
`x_(A)` is the mole fraction of gas A present in the mixture and P is the total pressure of the mixture.

The mole fraction of gas A in a mixture of gas A and C is =
`(moles of gas A)/(moles of gas A + moles of gas C)` and
`(moles of gas C)/(moles of gas A + moles of gas C)` respectively.

Thus on addition of 0.25 moles of C gas, the mole fraction of gas A will be
`(moles of gas A)/(moles of gas A + 0.25)`.

Which is different from the initial state.