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

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  $\frac{moles of gas A}{moles of gas A + 0.25}$.

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 = $\frac{moles of gas A}{moles of gas A + moles of gas C}$ and $\frac{moles of gas C}{moles of gas A + moles of gas C}$ respectively.

Thus on addition of 0.25 moles of C gas, the mole fraction of gas A will be  $\frac{moles of gas A}{moles of gas A + 0.25}$.

Which is different from the initial state.

Answer 2

the answer is decreases it