At a given temperature and pressure, the volume fraction of an ideal gas is equal to its mole fraction in a mixture of ideal gases- it is true or false?

Explanation

Suppose, at a given temperature (T) and pressure (P), the volumes of two ideal gases are `V_(A) and V_(B)`, respectively, and `n_(A) and n_(B)` are their respective number of moles. Let at the same temperature and pressure, the total volume of the mixture of these two gases is V.
`therefore`According to Amagat's law of partial volume, `V=V_(A)+V_(B)`
So, the partial volume of A in the mixture `=(V_(A))/(V)` and its mole fraction`=(n_(A))/(n_(A)+n_(B))` ltBrgt now, applying ideal gas equation to each component gas as well as the gas mixture, we get
`PV_(A)=n_(A)RT` . . . [1]
`PV_(B)=n_(B)RT` . . .[2]
`PV=(n_(A)+n_(B))RT` . . . [3]
Dividing equaton no. [1] by equation no. [3], we have ,brgt `(PV_(A))/(PV)=(n_(A))/(n_(A)+n_(B))" "therefore (V_(A))/(V)=(n_(A))/(n_(A)+n_(B))`
That is, the volume fraction of A=the mole fraction of A. Similarly, `(V_(B))/(V)=(n_(B))/(n_(A)+n_(B))`
[Dividing equation no. [2] by equation no[3] ].
tha is, the volume fraction of B=the mole fraction of B. Therefore, at a given temperature and pressure, the volume fraction of an ideal gas in a mixture of ideal gases is equal to its mole fraction. hence, the given statement is true.