AL chem about equilibrium

Consider the equilibrium: 2SO2(g) + O2(g) = 2SO3(g)

When helium is injected into the equilibrium system with the volume of system kept unchanged, the equilibrium would not be shifted. This is because the partial pressures of all components (i.e. SO2, O2 and SO3) remained unchanged. Partial pressure = (Mole fraction) * (Total pressure). The total pressure increases when helium is injected, but the mole fraction of each component decreases simultaneously. These two effects cancel each and thus the partial pressure of each component remains unchanged.

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For example, consider an equilibrium system at constant temperature T contains x mol of SO2, y mol of O2 and z mol of SO3, and the total pressure of the system is PT.

Original partial pressure of SO2 = (Mole fraction of SO2) x (Total pressure)
Original partial pressure of SO2 = [x/(x + y + z)] x PT
Original partial pressure of SO2 = xPT/(x + y + z)

w mol of He is then injected into the system.

According to Avogadro’s law, when P and T are constant, the volume of the system is directly proportional to the number of moles of gaseous in the system.
(New total pressure) : (Original total pressure) = (x + y + z + w) : (x + y + z)
(New total pressure) : PT = (x + y + z + w) : (x + y + z)
Then, new total pressure = PT(x + y + z + w)/(x + y + z)

The new mole fraction of SO2 = x/(x + y + z + w)

New partial pressure of SO2 = (New mole fraction of SO2) x (New total pressure)
New partial pressure of SO2 = [x/(x+ y + z + w)] x [PT(x + y + z + w)/(x + y + z)]
New partial pressure of SO2 = xPT/(x + y + z)
New partial pressure of SO2 = Original partial pressure of SO2
The partial pressure of SO2 remains unchanged.

Using the same method, you can show that the partial pressure of O2 and that of SO3 remain unchanged after the injection of helium into the system.
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