Effect on equilibrium, addition of inert gas

[Titan97]

If inert gases are added at constant volume to an equilibrium mixture, the total pressure increases. But the mole fraction of the gaseous reactants or products decrease.
Partial pressure $p=x\cdot P$ where P is total pressure and $x$ is mole fraction.
How can you say that $p$ remains constant?
(I read that the equilibrium is not affected since partial pressure of the participants of equilibrium remains constant)

 

[Borek]

Assume initial pressure of a gas to be p₀. Assume 1 mole of the gas to make calculations easier. Assume volume V. Assume you have added n moles of an inert gas. What is the new total pressure? What is the mole fraction of the original gas? What is its partial pressure? Don't guess - calculate!

 

[Chestermiller]

The mole fractions decrease by the same factor that the total pressure increases, so that the partial pressures remain the same. True or false: each gas in an ideal gas mixture exhibits PVT behavior as if the other gases are not present.

Chet

 

[Titan97]

Initial mole fraction = $\frac{1}{N}$
Initial total pressure = $P_0$
$P_0=\frac{NRT}{V}$
$p=x.P_0=RT/V$
Let n moles of argon be added.
New mole fraction = $\frac{1}{n+N}$
New total pressure = $\frac{(n+N)RT}{V}$
$p'=\frac{RT}{V}$
Hence there is no change in partial pressure.

 

[Titan97]

@Chestermiller Let the volume occupied by one gas be $v$. Won't $v$ change if more gases are added? Since adding more gases means lesser volume that can be occupied.

 

[Borek]

How can you say that pp remains constant?

Hence there is no change in partial pressure.

I assume you have just answered your own question

 

[Chestermiller]

@Chestermiller Let the volume occupied by one gas be $v$. Won't $v$ change if more gases are added? Since adding more gases means lesser volume that can be occupied.

The molecules of each gas occupy the entire volume. The partial pressures of the various gases add up to the total pressure.

 

[Titan97]

What if noble gases are added under constant pressure?
Let the equation be A(g)↔B(g)+C(g) (let ↔ be reversibility sign)
In this case, mole fraction decreases while total pressure remains constant. Volume is also constant.
Let moles of A be $n_1$,
moles of B=$n_2$
and moles of C=$n_3$.
Let moles of argon be $a$ and total number of moles before addition of argon be $n$.
Initially, $K=\frac{n_2.n_3}{n_1(n)}P$
After addition of argon, $K'=\frac{n_2.n_3}{n(n+a)}P$
(Pressure remains constant).
Hence K decreases which means products are favoured.
Is this correct?

 

[Borek]

(Pressure remains constant).

If you add argon, pressure doesn't stay constant.

 

[Titan97]

The vessel is maintained at constant pressure.

 

[Borek]

Then you are just diluting reacting gases, lowering their partial pressures - and that's a completely different case than the one you started with (you have mentioned constant volume in the very first post).

 

[Titan97]

Yes. I was just thinking about a new situation. But in post 8, I found that K changed. But K can only change if temperature changes.

 

[Borek]

in post 8, I found that K changed

No, you found that Q (reaction quotient) has changed, not K. After you change volume, equilibrium will shift till Q=K.

 

[Titan97]

That solves all my doubts on Le Chatelier's principles.