# Effect on equilibrium, addition of inert gas

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1. Aug 27, 2015

### 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)

2. Aug 27, 2015

### Staff: Mentor

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!

3. Aug 27, 2015

### Staff: Mentor

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

4. Aug 27, 2015

### 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.

5. Aug 27, 2015

### 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.

6. Aug 27, 2015

### Staff: Mentor

7. Aug 27, 2015

### Staff: Mentor

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

8. Aug 28, 2015

### 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?

9. Aug 28, 2015

### Staff: Mentor

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

10. Aug 28, 2015

### Titan97

The vessel is maintained at constant pressure.

11. Aug 28, 2015

### Staff: Mentor

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).

12. Aug 28, 2015

### 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.

13. Aug 29, 2015

### Staff: Mentor

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

14. Aug 29, 2015

### Titan97

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