SCIENCE · INVESTIGATE AND EXPLAIN

Recalculate partial pressures after a gas-volume perturbation

An immediate pressure change and a later compositional relaxation are distinct states.

Learning outcome: Compare volume settings at one Kp. Compute both partial pressures from final amounts, recover the normalized quotient and contrast the immediate dilution with final relaxation.

Starting recap

Refresh dimensionless activities, component balances, admissible roots, normalized partial pressures and independent versus shared measurement errors.

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Try

For .100 mol A2 at 300 K in 2 L with Kp=1 and p°=100 kPa, solve 4x²RT/(Vp°)=.100−x rather than holding final mole fractions fixed.

Open the experiment

Predict

Will increasing volume increase the dissociated fraction without changing the temperature-dependent Kp?

Your prediction and reason

Experiment

Compare volume settings at one Kp. Compute both partial pressures from final amounts, recover the normalized quotient and contrast the immediate dilution with final relaxation.

Separate gas-volume response from a changing constant

Worked teaching model · synthetic values · Fictional ideal-gas A2⇌2A, initial .100 mol A2 and no A at 300 K. R=8.314 kPa L/(mol K), pressure standard100 kPa. Extent is mol; Q=(pA/p°)²/(pA2/p°). Closed isothermal vessel; no real pressure experiment.

Vessel volume: 2 L Dimensionless Kp: 1 ratio

Reset keeps your prediction, captures and practice history.

Dissociated fraction versus volume: solid

Axes rescale. A quotient comparison, inventory intersection and relaxation trajectory have different horizontal meanings; inspect labels rather than treating every curve as elapsed time.

All calculated plot coordinates

Dissociated fraction versus volume
fraction of volume range 1 to 5 L	fraction of original A2 dissociated
0	0.2704211
0.01666667	0.2778618
0.03333333	0.2849953
0.05	0.2918488
0.06666667	0.2984462
0.08333333	0.3048079
0.1	0.3109519
0.1166667	0.3168941
0.1333333	0.3226485
0.15	0.3282278
0.1666667	0.3336431
0.1833333	0.3389046
0.2	0.3440214
0.2166667	0.3490018
0.2333333	0.3538533
0.25	0.3585829
0.2666667	0.3631969
0.2833333	0.3677011
0.3	0.3721008
0.3166667	0.3764011
0.3333333	0.3806064
0.35	0.3847211
0.3666667	0.3887491
0.3833333	0.392694
0.4	0.3965594
0.4166667	0.4003485
0.4333333	0.4040642
0.45	0.4077095
0.4666667	0.4112869
0.4833333	0.4147989
0.5	0.418248
0.5166667	0.4216364
0.5333333	0.4249662
0.55	0.4282393
0.5666667	0.4314577
0.5833333	0.4346231
0.6	0.4377372
0.6166667	0.4408018
0.6333333	0.4438182
0.65	0.4467879
0.6666667	0.4497125
0.6833333	0.4525931
0.7	0.455431
0.7166667	0.4582275
0.7333333	0.4609838
0.75	0.4637009
0.7666667	0.4663798
0.7833333	0.4690217
0.8	0.4716275
0.8166667	0.4741981
0.8333333	0.4767344
0.85	0.4792373
0.8666667	0.4817076
0.8833333	0.4841462
0.9	0.4865537
0.9166667	0.488931
0.9333333	0.4912787
0.95	0.4935975
0.9666667	0.4958881
0.9833333	0.4981511
1	0.5003872

Calculated under the declared model
Quantity	Value	Unit
Dissociation extent	0.03585829	mol
A2 remaining	0.06414171	mol
A formed	0.07171659	mol
A2 partial pressure	79.99112	kPa
A partial pressure	89.43776	kPa
Recovered dimensionless Kp	1	dimensionless

Larger volume favors the more numerous gas particles here without changing Kp at the fixed temperature.

Amounts change after relaxation, so simply dividing the old partial pressures by a volume factor gives the immediate disturbed state, not the final equilibrium. The pressure standard is explicit: substituting unnormalized pascals into a kPa-based expression would change the numerical quotient incorrectly. Inert gas at fixed volume does not change these ideal partial pressures.

These calculated packets are not performed chemical measurements. Keep feasible inventories, activity conventions, fitting observations and independent checks distinct.

Larger volume favors the more numerous gas particles here without changing Kp at the fixed temperature.

Observation to attach to a capture

Captures preserve recipes and replay their calculated results. They are synthetic model evidence, not physical measurements.

Understand

Build the inventory before the pressure quotient

Begin with .100 mol of fictional A2 and no A. For molar extent x, remaining A2 is .100−x and produced A is 2x. At fixed T and V, partial pressures are their amounts times RT/V. The atom-unit balance 2n(A2)+n(A)=.200 mol is preserved.

Using p°=100 kPa, the dimensionless quotient is (pA/p°)²/(pA2/p°). Substitute the inventories to obtain Kp=4x²RT/[Vp°(.100−x)]. Here R=8.314 kPa L/(mol K) and T=300 K. Using pascals for one pressure and kilopascals for another would not be a harmless change of notation.

Distinguish disturbance from re-equilibration

If a previously equilibrated vessel instantly doubles volume at fixed amounts, each partial pressure halves and this quotient halves. It falls below Kp, favoring dissociation. Subsequent reaction changes the amounts, so the immediate halved pressures are not the final answer.

A larger final volume yields a larger dissociated fraction within this ideal model. The same Kp can describe both final states because their pressures and compositions differ. By contrast, adding ideal inert gas at unchanged volume and temperature leaves each reacting species partial pressure unchanged.

Transfer and check

Consider a different instantaneous disturbance: halve an equilibrated vessel volume at the same temperature. Partial pressures double, so Q doubles, favoring association. State which quantities change immediately and which only change as reaction proceeds. Then explain why heating cannot be analyzed from this volume argument alone: temperature may change Kp, and the reaction enthalpy information is not supplied.

Source and evidence boundary

Conceptual reference checked 30 October 2026: OpenStax Chemistry 2e. These original hypothetical packets use explicitly defined activities, inventories and response assumptions. No material is mixed, heated, pressurized or measured. Numerical compatibility does not certify practical technique or uniquely establish a real chemical mechanism. Retain your prediction, feasible inventory, quotient check and changed-case interpretation.

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Public practice checks a specific claim. It does not establish independent mastery.

Why does ideal inert gas added at fixed volume not produce the same effect as vessel expansion?

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Previous: Solve association while rejecting impossible roots Next: Close a shared inventory across two equilibrium relations

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