JEE (Advanced) 2026, Paper 1, Chemistry, Section 3 — the answer is a numerical value, truncated or rounded to two decimal places: +4 if only the correct value is entered, 0 in all other cases.
Q.9 Two cylinders, both fitted with frictionless pistons, are filled with mixtures of He and Ar gases. In the first cylinder, the masses of He and Ar are m1 and m2, respectively. In the second cylinder, the masses of He and Ar are m2 and m1, respectively. The molar mass of Ar is 10 times the molar mass of He. The external pressure applied by the piston on the first cylinder needs to be 5 times that on the second cylinder so that the volume of the gas mixtures in both the cylinders are equal at the same temperature. Assuming He and Ar behave like ideal gases, the value of (m1/m2) is ____.
Two cylinders, two gases, one unknown ratio. Press START EXPERIMENT — the whole investigation runs by itself: load the masses → count the moles → apply PV = nRT → scan the ratio → lock the pistons → check → change the temperature → audit the traps → answer.
Two cylinders, swapped masses, one unknown ratio. The question's values are loaded — press START EXPERIMENT.
Moles are shown in units of m₂/(10 MHe), so the numbers stay simple: a mass m₂ of He counts 10, the same mass of Ar counts 1.
computed from n₁/n₂ at equal V and T — the lab looks for where the curve meets 5
He and Ar moles, stacked, at the current trial ratio
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Nothing investigated yet.
Change any number and watch the balance move: the mass ratio, the molar-mass ratio and the pressure ratio the piston must hold. The general answer is m₁/m₂ = (p·k − 1)/(k − p), where k = MAr/MHe and p = P₁/P₂ — it exists only when 1/k < p < k.
Run the experiment — the solution opens with the answer.
The experiment has not finished yet. Press START and let it run to the end — or use Reveal anyway in the header if you are revising.
Let MHe = M, so MAr = 10M. A gram of He is ten times as many moles as a gram of Ar.
For ideal gases PV = nRT. With V and T equal in both cylinders, the pressure is simply proportional to the total moles.
n₁ = m₁/M + m₂/10M and n₂ = m₂/M + m₁/10M. Put n₁ = 5 n₂.
5m₁ = 49m₂, so m₁/m₂ = 49/5.
R, T and V cancel — the temperature does not matter as long as it is the same for both.
n₁/n₂ = (98 + 1)/(10 + 9.8) = 99/19.8 = 5 ✓. Reading the pressure ratio backwards (P₂ = 5P₁) gives 1/9.8 = 0.10; equating masses instead of moles gives no solution at all; taking the pressure ratio itself as the answer gives 5 — all wrong.
For a student opening it for the first time. The whole investigation runs by itself after one press.
Two cylinders hold He and Ar with their masses swapped. Find m₁/m₂ so that the first needs 5 times the pressure at the same volume.
The question's values are loaded — MAr = 10 MHe, P₁ = 5 P₂. You do not need to change anything.
Try the real molar masses, another pressure ratio or another temperature. The lab tells you honestly when a change takes you off the question — or when no ratio can work. ↺ RESTORE QUESTION CONDITIONS puts everything back.
Press ▶ START EXPERIMENT. There is nothing else to click (about a minute; 2× makes it faster).
Masses turn into molecules, the molecules push on the pistons, the lab scans the ratio until P₁/P₂ reaches 5, locks the pistons level and checks the result.
When the audit is complete the lab says ✓ EXPERIMENT COMPLETE and reveals m₁/m₂. Badges unlock on the way.
In the Pressure explorer drag the three sliders and press SOLVE. Then read the Detailed solution.
Press ▶ REPLAY EXPERIMENT, or change a value and press ⚙ RUN WITH CUSTOM VALUES. CLASSROOM MODE enlarges everything for a projector.
| Exam | JEE Advanced 2026, Paper 1, Chemistry, Section 3, Q.9 |
|---|---|
| Chapter | States of Matter: Gases and Liquids |
| Topic | Ideal gas equation - pressure, moles and gas mixtures at equal volume and temperature |
| Difficulty | Easy |
| Answer | — run the experiment |