JEE Main 2016ChemistryStates of MatterHardMCQ

JEE Main 2016States of Matter Question with Solution

JEE Main 2016 (03 Apr)

Question

Two closed bulbs of equal volume V containing an ideal gas initially at pressure pi and temperature T1 are connected through a narrow tube of negligible volume, as shown in the figure below. The temperature of one of the bulbs is then raised to T2. The final pressure Pf is:

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Show full solutionCorrect option: A
Correct answer
A2piT2T1+T2

Step-by-step explanation

Given, two closed bulbs of equal volumeV containing ideal gas initially of pressure pi  and  temperature T1, which are connected by a narrow tube of negligible volume.

To find the final pressure Pf, when one raised to T2:
Number of moles of gas does not change,
  nTi=nTf
P i V RT 1 + P i V RT 1 = P f V RT 1 + P f V RT 2
2 P i T 1 = P f T 1 + P f T
2PiT1=Pf(T1+T2)T1×T2Pf=2Pi(T2T1+T2)

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About this question

This is a previous-year question from JEE Main 2016, covering the States of Matter chapter of Chemistry. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.