Verified Solution • 2025–26 Session moderate

QJEEM16Ch895c: JEE Main 2016 Chemistry - Class 11 JEEM16Ch895c Solution: Two closed bulbs of equal volume containing an ideal gas…

Two closed bulbs of equal volume (V)(V) containing an ideal gas initially at pressure pi{p_i} and temperature T1{T_1} 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.{T_2}. The final pressure pf{p_f} is :

JEE Main 2016 (Offline) Chemistry - Gaseous State Question 52 English

Multiple Choice Options

  • A 2pi(T2T1+T2)2{p_i}\left( {{{{T_2}} \over {{T_1} + {T_2}}}} \right)
  • B 2pi(T1T2T1+T2)2{p_i}\left( {{{{T_1}{T_2}} \over {{T_1} + {T_2}}}} \right)
  • C pi(T1T2T1+T2){p_i}\left( {{{{T_1}{T_2}} \over {{T_1} + {T_2}}}} \right)
  • D 2pi(T1T1+T2)2{p_i}\left( {{{{T_1}} \over {{T_1} + {T_2}}}} \right)

Step-by-Step Verified Solution

For a given mass of an ideal gas, the volume and amount (moles) of the gas are directly proportional if the temperature and pressure are constant. i.ei.e

V∝nV \propto n

Hence in the given case.

Initial moles and final moles are equal (nT)i=(nT)f{\left( {{n_T}} \right)_i} = {\left( {{n_T}} \right)_f}

PiVRT1+PiVRT1{{{P_i}V} \over {R{T_1}}} + {{{P_i}V} \over {R{T_1}}}

=PfVRT1+PfVRT2= {{{P_f}V} \over {R{T_1}}} + {{{P_f}V} \over {R{T_2}}}

2PiT1=PfT1+PfT22{{{P_i}} \over {{T_1}}} = {{{P_f}} \over {{T_1}}} + {{{P_f}} \over {{T_2}}}

⇒Pf=2 PiT2T1+T2\Rightarrow {P_f} = {{2\,{P_i}{T_2}} \over {{T_1} + {T_2}}}

Related Learning Resources & Exam Preparation