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JEE Main 20163 April
Exam
JEE Main
Year
2016
Date
3 April
Subject
Chemistry
Chapter
States of Matter

JEE Main 2016 · 3 April · Chemistry · States of Matter

Two closed bulbs of equal volume (V) containing an ideal gas...

Two closed bulbs of equal volume (V)\left(V\right) 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:

Two closed bulbs of equal volume (V) containing an ideal gas initially at pressure p _ i and temperature T _ 1 are

Options

  1. A2pi(T2T1+T2)2{p}_{i}\left(\dfrac{{T}_{2}}{{T}_{1} + {T}_{2}}\right)
  2. B2pi(T1T2T1+T2)2{p}_{i}\left(\dfrac{{T}_{1}{T}_{2}}{{T}_{1} + {T}_{2}}\right)
  3. Cpi(T1T2T1+T2){p}_{i}\left(\dfrac{{T}_{1}{T}_{2}}{{T}_{1} + {T}_{2}}\right)
  4. D2pi(T1T1+T2)2{p}_{i}\left(\dfrac{{T}_{1}}{{T}_{1} + {T}_{2}}\right)

Correct answer

A. 2pi(T2T1+T2)2{p}_{i}\left(\dfrac{{T}_{2}}{{T}_{1} + {T}_{2}}\right)

Step-by-step solution

Given, two closed bulbs of equal volume(V)\left(V\right) containing ideal gas initially of pressure piand{p}_{i}and temperature T1{T}_{1}, which are connected by a narrow tube of negligible volume.

To find the final pressure Pf{P}_{f}, when one raised to T2{T}_{2}:
Number of moles of gas does not change,
(nT)i=(nT)f{\left({n}_{\text{T}}\right)}_{\text{i}} = {\left({n}_{\text{T}}\right)}_{\text{f}}
PiVRT1+PiVRT1=PfVRT1+PfVRT2\dfrac{{\text{P}}_{\text{i}} \text{V}}{{\text{RT}}_{1}} + \dfrac{{\text{P}}_{\text{i}} \text{V}}{{\text{RT}}_{\text{1}}} = \dfrac{{\text{P}}_{\text{f}} \text{V}}{{\text{RT}}_{\text{1}}} + \dfrac{{\text{P}}_{\text{f}} \text{V}}{{\text{RT}}_{\text{2}}}
2PiT1=PfT1+PfT22 \dfrac{{\text{P}}_{\text{i}}}{{\text{T}}_{\text{1}}} = \dfrac{{\text{P}}_{\text{f}}}{{\text{T}}_{\text{1}}} + \dfrac{{\text{P}}_{\text{f}}}{{\text{T}}_{\text{2}}}
2PiT1=Pf(T1+T2)T1×T2Pf=2Pi(T2T1+T2)\begin{array}{ll}\dfrac{2Pi}{T1} = {P}_{f}\dfrac{( {T}_{1} + {T}_{2} )}{{T}_{1} \times {T}_{2}} \\ {P}_{f} = 2{P}_{i}( \dfrac{{T}_{2}}{{T}_{1} + {T}_{2}} )\end{array}
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