Discussion Forum

1)

A closed tank has two compartments  A and B , both filled with oxygen (assumed to be ideal gas) . The partition separating the two compartments is fixed and is a perfect heat insulator  (Fig .1) . If the old partition is replaced  by a new partition which  can slide and conduct  heat but does not allow the gas to leak across  (Fig. 2) , the volume ( in m³ ) of the compartment  A after the system attains equilibrium is................

Answer:

Option B

Explanation:

Given p₁=5 bar, V₁ =1 m³, T₁=400K

   So,          $n_{1}=\frac{5}{400R}$               (from pV=nRT)

          Similarly , p₂= 1 bar, V₂ = 3 m³, T₂ =300K

        $n_{2}=\frac{3}{300R}$

  Let  at equilibrium the new volume of A will be (1+x)

  So,  the new volume of B  will be (3-x) 

 Now, from the ideal gas equation

       $\frac{p_{1}V_{1}}{n_{1}RT_{1}}=\frac{P_{2}V_{2}}{n_{2}RT_{2}}$

     and at equilibrium (due to conduction of heat )

           $\frac{p_{1}}{T_{1}}=\frac{p_{2}}{T_{2}}$

 So     $\frac{V_{1}}{n_{1}}=\frac{V_{2}}{n_{2}}$ or V₁n₂ = V₂n₁

After putting the values

          $(1+x)\times \frac{3}{300R}=(3-x)\times \frac{5}{400R}$

 or       $(1+x)= \frac{(3-x)5}{4} or 4(1+x)= 15-5x$

 

  or $4+4x=15-5x$    or   $x=\frac{11}{9}$

 Hence, new volume of A i.e, (1+x) will comes as $1+\frac{11}{9}=\frac{20}{9} or 2.22$