Discussion Forum

1)

In the figure, a container is shown to have a movable (without friction) position on top. The container and the piston are all made of perfectly insulating material allowing no heat transfer between outside and inside the container. The container is divided into two compartments by a rigid partition made of a thermally conducting material that allows slow transfer of heat. The lower compartment of the container is filled with 2 moles of an ideal monoatomic gas at 700K and the upper compartment is filled with 2 moles of an ideal diatomic gas at 400K.  The heat capacities per mole of an ideal monoatomic gas are   $C_{v}=\frac{3}{2}R,C_{p}=\frac{5}{2}R,$  , and those for an ideal diatomic  gas are  $C_{v}=\frac{5}{2}R,C_{p}=\frac{7}{2}R,$

Now consider the partition to be free to move without friction so that the pressure of gases in both compartments in the same. Then total work done by the gases till the time they achieve  equilibrium will be

Answer:

Option D

Explanation:

  $\triangle W_{1}+\triangle U_{1}=\triangle Q_{1}$       ...........(i)

                   $\triangle W_{2}+\triangle U_{2}=\triangle Q_{2}$

               $\triangle Q_{1}+\triangle Q_{2}=0$

   $\therefore$     $ (nC_{p}\triangle T)_{1}+(nC_{p}\triangle T)_{2}=0$

      But  n₁ =n₂=n

$\therefore $    $\frac{5}{2} R(T-700)+\frac{7}{2} R(T-400)=0$

 Solving , we get T=525 k

Now, from equatins (i) and (ii) , we get

               $\triangle W_{1}+\triangle W_{2}=-\triangle U_{1}- \triangle U_{2}$

        as   $\triangle Q_{1}+\triangle Q_{2}=0$

  $\therefore$     $ \triangle W_{1}+\triangle W_{2}=-[(nC_{v}\triangle T)_{1}+(nC_{v}\triangle T)_{2}]$

     =  $-\left[2 \times\frac{3}{2}R\times (525-700)+2\times\frac{5}{2}R\times  (525-400)\right]$

           =-100 R