Discussion Forum



1)

A reversible cyclic process for an ideal gas is shown below. Hence , P , V and T are pressure, volume and temperature respectively , The thermodynamic parameters q,w, H and U are heat, work, enthalpy and internal energy, respectively.

2782019657_vol.JPG

The correct options is (are)


A) $q_{AC}=\triangle U_{BC}$ and $w_{AB}=p_{2}(V_{2}-V_{1})$

B) $w_{BC}=p_{2}(V_{2}-V_{1})$ and $q_{BC}=\triangle H_{AC}$

C) $\triangle H_{CA}<\triangle U_{CA}$ and $q_{AC}=\triangle U_{BC}$

D) $q_{BC}=\triangle H_{AC}$ and $\triangle H_{CA}>\triangle U_{CA}$

Answer:

Option (B,C)

Explanation:

In the given curve AC represents isochoric process as volume at both the points is same i.e., V1

Similarly , AB represents isothermal process ( as both the points are at T1 temperature )  and BC represents  isobaric process as both the points are at p2 pressure.

       Now (i) for option (a)

                          $q_{AC}=\triangle U_{BC}=nC_{v}(T_{2}-T_{1})$

where, n= number of moles,

     Cv= specific heat capacity at constant volume

   However, WAB ≠ P2 ( V2 -V1) instead 

            $w_{AB}=nRT_{1}ln(\frac{V_{2}}{V_{1}})$

   So, this option is incorrect.

(ii) For option (b)

           $q_{BC}=\triangle H_{AC}=nC_{p}(T_{2}-T_{1})$

   where , Cp= specific heat capacity at constant  pressure.

Likewise,

     $w_{BC}=-P_{2}(V_{1}-V_{2})$

Hence, the option is correct.

(iii)  For option (c)

        as $nC_{p}(T_{2}-T_{1})<nC_{v}(T_{2}-T_{1})$

So         $\triangle H_{CA}<\triangle U_{CA}$

 and    $q_{AC}=\triangle U_{BC}$

Hence, this option is also correct.

(IV)   For option (d)

         Although    $q_{BC}=\triangle H_{AC}$

    but  $\triangle H_{CA}>\triangle U_{CA}$

  Hence, this option is incorrect.