Question

Match the thermodynamic processes taking place in a system with the correct conditions.
In the table, $\Delta Q$ is the heat supplied, $\Delta W$ is the work done and $\Delta U$ is the change in internal energy of the system.

Process	Condition
(i) Adiabatic	(A) $\Delta W=0$
(ii) Isothermal	(B) $\Delta Q=0$
(iii) Isochoric	(C) $\Delta U\ne 0$, $\Delta W\ne 0$, $\Delta Q\ne 0$
(iv) Isobaric	(D) $\Delta U=0$

• A. (I)-(A), (II)-(B), (III)-(D), (IV)-(D)
• B. (I)-(B), (II)-(A), (III)-(D), (IV)-(C)
• C. (I)-(A), (II)-(A), (III)-(B), (IV)-(C)
• D. (I)-(B), (II)-(D), (III)-(A), (IV)-(C)

Answer 1

The correct option is D (I)-(B), (II)-(D), (III)-(A), (IV)-(C)

(I) Adiabatic process : No exchange of heat takes place with surroundings.
$\Rightarrow \Delta Q=0$

(II) Isothermal process : Temperature remains constant
$\therefore \Delta T=0\Rightarrow \Delta U=\frac{f}{2}nRT\Delta T\Rightarrow \Delta U=0$
No change in internal energy$[\Delta U=0]$

(III) Isochoric process volume remains constant
$\Delta V=0\Rightarrow W=\int P.dV=0$
Hence, work done is zero.

(IV) In isobaric process pressure remains constant.
$W=P.\Delta V\ne 0$
$\Delta U=\frac{f}{2}nRT\Delta T=\frac{f}{2}\left[P\Delta V\right]\ne 0$
$\therefore \Delta Qn{C}_p\Delta T\ne 0$.

Hence, $\left(D\right)$ is the correct answer.