Question

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

A) (I) – (1), (II) – (1), (III) – (2), (IV) – (3)

B) (I) – (1), (II) – (2), (III) – (4), (IV) – (4)

C) (I) – (2), (II) – (4), (III) – (1), (IV) – (3)

D) (I) – (2), (II) – (1), (III) – (4), (IV) – (3)

Column 1	Column 2
(I) Adiabatic	(1)$∆W=0$
(II)Isothermal	(2)$∆Q=0$
(III)Isochoric	(3)$∆U\ne 0,∆W\ne 0,∆Q\ne 0$
(IV)Isobaric	(4)$∆U=0$

Answer 1

Part I:

1. In the adiabatic process the system remains enclosed by an insulated wall.
2. There is no heat exchange between the system and the surroundings.
3. So the change in heat energy is zero as$∆Q=0$.

Hence, (I) - (2)

Part II:

1. In the isothermal process the temperature of the system remains constant.
2. There is no change in temperature, for that change in internal energy is also zero.
3. so, $∆U=0$ and$∆T=0$

Hence, (II) - (4)

Part III:

1. In the isochoric process the total volume of the system remains constant.
2. There is no change in volume$∆V=0$.
3. We know the work done
   $\Rightarrow ∆W=P∆V(P=$The pressure of the system)
   $\Rightarrow ∆W=0$

Hence, (III) - (1)

Part IV:

1. In the isobaric process the total pressure of the system remains constant.
2. The change in internal energy, heat exchange, and work done does not get affected for that purpose.
3. So,$∆U\ne 0,∆W\ne 0,∆Q\ne 0$ .

Thus, (IV) - (3)

Hence, the correct answer is an option (C) (I) – (2), (II) – (4), (III) – (1), (IV) – (3).