JEE PYQ: Thermodynamics Question 21
Question 21 - 2020 (04 Sep Shift 2)
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 change in internal energy of the system.
| Process | Condition |
|---|---|
| (I) Adiabatic | (A) $\Delta W = 0$ |
| (II) Isothermal | (B) $\Delta Q = 0$ |
| (III) Isochoric | (C) $\Delta U \neq 0, \Delta W \neq 0, \Delta Q \neq 0$ |
| (IV) Isobaric | (D) $\Delta U = 0$ |
(1) (I)-(A), (II)-(B), (III)-(D), (IV)-(D)
(2) (I)-(B), (II)-(A), (III)-(D), (IV)-(C)
(3) (I)-(A), (II)-(A), (III)-(B), (IV)-(C)
(4) (I)-(B), (II)-(D), (III)-(A), (IV)-(C)
Show Answer
Answer: (4)
Solution
(I) Adiabatic process: No exchange of heat with surroundings. $\Rightarrow \Delta Q = 0$ matches (B).
(II) Isothermal process: Temperature remains constant. $\Delta T = 0 \Rightarrow \Delta U = \frac{f}{2}nR\Delta T \Rightarrow \Delta U = 0$. Matches (D).
(III) Isochoric process: Volume remains constant. $\Delta V = 0 \Rightarrow W = \int P \cdot dV = 0$. Hence work done is zero. Matches (A).
(IV) Isobaric process: Pressure remains constant. $W = P \cdot \Delta V \neq 0$, $\Delta U = \frac{f}{2}nR\Delta T = \frac{f}{2}(P\Delta V) \neq 0$, $\therefore \Delta Q = nC_p\Delta T \neq 0$. Matches (C).
BETA
